Phyciscs 第五版 (Halliday Resnick Krane 著) 加州路德大学 课后答案 334页

'课后答案网，用心为你服务！大学答案---中学答案---考研答案---考试答案最全最多的课后习题参考答案，尽在课后答案网（www.khdaw.com）！Khdaw团队一直秉承用心为大家服务的宗旨，以关注学生的学习生活为出发点，旨在为广大学生朋友的自主学习提供一个分享和交流的平台。爱校园（www.aixiaoyuan.com）课后答案网（www.khdaw.com）淘答案（www.taodaan.com） 课后答案网www.khdaw.comkhdaw.comInstructorSolutionsManualforPhysicsbyHalliday,Resnick,andKranewww.khdaw.comPaulStanleyBeloitCollegeVolume2课后答案网khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comANoteToTheInstructor...Thesolutionsherearesomewhatbrief,astheyaredesignedfortheinstructor,notforthestudent.Checkwiththepublishersbeforeelectronicallypostinganypartofthesesolutions;website,ftp,orserveraccessmustberestrictedtoyourstudents.Ihavebeensomewhatcasualaboutsubscriptswheneveritisobviousthataproblemisonedimensional,orthatthechoiceofthecoordinatesystemisirrelevanttothenumericalsolution.Althoughthisdoesnotchangethevalidityoftheanswer,itwillsometimesobfuscatetheapproachifviewedbyanovice.Therearesometraditionalformula,suchasv2=v2+2ax;x0xxwhicharenotusedinthetext.Theworkedsolutionsuseonlymaterialfromthetext,sotheremaybetimeswhenthesolutionhereseemsunnecessarilyconvolutedanddrawnout.Yes,Iknowaneasierapproachexisted.Butifitwasnotinthetext,Ididnotuseithere.khdaw.comIalsotriedtoavoidreinventingthewheel.Therearesomeexercisesandproblemsinthetextwhichbuilduponpreviousexercisesandproblems.Insteadofrederivingexpressions,Isimplyreferyoutotheprevioussolution.Iadoptadierentapproachforroundingofsignicantguresthanpreviousauthors;inpartic-ular,Iusuallyroundintermediateanswers.Assuch,someofmyanswerswilldierfromthoseinthebackofthebook.ExercisesandProblemswhichareenclosedinaboxalsoappearintheStudent"sSolutionManualwithconsiderablymoredetailand,whenappropriate,includediscussiononanyphysicalimplicationsoftheanswer.Thesestudentsolutionscarefullydiscussthestepsrequiredforsolvingproblems,pointouttherelevantequationnumbers,orevenspecifywhereinthetextadditionalinformationcanbefound.Whentwoalmostequivalentmethodsofsolutionexist,oftenbotharepresented.Youarewww.khdaw.comencouragedtoreferstudentstotheStudent"sSolutionManualfortheseexercisesandproblems.However,thematerialfromtheStudent"sSolutionManualmustnotbecopied.PaulStanleyBeloitCollegestanley@clunet.edu课后答案网1khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE25-1ThechargetransferredisQ=(2:5104C=s)(20106s)=5:0101C:E25-2UseEq.25-4:s(8:99109N
m2=C2)(26:3106C)(47:1106C)r==1:40m(5:66N)E25-3UseEq.25-4:(8:99109N
m2=C2)(3:12106C)(1:48106C)F==2:74N:(0:123m)2E25-4(a)Theforcesareequal,som1a1=m2a2,orm=(6:31107kg)(7:22m=s2)=(9:16m=s2)=4:97107kg:khdaw.com2(b)UseEq.25-4:s(6:31107kg)(7:22m=s2)(3:20103m)211q==7:2010C(8:99109N
m2=C2)E25-5(a)UseEq.25-4,1q1q21(21:3C)(21:3C)F===1:77N40r1224(8:851012C2www.khdaw.com=N
m2)(1:52m)2(b)Inpart(a)wefoundF12;tosolvepart(b)weneedtorstndF13.Sinceq3=q2andr13=r12,wecanimmediatelyconcludethatF13=F12.Wemustassessthedirectionoftheforceofq3onq1;itwillbedirectedalongthelinewhichconnectsthetwocharges,andwillbedirectedawayfromq3.Thediagrambelowshowsthedirections.F23Fq课后答案网12FF23F12netFromthisdiagramwewanttondthemagnitudeofthenetforceonq1.Thecosinelawisappropriatehere:F2=F2+F22FFcos;net12131213=(1:77N)2+(1:77N)22(1:77N)(1:77N)cos(120);2=9:40N;Fnet=3:07N:khdaw.com2若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE25-6OriginallyF=CQ2=0:088N,whereCisaconstant.Whensphere3touches1the00chargeonbothbecomesQ0=2.Whensphere3thetouchessphere2thechargeoneachbecomes(Q0+Q0=2)=2=3Q0=4.Theforcebetweensphere1and2isthenF=C(Q=2)(3Q=4)=(3=8)CQ2=(3=8)F=0:033N:0000E25-7Theforcesonq3areF~31andF~32.TheseforcesaregivenbythevectorformofCoulomb"sLaw,Eq.25-5,F~=1q3q1^r=1q3q1^r;314r2314(2d)2310310F~=1q3q2^r=1q3q2^r:324r2324(d)2320320Thesetwoforcesaretheonlyforceswhichactonq3,soinordertohaveq3inequilibriumtheforcesmustbeequalinmagnitude,butoppositeindirection.Inshort,khdaw.comF~31=F~32;1q3q11q3q2^r31=^r32;40(2d)240(d)2q1q2^r31=^r32:41Notethat^r31and^r32bothpointinthesamedirectionandarebothofunitlength.Wethengetq1=4q2:E25-8Thehorizontalandverticalcontributionsfromtheupperleftchargeandlowerrightchargearestraightforwardtond.Thecontributionsfromtheupperleftchargerequireslightlymorework.pwww.khdaw.compThediagonaldistanceis2a;thecomponentswillbeweightedbycos45=2=2.Thediagonalchargewillcontributepp1(q)(2q)22q2Fx=p^i=^i;40(2a)2280a2pp1(q)(2q)22q2Fy=p^j=^j:40(2a)2280a2(a)Thehorizontalcomponentofthenetforceisthenp课后答案网1(2q)(2q)2q2Fx=^i+^i;40a280a2p4+2=2q2=^i;40a2=(4:707)(8:99109N
m2=C2)(1:13106C)2=(0:152m)2^i=2:34N^i:(b)Theverticalcomponentofthenetforceisthenp1(q)(2q)2q2Fy=^j+^j;40a280a2p2+2=2q2=^j;80a2=(1:293)(8:99109N
m2=C2)(1:13106C)2=(0khdaw.com:152m)2^j=0:642N^j:3若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE25-9ThemagnitudeoftheforceonthenegativechargefromeachpositivechargeisF=(8:99109N
m2=C2)(4:18106C)(6:36106C)=(0:13m)2=14:1N:Theforcefromeachpositivechargeisdirectedalongthesideofthetriangle;butfromsymmetryonlythecomponentalongthebisectorisofinterest.Thismeansthatweneedtoweighttheaboveanswerbyafactorof2cos(30)=1:73.Thenetforceisthen24:5N.E25-10Letthechargeononespherebeq,thenthechargeontheothersphereisQ=(52:6106C)q.Then1qQ=F;40r2(8:99109N
m2=C2)q(52:6106Cq)=(1:19N)(1:94m)2:Solvethisquadraticexpressionforqandgetanswersq=4:02105Candq=1:24106N.khdaw.com12E25-11ThisproblemissimilartoEx.25-7.Therearesomeadditionalissues,however.ItiseasyenoughtowriteexpressionsfortheforcesonthethirdchargeF~=1q3q1^r;314r231031F~=1q3q2^r:324r232032ThenF~31=www.khdaw.comF~32;1q3q11q3q22^r31=2^r32;40r3140r32q1q22^r31=2^r32:r31r32Theonlywaytosatisfythevectornatureoftheaboveexpressionistohave^r31=^r32;thismeansthatq3mustbecollinearwithq1andq2.q3couldbebetweenq1andq2,oritcouldbeoneitherside.Let"sresolvethisissuenowbyputtingthevaluesforq1andq2intotheexpression:(1:07C)(3:28C)课后答案网2^r31=2^r32;r31r32r2^r=(3:07)r2^r:32313132Sincesquaredquantitiesarepositive,wecanonlygetthistoworkif^r31=^r32,soq3isnotbetweenq1andq2.Wearethenleftwithr2=(3:07)r2;3231sothatq3isclosertoq1thanitistoq2.Thenr32=r31+r12=r31+0:618m,andifwetakethesquarerootofbothsidesoftheaboveexpression,pr31+(0:618m)=(3:07)r31;p(0:618m)=(3:07)r31r31;(0:618m)=0:752r31;0:822m=r31khdaw.com4若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE25-12Themagnitudeofthemagneticforcebetweenanytwochargesiskq2=a2,wherea=0:153m.Theforcebetweeneachchargeisdirectedalongthesideofthetriangle;butfromsymmetryonlythecomponentalongthebisectorisofinterest.Thismeansthatweneedtoweighttheaboveanswerbyafactorof2cos(30)=1:73.Thenetforceonanychargeisthen1:73kq2=a2.Thelengthoftheanglebisector,d,isgivenbyd=acos(30).Thedistancefromanychargetothecenteroftheequilateraltriangleisx,givenbyx2=(a=2)2+(dx)2.Thenx=a2=8d+d=2=0:644a:Theanglebetweenthestringsandtheplaneofthechargesis,givenbysin=x=(1:17m)=(0:644)(0:153m)=(1:17m)=0:0842;or=4:83.Theforceofgravityoneachballisdirectedverticallyandtheelectricforceisdirectedhorizontally.Thetwomustthenberelatedbykhdaw.comtan=FE=FG;so1:73(8:99109N
m2=C2)q2=(0:153m)2=(0:0133kg)(9:81m=s2)tan(4:83);orq=1:29107C:E25-13Onanycornerchargetherearesevenforces;onefromeachoftheothersevencharges.Thenetforcewillbethesum.Sincealleightchargesarethesamealloftheforceswillberepulsive.Weneedtosketchadiagramtoshowhowthechargesarelabeled.2www.khdaw.com14673课后答案网85Themagnitudeoftheforceofcharge2oncharge1is1q2F12=2;40r12wherer=a,thelengthofaside.Sincebothchargesarethesamewewroteq2.Bysymmetrywe12expectthatthemagnitudesofF12,F13,andF14willallbethesameandtheywillallbeatrightanglestoeachotherdirectedalongtheedgesofthecube.Writtenintermsofvectorstheforces5khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comwouldbe1q2F~12=^i;40a21q2F~13=^j;40a21q2F~14=k^:40a2Theforcefromcharge5is1q2F15=2;40r15andisdirectedalongthesidediagonalawayfromcharge5.Thedistancer15isalsothesidediagonaldistance,andcanbefoundfromr2=a2+a2=2a2;15then1q2khdaw.comF15=:402a2BysymmetryweexpectthatthemagnitudesofF15,F16,andF17willallbethesameandtheywillallbedirectedalongthediagonalsofthefacesofthecube.Intermsofcomponentswewouldhave1q2ppF~15=^j=2+k^=2;402a21q2ppF~16=^i=2+k^=2;402a21q2ppF~17=^i=2+^j=2:402a2www.khdaw.comThelastforceistheforcefromcharge8oncharge1,andisgivenby1q2F18=2;40r18andisdirectedalongthecubediagonalawayfromcharge8.Thedistancer18isalsothecubediagonaldistance,andcanbefoundfromr2=a2+a2+a2=3a2;18thenintermofcomponents课后答案网1q2pppF~18=^i=3+^j=3+k^=3:403a2Wecanaddthecomponentstogether.Bysymmetryweexpectthesameanswerforeachcom-ponents,sowe"lljustdoone.Howabout^i.Thiscomponenthascontributionsfromcharge2,6,7,and8:1q2121+p+p;40a212233or1q2(1:90)40a2pThethreecomponentsaddaccordingtoPythagorastopickupanalfactorof3,soq2Fnet=(0:262):0a2khdaw.com6若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE25-14(a)Yes.ChangingthesignofywillchangethesignofFy;sincethisisequivalenttoputtingthechargeq0ontheother"side,wewouldexpecttheforcetoalsopushintheother"direction.(b)TheequationshouldlookEq.25-15,exceptally"sshouldbereplacedbyx"s.Then1q0qFx=p:40xx2+L2=4(c)Settingtheparticleadistancedawayshouldgiveaforcewiththesamemagnitudeas1q0qF=p:40dd2+L2=4Thisforceisdirectedalongthe45line,soF=Fcos45andF=Fsin45.pxy(d)Letthedistancebed=x2+y2,andthenusethefactthatFx=F=cos=x=d.Thenx1xq0qFx=F=:khdaw.comd40(x2+y2+L2=4)3=2andy1yq0qFy=F=:d40(x2+y2+L2=4)3=2E25-15(a)Theequationisvalidforbothpositiveandnegativez,soinvectorformitwouldreadF~=Fk^=1q0qzk^:z4(z2+R2)3=20(b)Theequationisnotvalidforbothpositiveandnegativez.ReversingthesignofzshouldpreversethesignofFz,andonewaytoxthisistowrite1=www.khdaw.comz=z2.ThenF~=Fk^=12q0qzp1p1k^:z240Rz2z2E25-16Dividetherodintosmalldierentiallengthsdr,eachwithchargedQ=(Q=L)dr.Eachdierentiallengthcontributesadierentialforce1qdQ1qQdF==dr:40r240r2LIntegrate:ZZx+L课后答案网1qQF=dF=dr;x40r2L1qQ11=40Lxx+LE25-17YoumustsolveEx.16beforesolvingthisproblem!q0referstothechargethathadbeencalledqinthatproblem.Ineithercasethedistancefromq0willbethesameregardlessofthesignofq;ifq=Qthenqwillbeontheright,whileifq=Qthenqwillbeontheleft.Settingtheforcesequaltoeachotheronegets1qQ111qQ=;40Lxx+L40r2orpr=x(x+L):khdaw.com7若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE25-18YoumustsolveEx.16andEx.17beforesolvingthisproblem.Ifallchargesarepositivethenmovingq0oaxiswillresultinanetforceawayfromtheaxis.That"sunstable.Ifq=QthenbothqandQareonthesamesideofq0.Movingq0closertoqwillresultintheattractiveforcegrowingfasterthantherepulsiveforce,soq0willmoveawayfromequilibrium.E25-19WecanstartwiththeworkthatwasdoneforusonPage577,exceptsinceweareconcernedwithsin=z=rwewouldhave1q0dzzdFx=dFsin=p:40(y2+z2)y2+z2Wewillneedtotakeintoconsiderationthatchangessignforthetwohalvesoftherod.ThenZZ!0L=2q0zdz+zdzFx=+;40L=2(y2+z2)3=20(y2+z2)3=2ZL=2khdaw.comq0zdz=;200(y2+z2)3=2L=2q01=p;20y2+z20!q011=p:20yy2+(L=2)2E25-20UseEq.25-15tondthemagnitudeoftheforcefromanyonerod,butwriteitas1www.khdaw.comqQF=p;40rr2+L2=4wherer2=z2+L2=4.ThecomponentofthisalongthezaxisisF=Fz=r.Sincethereare4rods,zwehave1qQz1qQzF=p;=p;0r2r2+L2=40(z2+L2=4)z2+L2=2EquatingtheelectricforcewiththeforceofgravityandsolvingforQ,0mg22pQ=(z+L=4)z2+L2=2;qz课后答案网puttinginthenumbers,(8:851012C2=N
m2)(3:46107kg)(9:8m=s2)p((0:214m)2+(0:25m)2=4)(0:214m)2+(0:25m)2=2(2:451012C)(0:214m)soQ=3:07106C:E25-21Ineachcaseweconservechargebymakingsurethatthetotalnumberofprotonsisthesameonbothsidesoftheexpression.Wealsoneedtoconservethenumberofneutrons.(a)Hydrogenhasoneproton,Berylliumhasfour,soXmusthaveveprotons.ThenXmustbeBoron,B.(b)Carbonhassixprotons,Hydrogenhasone,soXmusthaveseven.ThenXisNitrogen,N.(c)Nitrogenhassevenprotons,Hydrogenhasone,butHeliumhastwo,soXhas7+12=6protons.ThismeansXisCarbon,C.khdaw.com8若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE25-22(a)UseEq.25-4:(8:99109N
m2=C2)(2)(90)(1:601019C)2F==290N:(121015m)2(b)a=(290N)=(4)(1:661027kg)=4:41028m=s2.E25-23UseEq.25-4:(8:99109N
m2=C2)(1:601019C)2F==2:89109N:(2821012m)2E25-24(a)UseEq.25-4:s(3:7109N)(5:01010m)219q==3:2010C:khdaw.com(8:99109N
m2=C2)(b)N=(3:201019C)=(1:601019C)=2.E25-25UseEq.25-4,1qq(11:61019C)(11:61019C)F=12=33=3:8N:40r24(8:851012C2=N
m2)(2:61015m)212E25-26(a)N=(1:15107C)=(1:601019C)=7:191011.(b)Thepennyhasenoughelectronstomakeatotalchargeofwww.khdaw.com1:37105C.Thefractionisthen(1:15107C)=(1:37105C)=8:401013:E25-27Equatethemagnitudesoftheforces:1q2=mg;40r2sos(8:99109N
m2=C2)(1:601019C)2r==5:07m课后答案网(9:111031kg)(9:81m=s2)E25-28Q=(75:0kg)(1:601019C)=(9:111031kg)=1:31013C.E25-29Themassofwateris(250cm3)(1:00g/cm3)=250g.Thenumberofmolesofwateris(250g)=(18:0g/mol)=13:9mol.Thenumberofwatermoleculesis(13:9mol)(6:021023mol1)=8:371024.Eachmoleculehastenprotons,sothetotalpositivechargeisQ=(8:371024)(10)(1:601019C)=1:34107C:9khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE25-30Thetotalpositivechargein0:250kgofwateris1:34107C.Mary"simbalanceisthenq=(52:0)(4)(1:34107C)(0:0001)=2:79105C;1whileJohn"simbalanceisq=(90:7)(4)(1:34107C)(0:0001)=4:86105C;2Theelectrostaticforceofattractionisthen1qq(2:79105)(4:86105)F=12=(8:99109N
m2=C2)=1:61018N:40r2(28:0m)2E25-31(a)ThegravitationalforceofattractionbetweentheMoonandtheEarthisGMEMMFG=;khdaw.comR2whereRisthedistancebetweenthem.IfboththeEarthandthemoonareprovidedachargeq,thentheelectrostaticrepulsionwouldbe1q2FE=:40R2Settingthesetwoexpressionequaltoeachother,q2=GMEMM;40whichhassolutionwww.khdaw.compq=40GMEMM;q=4(8:851012C2=Nm2)(6:671011Nm2=kg2)(5:981024kg)(7:361022kg);=5:711013C:(b)Weneed(5:711013C)=(1:601019C)=3:571032protonsoneachbody.Themassofprotonsneededisthen课后答案网(3:571032)(1:671027kg)=5:971065kg:Ignoringthemassoftheelectron(whynot?)wecanassumethathydrogenisallprotons,soweneedthatmuchhydrogen.P25-1Assumethatthespheresinitiallyhavechargesq1andq2.Theforceofattractionbetweenthemis1q1q2F1=2=0:108N;40r12wherer12=0:500m.Thenetchargeisq1+q2,andaftertheconductingwireisconnectedeachspherewillgethalfofthetotal.Thesphereswillhavethesamecharge,andrepelwithaforceof1(q+q)1(q+q)1212212F2=2=0:0360N:40r12khdaw.com10若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comSinceweknowtheseparationofthesphereswecanndq1+q2quickly,qq1+q2=240r122(0:0360N)=2:00CWe"llputthisbackintotherstexpressionandsolveforq2.1(2:00Cq2)q20:108N=;40r2123:001012C2=(2:00Cq)q;220=q2+(2:00C)q+(1:73C)2:22Thesolutionisq2=3:0Corq2=1:0C.Thenq1=1:0Corq1=3:0C.P25-2TheelectrostaticforceonQfromeachqhasmagnitudeqQ=4a2,whereaisthelength0ofthesideofthesquare.Themagnitudeofthevertical(horizontal)componentoftheforceofQonpQis2Q2=16a2.khdaw.com0(a)InordertohaveazeronetforceonQthemagnitudesofthetwocontributionsmustbalance,sop2Q2qQ=;160a240a2porq=2Q=4.Thechargesmustactuallyhaveoppositecharge.(b)No.P25-3(a)Thethirdcharge,q3,willbebetweenthersttwo.Thenetforceonthethirdchargewillbezeroif1qq314qq3=www.khdaw.com;40r31240r322whichwilloccurif12=r31r32ThetotaldistanceisL,sor31+r32=L,orr31=L=3andr32=2L=3.Nowthatwehavefoundthepositionofthethirdchargeweneedtondthemagnitude.Thesecondandthirdchargesbothexertaforceontherstcharge;wewantthisnetforceontherstchargetobezero,so1qq31q4q=;课后答案网40r13240r122orq34q=;(L=3)2L2whichhassolutionq3=4q=9.Thenegativesignisbecausetheforcebetweentherstandsecondchargemustbeintheoppositedirectiontotheforcebetweentherstandthirdcharge.(b)Considerwhathappenstothenetforceonthemiddlechargeifisisdisplacedasmalldistancez.Ifthecharge3ismovedtowardcharge1thentheforceofattractionwithcharge1willincrease.Butmovingcharge3closertocharge1meansmovingcharge3awayfromcharge2,sotheforceofattractionbetweencharge3andcharge2willdecrease.Socharge3experiencesmoreattractiontowardthechargethatitmovestoward,andlessattractiontothechargeitmovesawayfrom.Soundsunstabletome.11khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP25-4(a)Theelectrostaticforceonthechargeontherighthasmagnitudeq2F=;40x2TheweightoftheballisW=mg,andthetwoforcesarerelatedbyF=W=tansin=x=2L:Combining,2Lq2=4mgx3,so021=3qLx=:20(b)Rearrangeandsolveforq,s2(8:851012C2=N
m2)(0:0112kg)(9:81m=s2)(4:70102m)3q==2:28108C:khdaw.com(1:22m)P25-5(a)Originallytheballswouldnotrepel,sotheywouldmovetogetherandtouch;aftertouchingtheballswouldsplit"thechargeendingupwithq=2each.Theywouldthenrepelagain.(b)Thenewequilibriumseparationis21=31=30(q=2)L1x==x=2:96cm:20mg4P25-6TakethetimederivativeoftheexpressioninProblem25-4.Thendx2xdq2(4:70102m)www.khdaw.com==(1:20109C=s)=1:65103m=s:dt3qdt3(2:28108C)P25-7Theforcebetweenthetwochargesis1(Qq)qF=:40r212Wewanttomaximizethisforcewithrespecttovariationinq,thismeansndingdF=dqandsettingitequalto0.ThendFd1(Qq)q1Q2q课后答案网=2=2:dqdq40r1240r12ThiswillvanishifQ2q=0,orq=1Q.2P25-8Displacethechargeqadistancey.ThenetrestoringforceonqwillbeapproximatelyqQ1yqQ16F2=y:40(d=2)2(d=2)40d3SinceF=yiseectivelyaforceconstant,theperiodofoscillationisr331=2m0mdT=2=:kqQ12khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP25-9DisplacethechargeqadistancextowardoneofthepositivechargesQ.ThenetrestoringforceonqwillbeqQ11F=;40(d=2x)2(d=2+x)2qQ32x:40d3SinceF=xiseectivelyaforceconstant,theperiodofoscillationisr331=2m0mdT=2=:k2qQP25-10(a)Zero,bysymmetry.(b)RemovingapositiveCesiumionisequivalenttoaddingasinglychargednegativeionatthatsamelocation.ThenetforceisthenF=e2=4r2;khdaw.com0whereristhedistancebetweentheChlorideionandthenewlyplacednegativeion,orpr=3(0:20109m)2Theforceisthen(1:61019C)2F==1:92109N:4(8:851012C2=N
m2)3(0:20109m)2P25-11Wecanpretendthatthisproblemisinasingleplanecontainingallthreecharges.Thewww.khdaw.commagnitudeoftheforceonthetestchargeq0fromthechargeqontheleftis1qq0Fl=:40(a2+R2)Aforceofidenticalmagnitudeexistsfromthechargeontheright.weneedtoaddthesetwoforcesasvectors.OnlythecomponentsalongRwillsurvive,andeachforcewillcontributeanamountRFlsin=Flp;R2+a2sothenetforceonthetestparticlewillbe课后答案网2qq0Rp:40(a2+R2)R2+a2WewanttondthemaximumvalueasafunctionofR.Thismeanstakethederivative,andsetitequaltozero.Thederivativeis2qq13R20;40(a2+R2)3=2(a2+R2)5=2whichwillvanishwhena2+R2=3R2;pasimplequadraticequationwithsolutionsR=a=2.13khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE26-1E=F=q=ma=q.ThenE=(9:111031kg)(1:84109m=s2)=(1:601019C)=1:05102N=C:E26-2Theanswersto(a)and(b)arethesame!F=Eq=(3:0106N=C)(1:601019C)=4:81013N.E26-3F=W,orEq=mg,somg(6:641027kg)(9:81m=s2)E===2:03107N=C:q2(1:601019C)Thealphaparticlehasapositivecharge,thismeansthatitwillexperienceanelectricforcewhichisinthesamedirectionastheelectriceld.Sincethegravitationalforceisdown,theelectricforce,andconsequentlytheelectriceld,mustbedirectedup.E26-4(a)E=F=q=(3:0106N)=(2:0109C)=1:5103N=C.khdaw.com(b)F=Eq=(1:5103N=C)(1:601019C)=2:41016N:(c)F=mg=(1:671027kg)(9:81m=s2)=1:61026N:(d)(2:41016N)=(1:61026N)=1:51010:E26-5RearrangeE=q=4r2,0q=4(8:851012C2=N
m2)(0:750m)2(2:30N=C)=1:441010C:E26-6p=qd=(1:601019C)(4:30109)=6:881028C
m.E26-7UseEq.26-12forpointsalongtheperpendicularbisector.Thenwww.khdaw.com1p(3:561029C
m)E==(8:99109N
m2=C2)=1:95104N=C:40x3(25:4109m)3E26-8Ifthechargesonthelinex=awhere+qandqinsteadof+2qand2qthenatthecenterofthesquareE=0bysymmetry.ThissimpliestheproblemintondingEforacharge+qat(a;0)andqat(a;a).Thisisadipole,andtheeldisgivenbyEq.26-11.Forthisexercisewehavex=a=2andd=a,so1qaE=;40[2(a=2)2]3=2or,puttinginthenumbers,课后答案网E=1:11105N=C.E26-9Thechargesat1and7areoppositeandcanbeeectivelyreplacedwithasinglechargeof6qat7.Thesameistruefor2and8,3and9,onupto6and12.Bysymmetryweexpecttheeldtopointalongalinesothatthreechargesareaboveandthreebelow.Thatwouldmean9:30.E26-10IfbothchargesarepositivethenEq.26-10wouldreadE=2E+sin,andEq.26-11wouldlooklike1qxE=2p;40x2+(d=2)2x2+(d=2)21qx2p40x2x2whenxd.ThiscanbesimpliedtoE=2q=4x2.0khdaw.com14若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE26-11Treatthetwochargesontheleftasonedipoleandtreatthetwochargesontherightasaseconddipole.PointPisontheperpendicularbisectorofbothdipoles,sowecanuseEq.26-12tondthetwoelds.Forthedipoleontheleftp=2aqandtheelectriceldduetothisdipoleatPhasmagnitude12aqEl=40(x+a)3andisdirectedup.Forthedipoleontherightp=2aqandtheelectriceldduetothisdipoleatPhasmagnitude12aqEr=40(xa)3andisdirecteddown.ThenetelectriceldatPisthesumofthesetwoelds,butsincethetwocomponenteldspointinoppositedirectionswemustactuallysubtractthesevalues,E=ErEl;khdaw.com2aq11=;40(xa)3(x+a)3aq111=:20x3(1a=x)3(1+a=x)3Wecanusethebinomialexpansiononthetermscontaining1a=x,aq1E((1+3a=x)(13a=x));20x3aq1=(6a=x);20x3www.khdaw.com3(2qa2)=:20x4E26-12Doaseriesexpansiononthepartintheparentheses11R2R21p11=:1+R2=z22z22z2Substitutethisin,R2QEz=:课后答案网202z240z2E26-13Atthesurfacez=0andEz==20.Halfofthisvalueoccurswhenzisgivenby1z=1p;2z2+R2pwhichcanbewrittenasz2+R2=(2z)2.Solvethis,andz=R=3.E26-14LookatEq.26-18.Theelectriceldwillbeamaximumwhenz=(z2+R2)3=2isamaximum.Takethederivativeofthiswithrespecttoz,andget132z2z2+R23z2=:(z2+R2)3=22(z2+R2)5=2(z2+R2)5=2pThiswillvanishwhenthenumeratorvanishes,orwhenz=R=2.khdaw.com15若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE26-15(a)TheelectriceldstrengthjustabovethecentersurfaceofachargeddiskisgivenbyEq.26-19,butwithz=0,E=20Thesurfacechargedensityis=q=A=q=(R2).Combining,q=2R2E=2(8:851012C2=N
m2)(2:5102m)2(3106N=C)=1:04107C:0NoticeweusedanelectriceldstrengthofE=3106N=C,whichistheeldatairbreaksdownandsparkshappen.(b)Wewanttondouthowmanyatomsareonthesurface;ifaisthecrosssectionalareaofoneatom,andNthenumberofatoms,thenA=Naisthesurfaceareaofthedisk.ThenumberofatomsisA(0:0250m)2N===1:311017a(0:0151018m2)(c)Thetotalchargeonthediskis1:04107C,thiscorrespondstokhdaw.com(1:04107C)=(1:61019C)=6:51011electrons.(Weareignoringthesignofthechargehere.)Ifeachsurfaceatomcanhaveatmostoneexcesselectron,thenthefractionofatomswhicharechargedis(6:51011)=(1:311017)=4:96106;whichisn"tverymany.E26-16Imagineswitchingthepositiveandnegativecharges.Theelectriceldwouldalsoneedtoswitchdirections.Bysymmetry,then,theelectriceldcanonlypointverticallydown.Keepingonlythatcomponent,www.khdaw.comZ=21dE=2sin;040r22=:40r2But=q=(=2),soE=q=2r2.0E26-17WewanttotthedatatoEq.26-19,课后答案网zEz=1p:20z2+R2Thereareonlytwovariables,Randq,withq=R2.Wecanndveryeasilyifweassumethatthemeasurementshavenoerrorbecausethenatthesurface(wherez=0),theexpressionfortheelectriceldsimpliestoE=:20Then=2E=2(8:8541012C2=N
m2)(2:043107N=C)=3:618104C=m2.0Findingtheradiuswilltakealittlemorework.Wecanchooseonepoint,andmakethatthereferencepoint,andthensolveforR.StartingwithzEz=1p;20z2+R2khdaw.com16若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comandthenrearranging,20Ezz=1p;z2+R220Ez1=1p;1+(R=z)2120Ezp=1;1+(R=z)2211+(R=z)=;2(120Ez=)sR1=1:z(12E=)20zUsingz=0:03mandE=1:187107N=C,alongwithourvalueof=3:618104C=m2,wezndkhdaw.comsR1=
21;z12(8:8541012C2=Nm2)(1:187107N=C)=(3:618104C=m2)R=2:167(0:03m)=0:065m:(b)Andnowndthechargefromthechargedensityandtheradius,q=R2=(0:065m)2(3:618104C=m2)=4:80C:E26-18(a)=q=L.www.khdaw.com(b)Integrate:ZL+a12E=dxx;a4011=;40aL+aq1=;40a(L+a)since=q=L.(c)IfaLthen课后答案网Lcanbereplacedwith0intheaboveexpression.E26-19Asketchoftheeldlookslikethis.17khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comkhdaw.comE26-20(a)F=Eq=(40N=C)(1:601019C)=6:41018N(b)Linesaretwiceasfarapart,sotheeldishalfaslarge,orE=20N=C.www.khdaw.comE26-21Consideraviewofthediskonedge.课后答案网E26-22Asketchoftheeldlookslikethis.18khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comkhdaw.comE26-23Totheright.E26-24(a)Theelectriceldiszeronearertothesmallercharge;sincethechargeshaveoppositesignsitmustbetotherightofthe+2qcharge.Equatingthemagnitudesofthetwoelds,2q5q=;40x240(x+a)2orpp5x=www.khdaw.com2(x+a);whichhassolutionp2ax=pp=2:72a:52E26-25Thiscanbedonequicklywithaspreadsheet.E课后答案网xdE26-26(a)AtpointA,1q2q1qE==;40d2(2d)2402d219khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comwherethenegativesignindicatesthatE~isdirectedtotheleft.AtpointB,1q2q16qE==;40(d=2)2(d=2)240d2wherethepositivesignindicatesthatE~isdirectedtotheright.AtpointC,1q2q17qE=+=;40(2d)2d2404d2wherethenegativesignindicatesthatE~isdirectedtotheleft.E26-27(a)Theelectricelddoes(negative)workontheelectron.ThemagnitudeofthisworkisW=Fd,whereF=Eqisthemagnitudeoftheelectricforceontheelectronanddisthedistancethroughwhichtheelectronmoves.Combining,khdaw.comW=F~
~d=qE~
~d;whichgivestheworkdonebytheelectriceldontheelectron.TheelectronoriginallypossessedakineticenergyofK=1mv2,sincewewanttobringtheelectrontoarest,theworkdonemustbe2negative.Thechargeqoftheelectronisnegative,soE~and~darepointinginthesamedirection,andE~
~d=Ed.Bytheworkenergytheorem,12W=
K=0mv:2Weputallofthistogetherandndd,Wmv2(9:111031www.khdaw.comkg)(4:86106m=s)2d====0:0653m:qE2qE2(1:601019C)(1030N=C)(b)Eq=magivesthemagnitudeoftheacceleration,andvf=vi+atgivesthetime.Butvf=0.Combiningtheseexpressions,mv(9:111031kg)(4:86106m=s)t=i==2:69108s:Eq(1030N=C)(1:601019C)(c)Wewillapplytheworkenergytheoremagain,exceptnowwedon"tassumethenalkineticenergyiszero.Instead,课后答案网W=
K=KfKi;anddividingthroughbytheinitialkineticenergytogetthefractionlost,WKfKi==fractionalchangeofkineticenergy.KiKiButK=1mv2,andW=qEd,sothefractionalchangeisi2WqEd(1:601019C)(1030N=C)(7:88103m)===12:1%:Ki1mv21(9:111031kg)(4:86106m=s)222E26-28(a)a=Eq=m=(2:16104N=C)(1:601019C)=(1:671027kg)=2:071012m=s2.pp(b)v=2ax=2(2:071012m=s2)(1:22102m)=2:25105m=s:20khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE26-29(a)E=2q=4r2,or0(1:88107C)E==5:85105N=C:2(8:851012C2=N
m2)(0:152m=2)2(b)F=Eq=(5:85105N=C)(1:601019C)=9:361014N.E26-30(a)Theaveragespeedbetweentheplatesis(1:95102m)=(14:7109s)=1:33106m=s.Thespeedwithwhichtheelectronhitstheplateistwicethis,or2:65106m=s.(b)Theaccelerationisa=(2:65106m=s)=(14:7109s)=1:801014m=s2.TheelectriceldthenhasmagnitudeE=ma=q,orE=(9:111031kg)(1:801014m=s2)=(1:601019C)=1:03103N=C:E26-31Thedropisbalancediftheelectricforceisequaltotheforceofgravity,orEq=mg.Themassofthedropisgivenintermsofthedensitybykhdaw.com43m=V=r:3Combining,mg4r3g4(851kg=m3)(1:64106m)3(9:81m=s2)q====8:111019C:E3E3(1:92105N=C)Wewantthechargeintermsofe,sowedivide,andgetq(8:111019C)e=(1:601019www.khdaw.comC)=5:075:E26-32(b)F=(8:99109N
m2=C2)(2:16106C)(85:3109C)=(0:117m)2=0:121N:(a)E=F=q=(0:121N)=(2:16106C)=5:60104N=C:21E=F=q=(0:121N)=(85:3109C)=1:42106N=C:12E26-33IfeachvalueofqmeasuredbyMillikanwasamultipleofe,thenthedierencebetweenanytwovaluesofqmustalsobeamultipleofq.Thesmallestdierencewouldbethesmallestmultiple,andthismultiplemightbeunity.Thedierencesare1.641,1.63,1.60,1.63,3.30,3.35,3.18,3.24,alltimes1019C.Thisisaprettyclearindicationthatthefundamentalchargeisontheorderof1:61019课后答案网C.Ifso,thelikelynumberoffundamentalchargesoneachofthedropsisshownbelowinatablearrangedliketheoneinthebook:48125101471116Thetotalnumberofchargesis87,whilethetotalchargeis142:691019C,sotheaveragechargeperquantais1:641019C.21khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE26-34Becauseoftheelectriceldtheaccelerationtowardthegroundofachargedparticleisnotg,butgEq=m,wherethesigndependsonthedirectionoftheelectriceld.(a)Ifthelowerplateispositivelychargedthena=gEq=m.Replaceginthependulumperiodexpressionbythis,andthensLT=2:gEq=m(b)Ifthelowerplateisnegativelychargedthena=g+Eq=m.Replaceginthependulumperiodexpressionbythis,andthensLT=2:g+Eq=mE26-35Theinkdroptravelsanadditionaltimet0=d=v,wheredistheadditionalhorizontalxdistancebetweentheplatesandthepaper.Duringthistimeittravelsanadditionalverticaldistancey0=vt0,wherev=at=2y=t=2yv=L.Combining,khdaw.comyyx2yvt02yd2(6:4104m)(6:8103m)y0=x===5:44104m;LL(1:6102m)sothetotalde ectionisy+y0=1:18103m.E26-36(a)p=(1:48109C)(6:23106m)=9:221015C
m:(b)
U=2pE=2(9:221015C
m)(1100N=C)=2:031011J.E26-37Use=pEsin,whereistheanglebetween~pandE~.Forthisdipolep=qd=2edorp=2(1:61019C)(0:78109m)=2:510www.khdaw.com28C
m.ForallthreecasespE=(2:51028C
m)(3:4106N=C)=8:51022N
m:Theonlythingwecareaboutistheangle.(a)Fortheparallelcase=0,sosin=0,and=0.(b)Fortheperpendicularcase=90,sosin=1,and=8:51022N
m:.(c)Fortheanti-parallelcase=180,sosin=0,and=0.E26-38(c)Equalandopposite,or5:221016N.(d)UseEq.26-12andF=Eq.Then4x3F课后答案网0p=;q4(8:851012C2=N
m2)(0:285m)3(5:221016N)=;(3:16106C)=4:251022C
m:E26-39Thepoint-likenucleuscontributesanelectriceld1ZeE+=;40r2whiletheuniformsphereofnegativelychargedelectroncloudofradiusRcontributesanelectriceldgivenbyEq.26-24,1ZerE=:40R3khdaw.com22若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comThenetelectriceldisjustthesum,Ze1rE=40r2R3E26-40Theshelltheoremrstdescribedforgravitationinchapter14isapplicableheresincebothelectricforcesandgravitationalforcesfalloas1=r2.Thenetpositivechargeinsidethesphereofradiusd=2isgivenbyQ=2e(d=2)3=R3=ed3=4R3.Thenetforceoneitherelectronwillbezerowhene2eQ4e2d3e2d===;d2(d=2)2d24R3R3whichhassolutiond=R.P26-1(a)Letthepositivechargebelocatedclosertothepointinquestion,thentheelectriceldfromthepositivechargeiskhdaw.com1qE+=40(xd=2)2andisdirectedawayfromthedipole.Thenegativechargeislocatedfartherfromthepointinquestion,so1qE=40(x+d=2)2andisdirectedtowardthedipole.Thenetelectriceldisthesumofthesetwoelds,butsincethetwocomponenteldspointinoppositedirectionwemustactuallysubtractthesevalues,www.khdaw.comE=E+E;1q1q=;40(zd=2)240(z+d=2)21q11=40z2(1d=2z)2(1+d=2z)2Wecanusethebinomialexpansiononthetermscontaining1d=2z,1qE((1+d=z)(1d=z));课后答案网40z21qd=20z3(b)Theelectriceldisdirectedawayfromthepositivechargewhenyouareclosertothepositivecharge;theelectriceldisdirectedtowardthenegativechargewhenyouareclosertothenegativecharge.Inshort,alongtheaxistheelectriceldisdirectedinthesamedirectionasthedipolemoment.P26-2Thekeytothisproblemwillbetheexpansionof113zd1:(x2+(zd=2)2)3=2(x2+z2)3=22x2+z223khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.compfordx2+z2.Farfromthechargestheelectriceldofthepositivechargehasmagnitude1qE+=;40x2+(zd=2)2thecomponentsofthisare1qxEx;+=p;40x2+z2x2+(zd=2)21q(zd=2)Ez;+=p:40x2+z2x2+(zd=2)2Expandbothaccordingtotherstsentence,then1xq3zdEx;+1+;40(x2+z2)3=22x2+z21(zd=2)q3zdEz;+=1+:khdaw.com40(x2+z2)3=22x2+z2Similarexpressionexistforthenegativecharge,exceptwemustreplaceqwithqandthe+intheparentheseswitha,andzd=2withz+d=2intheEzexpression.Allthatisleftistoaddtheexpressions.Then1xq3zd1xq3zdEx=1++1;40(x2+z2)3=22x2+z240(x2+z2)3=22x2+z213xqzd=;40(x2+z2)5=21(zd=2)q3zd1(z+d=2)q3zdEz=1++1;40(x2+z2)3=22x2+z2www.khdaw.com40(x2+z2)3=22x2+z213z2dq1dq=;40(x2+z2)5=240(x2+z2)3=21(2z2x2)dq=:40(x2+z2)5=2pP26-3(a)Eachpointontheringisadistancez2+R2fromthepointontheaxisinquestion.Sinceallpointsareequaldistantandsubtendthesameanglefromtheaxisthenthetophalfoftheringcontributesq1zE1z=p;课后答案网40(x2+R2)z2+R2whilethebottomhalfcontributesasimilarexpression.Add,andq1+q2zqzEz==;40(z2+R2)3=240(z2+R2)3=2whichisidenticaltoEq.26-18.(b)Theperpendicularcomponentwouldbezeroifq1=q2.Itisn"t,soitmustbethedierenceq1q2whichisofinterest.Assumethischargedierenceisevenlydistributedonthetophalfofthering.Ifitisapositivedierence,thenE?mustpointdown.Weareonlyinterestedthenintheverticalcomponentasweintegratearoundthetophalfofthering.ThenZ1(q1q2)=E?=cosd;040z2+R2q1q21=:220z2+R2khdaw.com24若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP26-4Usetheapproximation1=(zd)2(1=z2)(12d=z+3d2=z2).Addthecontributions:1q2qqE=+;40(z+d)2z2(zd)2q2d3d22d3d21+2+1++;40z2zz2zz2q6d23Q==;40z2z240z4whereQ=2qd2.P26-5Amonopoleeldfallsoas1=r2.Adipoleeldfallsoas1=r3,andconsistsoftwooppositelychargemonopolesclosetogether.Aquadrupoleeld(seeExercise11aboveorreadProblem4)fallsoas1=r4and(can)consistoftwootherwiseidenticaldipolesarrangedwithanti-paralleldipolemoments.Justtakingaleapoffaithitseemsasifwecanconstructa1khdaw.com=r6eldbehaviorbyextendingthereasoning.Firstweneedanoctopolewhichisconstructedfromaquadrupole.Wewanttokeepthingsassimpleaspossible,sotheconstructionstepsare1.Themonopoleisacharge+qatx=0.2.Thedipoleisacharge+qatx=0andachargeqatx=a.We"llcallthisadipoleatx=a=23.Thequadrupoleisthedipoleatx=a=2,andaseconddipolepointingtheotherwayatx=a=2.Thechargesarethenqatx=a,+2qatx=0,andqatx=a.www.khdaw.com4.Theoctopolewillbetwostacked,osetquadrupoles.Therewillbeqatx=a,+3qatx=0,3qatx=a,and+qatx=2a.5.Finally,ourdistributionwithafareldbehaviorof1=r6.Therewillbe+qatx=2a,4qatx=a,+6qatx=0,4qatx=a,and+qatx=2a.P26-6TheverticalcomponentofE~issimplyhalfofEq.26-17.ThehorizontalcomponentisgivenbyavariationoftheworkrequiredtoderiveEq.26-16,1dzzdEz=dEsin=p;课后答案网40y2+z2y2+z2whichintegratestozeroifthelimitsare1to+1,butinthiscase,Z11Ez=dEz=:040zSincetheverticalandhorizontalcomponentsareequalthenE~makesanangleof45.P26-7(a)Swapallpositiveandnegativechargesintheproblemandtheelectriceldmustreversedirection.Butthisisthesameas ippingtheproblemover;consequently,theelectriceldmustpointparalleltotherod.ThisonlyholdstrueatpointP,becausepointPdoesn"tmovewhenyou iptherod.25khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com(b)Weareonlyinterestedintheverticalcomponentoftheeldascontributedfromeachpointontherod.Wecanintegrateonlyhalfoftherodanddoubletheanswer,sowewanttoevaluateZL=21dzzEz=2p;040y2+z2y2+z2p2(L=2)2+y2y=p:40y(L=2)2+y2(c)Thepreviousexpressionisexact.IfyL,thentheexpressionsimplieswithaTaylorexpansiontoL2Ez=;40y3whichlookssimilartoadipole.P26-8EvaluateZR1zdqkhdaw.comE=;040(z2+r2)3=2whereristheradiusofthering,zthedistancetotheplaneofthering,anddqthedierentialchargeonthering.Butr2+z2=R2,anddq=(2rdr),where=q=2R2.ThenZpRqR2r2rdrE=;040R5q1=:403R2P26-9Thekeystatementistheseconditalicizedparagraphonpage595;thenumberofeldwww.khdaw.comlinesthroughaunitcross-sectionalareaisproportionaltotheelectriceldstrength.Iftheexponentisn,thentheelectriceldstrengthadistancerfromapointchargeiskqE=;rnandthetotalcrosssectionalareaatadistanceristheareaofasphericalshell,4r2.Thenthenumberofeldlinesthroughtheshellisproportionaltokq22nEA=4r=4kqr:课后答案网rnNotethatthenumberofeldlinesvarieswithrifn6=2.Thismeansthataswegofartherfromthepointchargeweneedmoreandmoreeldlines(orfewerandfewer).Buttheeldlinescanonlystartoncharges,andwedon"thaveanyexceptforthepointcharge.Wehaveaproblem;wereallydoneedn=2ifwewantworkableeldlines.P26-10Thedistancetraveledbytheelectronwillbed=at2=2;thedistancetraveledbythe11protonwillbed=at2=2.aandaarerelatedbyma=ma,sincetheelectricforceisthe22121122same(samechargemagnitude).Thend+d=(a+a)t2=2isthe5.00cmdistance.Divideby1212theprotondistance,andthend1+d2a1+a2m2==+1:d2a2m1Thend=(5:00102m)=(1:671027=9:111031+1)=2:73105m:2khdaw.com26若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP26-11Thisismerelyafancyprojectilemotionproblem.vx=v0coswhilevy;0=v0sin.Thexandypositionsarex=vxtand1ax2y=at2+vt=+xtan:2y;02v2cos20TheaccelerationoftheelectronisverticallydownandhasamagnitudeofFEq(1870N=C)(1:61019C)a====3:2841014m=s2:mm(9:111031kg)Wewanttondouthowtheverticalvelocityoftheelectronatthelocationofthetopplate.Ifwegetanimaginaryanswer,thentheelectrondoesn"tgetashighasthetopplate.qvy=vy;02+2a
y;p=(5:83106m=s)2sin(39)2+2(3:2841014m=s2)(1:97102m);khdaw.com=7:226105m=s:Thisisarealanswer,sothismeanstheelectroneitherhitsthetopplate,oritmissesbothplates.Thetimetakentoreachtheheightofthetopplateis
v(7:226105m=s)(5:83106m=s)sin(39)t=y==8:972109s:a(3:2841014m=s2)Inthistimetheelectronhasmovedahorizontaldistanceofx=(5:83106m=s)cos(39)(8www.khdaw.com:972109s)=4:065102m:Thisisclearlyontheupperplate.P26-12NearthecenteroftheringzR,soaTaylorexpansionyieldszE=:20R2TheforceontheelectronisF=Ee,sotheeectivespring"constantisk=e=2R2.Thismeans0rrrkeeq!===:课后答案网m20mR240mR3P26-13U=pEcos,sotheworkrequiredto ipthedipoleisW=pE[cos(0+)cos0]=2pEcos0:P26-14Ifthetorqueonasystemisgivenbypjj=,whereisaconstant,thenthefrequencyofoscillationofthesystemisf==I=2.Inthiscase=pEsinpE,sopf=pE=I=2:27khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP26-15UsetheavariationoftheexactresultfromProblem26-1.Thetwochargearepositive,butsincewewilleventuallyfocusontheareabetweenthechargeswemustsubtractthetwoeldcontributions,sincetheypointinoppositedirections.Thenq11Ez=40(za=2)2(z+a=2)2andthentakethederivative,dEzq11=:dz20(za=2)3(z+a=2)3Applyingthebinomialexpansionforpointsza,dEz8q111=;dz20a3(2z=a1)3(2z=a+1)38q1((1+6z=a)(16z=a));khdaw.com20a38q1=:0a3Thereweresomefancysign ipsinthesecondline,soreviewthosestepscarefully!(b)Theelectrostaticforceonadipoleisthedierenceinthemagnitudesoftheelectrostaticforcesonthetwochargesthatmakeupthedipole.NearthecenteroftheabovechargearrangementtheelectriceldbehavesasdEzEzEz(0)+z+higherorderedterms.dzz=0www.khdaw.comThenetforceonadipoleisdEzdEzF+F=q(E+E)=qEz(0)+z+Ez(0)zdzdzz=0z=0wherethe+"and-"subscriptsrefertothelocationsofthepositiveandnegativecharges.ThislastlinecanbesimpliedtoyielddEzdEzq(z+z)=qd:dzdz课后答案网z=0z=028khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE27-1=(1800N=C)(3:2103m)2cos(145)=7:8103N
m2=C.EE27-2TherightfacehasanareaelementgivenbyA~=(1:4m)2^j.(a)=A~
E~=(2:0m2)^j
(6N=C)^i=0.E(b)=(2:0m2)^j
(2N=C)^j=4N
m2=C:E(c)=(2:0m2)^j
[(3N=C)^i+(4N=C)k^]=0:E(d)IneachcasetheeldisuniformsowecansimplyevaluateE=E~
A~,whereA~hassixparts,oneforeveryface.Thefaces,however,havethesamesizebutareorganizedinpairswithoppositedirections.Thesewillcancel,sothetotal uxiszeroinallthreecases.E27-3(a)The atbaseiseasyenough,sinceaccordingtoEq.27-7,ZE=E~
dA~:Therearetwoimportantfactstoconsiderinordertointegratethisexpression.khdaw.comE~isparalleltotheaxisofthehemisphere,E~pointsinwardwhiledA~pointsoutwardonthe atbase.E~isuniform,soitiseverywherethesameonthe atbase.SinceE~isanti-paralleltodA~,E~
dA~=EdA,thenZZE=E~
dA~=EdA:SinceE~isuniformwecansimplifythisasZZ=EdA=EdA=EA=R2E:EThelaststepsarejustsubstitutingtheareaofacircleforthe atsideofthehemisphere.www.khdaw.com(b)WemustrstsortoutthedotproductEdAR课后答案网qfWecansimplifythevectorpartoftheproblemwithE~
dA~=cosEdA,soZZE=E~
dA~=cosEdAOnceagain,E~isuniform,sowecantakeitoutoftheintegral,ZZE=cosEdA=EcosdAFinally,dA=(Rd)(Rsind)onthesurfaceofaspherecenteredonkhdaw.comR=0.29若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comWe"llintegratearoundtheaxis,from0to2.We"llintegratefromtheaxistotheequator,from0to=2.ThenZZ2Z=2=EcosdA=ER2cossindd:E00Pullingouttheconstants,doingtheintegration,andthenwriting2cossinassin(2),Z=2Z=2=2R2Ecossind=R2Esin(2)d;E00Changevariablesandlet=2,thenwehaveZ212E=REsind=RE:02E27-4khdaw.comThroughS1,E=q=0.ThroughS2,E=q=0.ThroughS3,E=q=0.ThroughS4,E=0.ThroughS5,E=q=0.E27-5ByEq.27-8,q(1:84C)52E==2=2:0810N
m=C:0(8:851012C=N
m2)E27-6Thetotal uxthroughthesphereis=(1+23+45+6)(103N
m2=C)=3103N
m2=C:Ewww.khdaw.comThechargeinsidethedieis(8:851012C2=N
m2)(3103N
m2=C)=2:66108C:E27-7Thetotal uxthroughacubewouldbeq=0.Sincethechargeisinthecenterofthecubeweexpectthatthe uxthroughanysidewouldbethesame,or1=6ofthetotal ux.Hencethe uxthroughthesquaresurfaceisq=60.E27-8Iftheelectriceldisuniformthentherearenofreechargesnear(orinside)thenet.The uxthroughthenettingmustbeequalto,butoppositeinsign,fromthe uxthroughtheopening.The uxthroughtheopeningisEa2,sothe uxthroughthenettingisEa2.课后答案网E27-9Thereisno uxthroughthesidesofthecube.The uxthroughthetopofthecubeis(58N=C)(100m)2=5:8105N
m2=C.The uxthroughthebottomofthecubeis(110N=C)(100m)2=1:1106N
m2=C:Thetotal uxisthesum,sothechargecontainedinthecubeisq=(8:851012C2=N
m2)(5:2105N
m2=C)=4:60106C:E27-10(a)Thereisonlya uxthroughtherightandleftfaces.Throughtherightface=(2:0m2)^j
(3N=C
m)(1:4m)^j=8:4N
m2=C:RThe uxthroughtheleftfaceiszerobecausey=0.khdaw.com30若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE27-11Thereareeightcubeswhichcanbewrapped"aroundthecharge.Eachcubehasthreeexternalfacesthatareindistinguishableforatotaloftwenty-fourfaces,eachwiththesame uxE.Thetotal uxisq=0,sothe uxthroughonefaceisE=q=240.Notethatthisisthe uxthroughfacesoppositethecharge;forfaceswhichtouchthechargetheelectriceldisparalleltothesurface,sothe uxwouldbezero.E27-12UseEq.27-11,=2rE=2(8:851012C2=N
m2)(1:96m)(4:52104N=C)=4:93106C=m:0E27-13(a)q=A=(2:0106C=m2)(0:12m)(0:42m)=3:17107C:(b)Thechargedensitywillbethesame!q=A=(2:0106C=m2)(0:08m)(0:28m)=1:41107C:E27-14TheelectriceldfromthesheetontheleftisofmagnitudeEl==20,andpointsdirectlyawayfromthesheet.Themagnitudeoftheelectriceldfromthesheetontherightisthesame,khdaw.combutitpointsdirectlyawayfromthesheetontheright.(a)Totheleftofthesheetsthetwoeldsaddsincetheypointinthesamedirection.ThismeansthattheelectriceldisE~=(=0)^i.(b)Betweenthesheetsthetwoelectriceldscancel,soE~=0.(c)Totherightofthesheetsthetwoeldsaddsincetheypointinthesamedirection.ThismeansthattheelectriceldisE~=(=0)^i.E27-15TheelectriceldfromtheplateontheleftisofmagnitudeEl==20,andpointsdirectlytowardtheplate.Themagnitudeoftheelectriceldfromtheplateontherightisthesame,butitpointsdirectlyawayfromtheplateontheright.www.khdaw.com(a)Totheleftoftheplatesthetwoeldscancelsincetheypointintheoppositedirections.ThismeansthattheelectriceldisE~=0.(b)Betweentheplatesthetwoelectriceldsaddsincetheypointinthesamedirection.ThismeansthattheelectriceldisE~=(=0)^i.(c)Totherightoftheplatesthetwoeldscancelsincetheypointintheoppositedirections.ThismeansthattheelectriceldisE~=0.E27-16ThemagnitudeoftheelectriceldisE=mg=q.Thesurfacechargedensityontheplatesis=0E=0mg=q,or(8:85课后答案网1012C2=N
m2)(9:111031kg)(9:81m=s2)==4:941022C=m2:(1:601019C)E27-17Wedon"treallyneedtowriteanintegral,wejustneedthechargeperunitlengthinthecylindertobeequaltozero.Thismeansthatthepositivechargeincylindermustbe+3:60nC=m.ThispositivechargeisuniformlydistributedinacircleofradiusR=1:50cm,so3:60nC=m3:60nC=m3===5:09C=m:R2(0:0150m)231khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE27-18Theproblemhassphericalsymmetry,souseaGaussiansurfacewhichisasphericalshell.TheE~eldwillbeperpendiculartothesurface,soGauss"lawwillsimplifytoIIIq==E~
dA~=EdA=EdA=4r2E:enc0(a)ForpointPthechargeenclosedisq=1:26107C,so1enc(1:26107C)E==3:38106N=C:4(8:851012C2=N
m2)(1:83102m)2(b)InsideaconductorE=0.E27-19Theprotonorbitswithaspeedv,sothecentripetalforceontheprotonisF=mv2=r.CThiscentripetalforceisfromtheelectrostaticattractionwiththesphere;solongastheprotonisoutsidethespheretheelectriceldisequivalenttothatofapointchargeQ(Eq.27-15),khdaw.com1QE=:40r2IfqisthechargeontheprotonwecanwriteF=Eq,ormv21Q=qr40r2SolvingforQ,4mv2r0Q=;qwww.khdaw.com4(8:851012C2=N
m2)(1:671027kg)(294103m=s)2(0:0113m)=;(1:601019C)=1:13109C:E27-20Theproblemhassphericalsymmetry,souseaGaussiansurfacewhichisasphericalshell.TheE~eldwillbeperpendiculartothesurface,soGauss"lawwillsimplifytoIIIq==E~
dA~=EdA=EdA=4r2E:enc0课后答案网8(a)Atr=0:120mqenc=4:0610C.Then(4:06108C)E==2:54104N=C:4(8:851012C2=N
m2)(1:20101m)2(b)Atr=0:220mq=5:99108C.Thenenc(5:99108C)E==1:11104N=C:4(8:851012C2=N
m2)(2:20101m)2(c)Atr=0:0818mqenc=0C.ThenE=0.32khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE27-21Theproblemhascylindricalsymmetry,souseaGaussiansurfacewhichisacylindricalshell.TheE~eldwillbeperpendiculartothecurvedsurfaceandparalleltotheendsurfaces,soGauss"lawwillsimplifytoIZZqenc=0=E~
dA~=EdA=EdA=2rLE;whereListhelengthofthecylinder.Notethat=q=2rLrepresentsasurfacechargedensity.(a)r=0:0410misbetweenthetwocylinders.Then(24:1106C=m2)(0:0322m)E==2:14106N=C:(8:851012C2=N
m2)(0:0410m)Itpointsoutward.(b)r=0:0820misoutsidethetwocylinders.Then(24:1106C=m2)(0:0322m)+(18:0106C=m2)(0:0618m)E==4:64105N=C:khdaw.com(8:851012C2=N
m2)(0:0820m)Thenegativesignisbecauseitispointinginward.E27-22Theproblemhascylindricalsymmetry,souseaGaussiansurfacewhichisacylindricalshell.TheE~eldwillbeperpendiculartothecurvedsurfaceandparalleltotheendsurfaces,soGauss"lawwillsimplifytoIZZqenc=0=E~
dA~=EdA=EdA=2rLE;whereListhelengthofthecylinder.Thechargeenclosediswww.khdaw.comZ
q=dV=Lr2R2encTheelectriceldisgivenby

Lr2R2r2R2E==:20rL20rAtthesurface,
(2R)2R23REs==:课后答案网202R40SolveforrwhenEishalfofthis:3Rr2R2=;82r3rR=4r24R2;0=4r23rR4R2:Thesolutionisr=1:443R.That"s(2R1:443R)=0:557Rbeneaththesurface.E27-23Theelectriceldmustdoworkontheelectrontostopit.TheelectriceldisgivenbyE==20.TheworkdoneisW=Fd=Eqd.disthedistanceinquestion,so2K2(8:851012C2=N
m2)(1:15105eV)0d===0:979mq(2:08106C=m2)ekhdaw.com33若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE27-24LetthesphericalGaussiansurfacehavearadiusofRandbecenteredontheorigin.Choosetheorientationoftheaxissothattheinnitelineofchargeisalongthezaxis.TheelectriceldisthendirectedradiallyoutwardfromthezaxiswithmagnitudeE==20,whereistheperpendiculardistancefromthezaxis.NowwewanttoevaluateIE=E~
dA~;overthesurfaceofthesphere.Insphericalcoordinates,dA=R2sindd,=Rsin,andE~
dA~=EAsin.ThenI2RE=sinRdd=:200E27-25(a)Theproblemhascylindricalsymmetry,souseaGaussiansurfacewhichisacylindricalshell.TheE~eldwillbeperpendiculartothecurvedsurfaceandparalleltotheendsurfaces,soGauss"lawwillsimplifytoIZZkhdaw.comqenc=0=E~
dA~=EdA=EdA=2rLE;whereListhelengthofthecylinder.Nowfortheqencpart.Ifthe(uniform)volumechargedensityis,thenthechargeenclosedintheGaussiancylinderisZZq=dV=dV=V=r2L:encCombining,r2L==E2rLorE=r=2:00(b)OutsidethechargedcylinderthechargeenclosedintheGaussiansurfaceisjustthechargeinthecylinder.ThenZZq=dV=dV=V=R2L:encwww.khdaw.comandR2L==E2rL;0andthennallyR2E=:20rE27-26(a)q=4(1:22m)2(8:13106C=m2)=1:52104C.(b)=q==(1:52104C)=(8:851012C2=N
m2)=1:72107N
m2=C:E0(c)E===(8:13106C=m2)=(8:851012C2=N
m2)=9:19105N=C0课后答案网E27-27(a)=(2:4106C)=4(0:65m)2=4:52107C=m2:(b)E===(4:52107C=m2)=(8:851012C2=N
m2)=5:11104N=C:0E27-28E===q=4r2.00E27-29(a)TheneareldisgivenbyEq.27-12,E==20,so(6:0106C)=(8:0102m)2E=5:3107N=C:2(8:851012C2=N
m2)(b)Veryfarfromanyobjectapointchargeapproximationisvalid.Then1q1(6:0106C)E===60N=C:40r24(8:851012C2=N
m2)(30m)khdaw.com234若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP27-1ForasphericallysymmetricmassdistributionchooseasphericalGaussianshell.ThenIII~g
dA~=gdA=gdA=4r2g:Thengr2g==m;4GGorGmg=:r2Thenegativesignindicatesthedirection;~gpointtowardthemasscenter.P27-2(a)The uxthroughallsurfacesexcepttherightandleftfaceswillbezero.Throughtheleftface,p=EA=baa2:lyThroughtherightface,khdaw.comp=EA=b2aa2:ryThenet uxisthenpp=ba5=2(21)=(8830N=C
m1=2)(0:130m)5=2(21)=22:3N
m2=C:(b)Thechargeenclosedisq=(8:851012C2=N
m2)(22:3N
m2=C)=1:971010C.P27-3Thenetforceonthesmallsphereiszero;thisforceisthevectorsumoftheforceofgravityW,theelectricforceFE,andthetensionT.www.khdaw.comqTFEW课后答案网TheseforcesarerelatedbyEq=mgtan:WealsohaveE==20,so20mgtan=;q2(8:851012C2=N
m2)(1:12106kg)(9:81m=s2)tan(27:4)=;(19:7109C)=5:11109C=m2:35khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP27-4Thematerialsareconducting,soallchargewillresideonthesurfaces.Theelectriceldinsideanyconductoriszero.Theproblemhassphericalsymmetry,souseaGaussiansurfacewhichisasphericalshell.TheE~eldwillbeperpendiculartothesurface,soGauss"lawwillsimplifytoIIIq==E~
dA~=EdA=EdA=4r2E:enc0Consequently,E=q=4r2.enc0(a)WithinthesphereE=0.(b)Betweenthesphereandtheshellq=q.ThenE=q=4r2.enc0(c)WithintheshellE=0.(d)Outsidetheshellqenc=+qq=0.ThenE=0.(e)SinceE=0insidetheshell,qenc=0,thisrequiresthatqresideontheinsidesurface.Thisisnochargeontheoutsidesurface.P27-5Theproblemhascylindricalsymmetry,souseaGaussiansurfacewhichisacylindricalshell.TheE~eldwillbeperpendiculartothecurvedsurfaceandparalleltotheendsurfaces,soGauss"lawwillsimplifytokhdaw.comIZZqenc=0=E~
dA~=EdA=EdA=2rLE;whereListhelengthofthecylinder.Consequently,E=qenc=20rL.(a)Outsidetheconductingshellqenc=+q2q=q.ThenE=q=20rL.Thenegativesignindicatesthattheeldispointinginwardtowardtheaxisofthecylinder.(b)SinceE=0insidetheconductingshell,qenc=0,whichmeansachargeofqisontheinsidesurfaceoftheshell.Theremainingqmustresideontheoutsidesurfaceoftheshell.(c)Intheregionbetweenthecylindersqenc=+q.ThenE=+q=20rL.Thepositivesignindicatesthattheeldispointingoutwardfromtheaxisofthecylinder.www.khdaw.comP27-6SubtractEq.26-19fromEq.26-20.ThenzE=p:20z2+R2P27-7ThisproblemiscloselyrelatedtoEx.27-25,exceptforthepartconcerningqenc.We"llsetuptheproblemthesameway:theGaussiansurfacewillbea(imaginary)cylindercenteredontheaxisofthephysicalcylinder.ForGaussiansurfacesofradiusrR,qenc=l.We"vealreadyworkedouttheintegral课后答案网ZE~
dA~=2rlE;tubeforthecylinder,andthenfromGauss"law,Zqenc=0E~
dA~=20rlE:tube(a)WhenrRthereisachargelenclosed,soE=:20rkhdaw.com36若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP27-8ThisproblemiscloselyrelatedtoEx.27-25,exceptforthepartconcerningqenc.We"llsetuptheproblemthesameway:theGaussiansurfacewillbea(imaginary)cylindercenteredontheaxisofthephysicalcylinders.ForGaussiansurfacesofradiusrr>a,qenc=l.We"vealreadyworkedouttheintegralZE~
dA~=2rlE;tubeforthecylinder,andthenfromGauss"law,Zqenc=0E~
dA~=20rlE:tube(a)Whenrr>athereisachargelenclosed,soE=:20rP27-9Uniformcircularorbitsrequireaconstantnetforcetowardsthecenter,soF=Eq=q=2r.ThespeedofthepositronisgivenbyF=mv2=r;thekineticenergyisK=mv2=2=0Fr=2.Combining,qK=;40(30109C=m)(1:61019C)=www.khdaw.com;4((8:851012C2=N
m2)=4:311017J=270eV:P27-10=20rE,soq=2(8:851012C2=N
m2)(0:014m)(0:16m)(2:9104N=C)=3:6109C:P27-11(a)PutasphericalGaussiansurfaceinsidetheshellcenteredonthepointcharge.Gauss"lawstatesI课后答案网E~
dA~=qenc:0SincethereissphericalsymmetrytheelectriceldisnormaltothesphericalGaussiansurface,anditiseverywherethesameonthissurface.ThedotproductsimpliestoE~
dA~=EdA,whilesinceEisaconstantonthesurfacewecanpullitoutoftheintegral,andweendupwithIqEdA=;0Hwhereqisthepointchargeinthecenter.NowdA=4r2,whereristheradiusoftheGaussiansurface,soqE=:40r2(b)Repeattheabovesteps,exceptputtheGaussiansurfaceoutsidetheconductingshell.Keepitcenteredonthecharge.Twothingsaredierentfromtheabovederivation:(1)khdaw.comrisbigger,and37若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com(2)thereisanunchargedsphericalconductingshellinsidetheGaussiansurface.Neitherchangewillaectthesurfaceintegralorqenc,sotheelectriceldoutsidetheshellisstillqE=;40r2(c)Thisisasubtlequestion.Withallthesymmetryhereitappearsasiftheshellhasnoeect;theeldjustlookslikeapointchargeeld.If,however,thechargeweremovedocentertheeldinsidetheshellwouldbecomedistorted,andwewouldn"tbeabletouseGauss"lawtondit.Sotheshelldoesmakeadierence.Outsidetheshell,however,wecan"ttellwhatisgoingoninsidetheshell.Sotheelectriceldoutsidetheshelllookslikeapointchargeeldoriginatingfromthecenteroftheshellregardlessofwhereinsidetheshellthepointchargeisplaced!(d)Yes,qinducessurfacechargesontheshell.Therewillbeachargeqontheinsidesurfaceandachargeqontheoutsidesurface.(e)Yes,asthereisanelectriceldfromtheshell,isn"tthere?(f)No,astheelectriceldfromtheoutsidechargewon"tmakeitthroughaconductingshell.Theconductoractsasashield.khdaw.com(g)No,thisisnotacontradiction,becausetheoutsidechargeneverexperiencedanyelectrostaticattractionorrepulsionfromtheinsidecharge.Theforceisbetweentheshellandtheoutsidecharge.P27-12Therepulsiveelectrostaticforcesmustexactlybalancetheattractivegravitationalforces.Then1q2m2=G;40r2r2porm=q=40G:Numerically,(1:601019C)m=qwww.khdaw.com=1:86109kg:4(8:851012C2=N
m2)(6:671011N
m2=kg2)P27-13Theproblemhassphericalsymmetry,souseaGaussiansurfacewhichisasphericalshell.TheE~eldwillbeperpendiculartothesurface,soGauss"lawwillsimplifytoIIIq==E~
dA~=EdA=EdA=4r2E:enc0Consequently,E=q=4r2.Renc0q=q+4r2dr,orenc课后答案网Zrq=q+4Ardr=q+2A(r2a2):encaTheelectriceldwillbeconstantifqbehavesasr2,whichrequiresq=2Aa2,orA=q=2a2.encP27-14(a)Theproblemhassphericalsymmetry,souseaGaussiansurfacewhichisasphericalshell.TheE~eldwillbeperpendiculartothesurface,soGauss"lawwillsimplifytoIIIq==E~
dA~=EdA=EdA=4r2E:enc0Consequently,E=q=4r2.Renc0q=4r2dr=4r3=3,soencE=r=30khdaw.com38若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comandisdirectedradiallyoutfromthecenter.ThenE~=~r=30.(b)TheelectriceldintheholeisgivenbyE~h=E~E~b,whereE~istheeldfrompart(a)andE~bistheeldthatwouldbeproducedbythematterthatwouldhavebeenintheholehadtheholenotbeenthere.ThenE~b=~b=30;where~bisavectorpointingfromthecenterofthehole.Then~r~bE~h==(~r~b):303030But~r~b=~a,soE~h=~a=30.P27-15Ifapointisanequilibriumpointthentheelectriceldatthatpointshouldbezero.Ifitisastablepointthenmovingthetestcharge(assumedpositive)asmalldistancefromtheequilibriumpointshouldresultinarestoringforcedirectedbacktowardtheequilibriumpoint.Inotherwords,therewillbeapointwheretheelectriceldiszero,andaroundthispointtherewillbekhdaw.comanelectriceldpointinginward.ApplyingGauss"lawtoasmallsurfacesurroundingourpointP,wehaveanetinward ux,sotheremustbeanegativechargeinsidethesurface.ButthereshouldbenothinginsidethesurfaceexceptanemptypointP,sowehaveacontradiction.P27-16(a)FollowtheexampleonPage618.BysymmetryE=0alongthemedianplane.Thechargeenclosedbetweenthemedianplaneandasurfaceadistancexfromtheplaneisq=Ax.ThenE=Ax=0A=A=0:(b)Outsidetheslabthechargeenclosedbetweenthemedianplaneandasurfaceadistancewww.khdaw.comxfromtheplaneisisq=Ad=2,regardlessofx.TheE=Ad=2=0A=d=20:P27-17(a)Thetotalchargeisthevolumeintegraloverthewholesphere,ZQ=dV:Thisisactuallyathreedimensionalintegral,anddV=Adr,whereA=4r2.Then课后答案网ZQ=dV;ZRSr2=4rdr;0R4S14=R;R4=R3:S(b)PutasphericalGaussiansurfaceinsidethespherecenteredonthecenter.WecanuseGauss"lawherebecausethereissphericalsymmetryintheentireproblem,bothinsideandoutsidetheGaussiansurface.Gauss"lawstatesIE~
dA~=qenc:039khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comSincethereissphericalsymmetrytheelectriceldisnormaltothesphericalGaussiansurface,anditiseverywherethesameonthissurface.ThedotproductsimpliestoE~
dA~=EdA,whilesinceEisaconstantonthesurfacewecanpullitoutoftheintegral,andweendupwithIqencEdA=;0HNowdA=4r2,whereristheradiusoftheGaussiansurface,soqencE=:40r2Wearen"tdoneyet,becausethechargeencloseddependsontheradiusoftheGaussiansurface.Weneedtodopart(a)again,exceptthistimewedon"twanttodothewholevolumeofthesphere,weonlywanttogooutasfarastheGaussiansurface.ThenZqenc=dV;khdaw.comZrSr2=4rdr;0R4S14=r;R4r4=S:RCombinetheselasttworesultsandr4SE=;4www.khdaw.com0r2Rr2S=;40RQr2=:40R4Inthelastlineweusedtheresultsofpart(a)toeliminateSfromtheexpression.P27-18(a)InsidetheconductorE=0,soaGaussiansurfacewhichisembeddedintheconductorbutcontainingtheholemusthaveanetenclosedchargeofzero.Thecavitywallmustthenhaveachargeof3:0C.课后答案网(b)Thenetchargeontheconductoris+10:0C;thechargeontheoutersurfacemustthenbe+13:0C.P27-19(a)InsidetheshellE=0,sothenetchargeinsideaGaussiansurfaceembeddedintheshellmustbezero,sotheinsidesurfacehasachargeQ.(b)StillQ;theoutsidehasnothingtodowiththeinside.(c)(Q+q);seereason(a).(d)Yes.40khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comThroughoutthischapterwewillusetheconventionthatV(1)=0unlessexplicitlystatedotherwise.ThenthepotentialinthevicinityofapointchargewillbegivenbyEq.28-18,V=q=40r.E28-1(a)LetU12bethepotentialenergyoftheinteractionbetweenthetwoup"quarks.Then(2=3)2e(1:601019C)U=(8:99109N
m2=C2)=4:84105eV:12(1:321015m)(b)LetU13bethepotentialenergyoftheinteractionbetweenanup"quarkandadown"quark.Then(1=3)(2=3)e(1:601019C)U=(8:99109N
m2=C2)=2:42105eV13(1:321015m)NotethatU13=U23.Thetotalelectricpotentialenergyisthesumofthesethreeterms,orzero.E28-2khdaw.comTherearesixinteractionterms,oneforeverychargepair.Numberthechargesclockwisefromtheupperlefthandcorner.ThenU=q2=4a;120U=q2=4a;230U=q2=4a;340U=q2=4a;410pU=(q)2=4(2a);130pU=q2=4(2a):24www.khdaw.com0Addthesetermsandget2q2q2U=p4=0:206240a0aTheamountofworkrequiredisW=U.E28-3(a)Webuildtheelectrononepartatatime;eachparthasachargeq=e=3.Movingtherstpartfrominnitytothelocationwherewewanttoconstructtheelectroniseasyandtakesnoworkatall.Movingthesecondpartinrequiresworktochangethepotentialenergyto1q1q2课后答案网U12=;40rwhichisbasicallyEq.28-7.Theseparationr=2:821015m.Bringinginthethirdpartrequiresworkagainsttheforceofrepulsionbetweenthethirdchargeandbothoftheothertwocharges.PotentialenergythenexistsintheformU13andU23,whereallthreechargesarethesame,andallthreeseparationsarethesame.ThenU12=U13=U12,sothetotalpotentialenergyofthesystemis1(e=3)23(1:601019C=3)2U=3==2:721014J40r4(8:851012C2=N
m2)(2:821015m)(b)Dividingouranswerbythespeedoflightsquaredtondthemass,2:721014Jm==3:021031kg:(3:00108m=s)2khdaw.com41若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE28-4Therearethreeinteractionterms,oneforeverychargepair.Numberthechargesfromtheleft;leta=0:146m.Then(25:5109C)(17:2109C)U12=;40a(25:5109C)(19:2109C)U13=;40(a+x)(17:2109C)(19:2109C)U23=:40xAddtheseandsetitequaltozero.Then(25:5)(17:2)(25:5)(19:2)(17:2)(19:2)=+;aa+xxwhichhassolutionkhdaw.comx=1:405a=0:205m.E28-5Thevolumeofthenuclearmaterialis4a3=3,wherea=8:01015m.Upondividinginp33315halfeachpartwillhavearadiusrwhere4r=3=4a=6.Consequently,r=a=2=6:3510m.Eachfragmentwillhaveachargeof+46e.(a)Theforceofrepulsionis(46)2(1:601019C)2F==3000N4(8:851012C2=N
m2)[2(6:351015m)]2(b)Thepotentialenergyis(46)2e(1:6010www.khdaw.com19C)U==2:4108eV4(8:851012C2=N
m2)2(6:351015m)E28-6Thisisawork/kineticenergyproblem:1mv2=q
V.Then20s2(1:601019C)(10:3103V)7v0==6:010m=s:(9:111031kg)E28-7(a)Theenergyreleasedisequaltothechargestimesthepotentialthroughwhichthechargewasmoved.Then课后答案网
U=q
V=(30C)(1:0109V)=3:01010J:(b)Althoughtheproblemmentionsacceleration,wewanttofocusonenergy.Theenergywillchangethekineticenergyofthecarfrom0toK=3:01010J.Thespeedofthecaristhenfrs2K2(3:01010J)v===7100m=s:m(1200kg)(c)TheenergyrequiredtomelticeisgivenbyQ=mL,whereListhelatentheatoffusion.ThenQ(3:01010J)m===90;100kg:L(3:33105J=kg)42khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE28-8(a)
U=(1:601019C)(1:23109V)=1:971010J:(b)
U=e(1:23109V)=1:23109eV.E28-9Thisisanenergyconservationproblem:1mv2=q
V;
V=q=4(1=r1=r).Com-2012bining,sq211v=;20mr1r2s(3:1106C)211=;2(8:851012C2=N
m2)(18106kg)(0:90103m)(2:5103m)=2600m=s:E28-10Thisisanenergyconservationproblem:1q21khdaw.comm(2v)2=mv2:240r2Rearrange,q2r=;60mv2(1:601019C)2==1:42109m:6(8:851012C2=N
m2)(9:111031kg)(3:44105m=s)2)E28-11(a)V=(1:601019C)=4(8:851012www.khdaw.comC2=N
m2)(5:291011m)=27:2V.(b)U=qV=(e)(27:2V)=27:2eV.(c)ForuniformcircularorbitsF=mv2=r;theforceiselectrical,orF=e2=4r2.Kinetic0energyisK=mv2=2=Fr=2,soe2(1:601019C)K===13:6eV:80r8(8:851012C2=N
m2)(5:291011m)(d)Theionizationenergyis(K+U),orEion=[(13:6eV)+(27:2eV)]=13:6eV:E28-12(a)Theelectricpotentialat课后答案网Ais1qq(5:0106C)(2:0106C)V=1+2=(8:99109N
m2=C)+=6:0104V:A40r1r2(0:15m)(0:05m)TheelectricpotentialatBis1qq(5:0106C)(2:0106C)V=1+2=(8:99109N
m2=C)+=7:8105V:B40r2r1(0:05m)(0:15m)(b)W=q
V=(3:0106C)(6:0104V7:8105V)=2:5J.(c)Sinceworkispositivethenexternalworkisconvertedtoelectrostaticpotentialenergy.43khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE28-13(a)ThemagnitudeoftheelectriceldwouldbefoundfromF(3:901015N)E===2:44104N=C:q(1:601019C)(b)ThepotentialdierencebetweentheplatesisfoundbyevaluatingEq.28-15,Zb
V=E~
d~s:aTheelectriceldbetweentwoparallelplatesisuniformandperpendiculartotheplates.ThenE~
d~s=Edsalongthispath,andsinceEisuniform,ZbZbZb
V=E~
d~s=Eds=Eds=E
x;aaawhere
khdaw.comxistheseparationbetweentheplates.Finally,
V=(2:44104N=C)(0:120m)=2930V.E28-14
V=E
x,so22(8:851012C2=N
m2)
x=0
V=(48V)=7:1103m(0:12106C=m2)E28-15TheelectriceldaroundaninnitelylongstraightwireisgivenbyE==20r.ThepotentialdierencebetweentheinnerwireandtheoutercylinderisgivenbyZb
V=(=20r)www.khdaw.comdr=(=20)ln(a=b):aTheelectriceldnearthesurfaceofthewireisthengivenby
V(855V)8E====1:3210V=m:20aaln(a=b)(6:70107m)ln(6:70107m=1:05102m)Theelectriceldnearthesurfaceofthecylinderisthengivenby
V(855V)3E====8:4310V=m:20aaln(a=b)(1:05102m)ln(6:70107m=1:05102m)课后答案网523E28-16
V=E
x=(1:9210N=C)(1:5010m)=2:8810V:E28-17(a)Thisisanenergyconservationproblem:1(2)(79)e2(2)(79)e(1:601019C)K==(8:99109N
m2=C)=3:2107eV40r(7:01015m)(b)ThealphaparticlesusedbyRutherfordnevercameclosetohittingthegoldnuclei.E28-18Thisisanenergyconservationproblem:mv2=2=eq=4r,or0s(1:601019C)(1:761015C)4v==2:1310m=s2(8:851012C2=N
m2)(1:22102m)(9:111031kg)44khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE28-19(a)WeevaluateVAandVBindividually,andthenndthedierence.1q1(1:16C)VA==2=5060V;40r4(8:851012C=N
m2)(2:06m)and1q1(1:16C)VB==2=8910V;40r4(8:851012C=N
m2)(1:17m)ThedierenceisthenVAVB=3850V.(b)Theansweristhesame,sincewhenconcerningourselveswithelectricpotentialweonlycareaboutdistances,andnotdirections.E28-20Thenumberofexcess"electronsoneachgrainis4rV4(8:851012C2=N
m)(1:0106m)(400V)05n===2:810khdaw.come(1:601019C)E28-21Theexcesschargeontheshuttleisq=4rV=4(8:851012C2=N
m)(10m)(1:0V)=1:1109C0E28-22q=1:37105C,so(1:37105C)V=(8:99109N
m2=C2)=1:93108V:(6:37106m)E28-23Theratiooftheelectricpotentialtotheelectriceldstrengthiswww.khdaw.comV1q1q===r:E40r40r2InthisproblemristheradiusoftheEarth,soatthesurfaceoftheEarththepotentialisV=Er=(100V=m)(6:38106m)=6:38108V:E28-24UseEq.28-22:(1:47)(3:341030C
m)V=(8课后答案网:99109N
m2=C2)=1:63105V:(52:0109m)2E28-25(a)WhenndingVAweneedtoconsiderthecontributionfromboththepositiveandthenegativecharge,so1qVA=qa+40a+dTherewillbeasimilarexpressionforVB,1qVB=qa+:40a+d45khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comNowtoevaluatethedierence.1q1qVAVB=qa+qa+;40a+d40a+dq11=;20aa+dqa+da=;20a(a+d)a(a+d)qd=:20a(a+d)(b)Doesitdowhatweexpectwhend=0?IexpectitthedierencetogotozeroasthetwopointsAandBgetclosertogether.Thenumeratorwillgotozeroasdgetssmaller.Thedenominator,however,staysnite,whichisagoodthing.Soyes,VaVB!0asd!0.E28-26(a)Sincebothchargesarepositivetheelectricpotentialfrombothchargeswillbepositive.Therewillbenokhdaw.comnitepointswhereV=0,sincetwopositivescan"taddtozero.(b)Betweenthechargestheelectriceldfromeachchargepointstowardtheother,soE~willppvanishwhenq=x2=2q=(dx)2.Thishappenswhendx=2x,orx=d=(1+2).pE28-27ThedistancefromCtoeitherchargeis2d=2=1:39102m.(a)VatCis2(2:13106C)V=(8:99109N
m2=C2)=2:76106V(1:39102m)(b)W=qV=(1:91106C)(2:76106V)=5:27J.(c)Don"tforgetaboutthepotentialenergyoftheoriginaltwocharges!www.khdaw.com62922(2:1310C)U0=(8:9910N
m=C)=2:08J(1:96102m)Addthistotheanswerfrompart(b)toget7:35J.E28-28ThepotentialisgivenbyEq.28-32;atthesurfaceVs=R=20,halfofthisoccurswhenpR2+z2z=R=2;R2+z2=R2=4+Rz+z2;3R=4=z:课后答案网E28-29Wecanndthelinearchargedensitybydividingthechargebythecircumference,Q=;2RwhereQreferstothechargeonthering.TheworkdonetomoveachargeqfromapointxtotheoriginwillbegivenbyW=q
V;W=q(V(0)V(x));1Q1Q=qpp;40R240R2+x2qQ11=p:40RR2+x2khdaw.com46若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comPuttinginthenumbers,!(5:931012C)(9:12109C)11p=1:861010J:4(8:851012C2=N
m2)1:48m(1:48m)2+(3:07m)2E28-30(a)TheelectriceldstrengthisgreatestwherethegradientofVisgreatest.Thatisbetweendande.(b)Theleastabsolutevalueoccurswherethegradientiszero,whichisbetweenbandcandagainbetweeneandf.E28-31Thepotentialonthepositiveplateis2(5:52V)=11:0V;theelectriceldbetweentheplatesisE=(11:0V)=(1:48102m)=743V=m.E28-32Takethederivative:E=@V=@z.E28-33khdaw.comTheradialpotentialgradientisjustthemagnitudeoftheradialcomponentoftheelectriceld,@VEr=@rThen@V1q=;@r40r2179(1:601019C)=;4(8:851012C2=N
m2)(7:01015m)2=2:321021V=m:www.khdaw.comE28-34Evaluate@V=@r,andZe1rE=+2:40r22R3E28-35E=@V=@x=2(1530V=m2)x.Atthepointinquestion,E=2(1530V=m2)(1:28x102m)=39:2V=m.E28-36Drawthewiressothattheyareperpendiculartotheplaneofthepage;theywillthen课后答案网comeoutof"thepage.Theequipotentialsurfacesarethenlineswheretheyintersectthepage,andtheylooklike47khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comkhdaw.comE28-37(a)jVVj=jW=qj=j(3:941019J)=(1:601019C)j=2:46V.TheelectriceldBAdidworkontheelectron,sotheelectronwasmovingfromaregionoflowpotentialtoaregionofhighpotential;orVB>VA.Consequently,VBVA=2:46V.(b)VCisatthesamepotentialasVB(bothpointsareonthesameequipotentialline),soVCVA=VBVA=2:46V.(c)VCisatthesamepotentialasVB(bothpointsareonthesameequipotentialline),soVCVB=0V.www.khdaw.comE28-38(a)Forpointchargesr=q=40V,sor=(8:99109N
m2=C2)(1:5108C)=(30V)=4:5m(b)No,sinceV/1=r.E28-39Thedottedlinesareequipotentiallines,thesolidarrowsareelectriceldlines.Notethattherearetwiceasmanyelectriceldlinesfromthelargercharge!课后答案网48khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comkhdaw.comE28-40Thedottedlinesareequipotentiallines,thesolidarrowsareelectriceldlines.www.khdaw.com课后答案网E28-41Thiscaneasilybedonewithaspreadsheet.Thefollowingisasketch;theelectriceldistheboldcurve,thepotentialisthethincurve.49khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comkhdaw.comsphereradiusrE28-42OriginallyV=q=40r,whereristheradiusofthesmallersphere.(a)Connectingthesphereswillbringthemtothesamepotential,orV1=V2.(b)q1+q2=q;V1=q1=40randV2=q2=402r;combiningalloftheaboveq2=2q1andq1=q=3andq2=2q=3.E28-43(a)q=4R2,soV=q=4R=R=www.khdaw.com,or00V=(1:601019C=m2)(6:37106m)=(8:851012C2=N
m2)=0:115V(b)PretendtheEarthisaconductor,thenE==epsilon0,soE=(1:601019C=m2)=(8:851012C2=N
m2)=1:81108V=m:E28-44V=q=40R,soV=(8:99109N
m2=C2)(15109C)=(0:16m)=850V:课后答案网E28-45(a)q=4RV=4(8:851012C2=N
m2)(0:152m)(215V)=3:63109C0(b)=q=4R2=(3:63109C)=4(0:152m)2=1:25108C=m2.E28-46Thedottedlinesareequipotentiallines,thesolidarrowsareelectriceldlines.50khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comkhdaw.comwww.khdaw.comE28-47(a)Thetotalcharge(Q=57:2nC)willbedividedupbetweenthetwospheressothattheyareatthesamepotential.Ifq1isthechargeononesphere,thenq2=Qq1isthechargeontheother.ConsequentlyV1=V2;1q11Qq1=;40r140r2q1r2=(Qq1)r1;Qr2q1=:r2+r1课后答案网Puttinginthenumbers,wendQr1(57:2nC)(12:2cm)q1===38:6nC;r2+r1(5:88cm)+(12:2cm)andq2=Qq1=(57:2nC)(38:6nC)=18:6nC.(b)Thepotentialoneachsphereshouldbethesame,soweonlyneedtosolveone.Then1q11(38:6nC)==2850V:40r14(8:851012C2=N
m2)(12:2cm)E28-48(a)V=(8:99109N
m2=C2)(31:5109C)=(0:162m)=1:75103V.(b)V=q=40r,sor=q=40V,andthenr=(8:99109N
m2=C2)(31:5109C)=(1:2010khdaw.com3V)=0:236m:51若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comThatis(0:236m)(0:162m)=0:074mabovethesurface.E28-49(a)Applythepointchargeformula,butsolveforthecharge.Then1q=V;40rq=40rV;q=4(8:851012C2=N
m2)(1m)(106V)=0:11mC:Nowthat"safairlysmallcharge.Butiftheradiusweredecreasedbyafactorof100,sowouldthecharge(1:10C).Consequently,smallermetalballscanberaisedtohigherpotentialswithlesscharge.(b)Theelectriceldnearthesurfaceoftheballisafunctionofthesurfacechargedensity,E==.Butsurfacechargedensitydependsonthearea,andvariesasr2.Foragivenpotential,0theelectriceldnearthesurfacewouldthenbegivenbyqVE===:khdaw.com040r2rNotethattheelectriceldgrowsastheballgetssmaller.Thismeansthatthebreakdowneldismorelikelytobeexceededwithalowvoltagesmallball;you"llgetsparking.E28-50AVolt"isaJouleperCoulomb.Thepowerrequiredbythedrivebeltistheproduct(3:41106V)(2:83103C=s)=9650W.P28-1(a)AccordingtoNewtonianmechanicswewantK=1mv2tobeequaltoW=q
V2whichmeansmv2(0:511MeV)
V===256kV:2qwww.khdaw.com2emc2istherestmassenergyofanelectron.(b)Let"sdosomerearrangingrst."#21K=mcp1;12K1=p1;mc212K1+1=p;mc212课后答案网1p=12;K+1mc212
2=1;K+1mc2andnally,s1=1
2K+1mc2Puttinginthenumbers,vu1ut12=0:746;(256keV)+1(511keV)sov=0:746c.khdaw.com52若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP28-2(a)ThepotentialofthehollowsphereisV=q=40r.TheworkrequiredtoincreasethechargebyanamountdqisdW=V=;dq.Integrating,Ze2qeW=dq=:040r80rThiscorrespondstoanelectricpotentialenergyofe(1:601019C)W==2:55105eV=4:081014J:8(8:851012C2=N
m2)(2:821015m)(b)Thiswouldbeamassofm=(4:081014J)=(3:00108m=s)2=4:531031kg.P28-3Thenegativechargeisheldinorbitbyelectrostaticattraction,ormv2qQ=:khdaw.comr40r2Thekineticenergyofthechargeis12qQK=mv=:280rTheelectrostaticpotentialenergyisqQU=;40rsothetotalenergyisqQE=www.khdaw.com:80rTheworkrequiredtochangeorbitisthenqQ11W=:80r1r2RP28-4(a)V=Edr,soZr2qrqrV=dr=:040R380R3课后答案网(b)
V=q=80R.(c)IfinsteadofV=0atr=0aswasdoneinpart(a)wetakeV=0atr=1,thenV=q=40Ronthesurfaceofthesphere.ThenewexpressionforthepotentialinsidethespherewilllooklikeV=V0+V,whereV0istheanswerfrompart(a)andVisaconstantsothatthesssurfacepotentialiscorrect.ThenqqR23qR2Vs=+=;40R80R380R3andthenqr23qR2q(3R2r2)V=+=:80R380R380R353khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP28-5Thetotalelectricpotentialenergyofthesystemisthesumofthethreeinteractionpairs.Oneofthesepairsdoesnotchangeduringtheprocess,soitcanbeignoredwhenndingthechangeinpotentialenergy.Thechangeinelectricalpotentialenergyisthenq2q2q211
U=22=:40rf40ri20rfriInthiscaseri=1:72m,whilerf=0:86m.Thechangeinpotentialenergyisthen9222118
U=2(8:9910N
m=C)(0:122C)=1:5610J(0:86m)(1:72m)Thetimerequiredist=(1:56108)=(831W)=1:87105s=2:17days:P28-6khdaw.com(a)Applyconservationofenergy:qQqQK=;ord=;40d40Kwheredisthedistanceofclosestapproach.(b)Applyconservationofenergy:qQ12K=+mv;40(2d)2pso,combiningwiththeresultsinpart(a),v=K=m.www.khdaw.comP28-7(a)FirstapplyEq.28-18,butsolveforr.Thenq(32:01012C)r===562m:40V4(8:851012C2=N
m2)(512V)(b)Iftwosuchdropsjointogetherthechargedoubles,andthevolumeofwaterdoubles,butthep3radiusofthenewdroponlyincreasesbyafactorof2=1:26becausevolumeisproportionaltotheradiuscubed.Thepotentialonthesurfaceofthenewdropwillbe课后答案网1qnewVnew=;40rnew12qold=p;4302rold2=31qold2=3=(2)=(2)Vold:40roldThenewpotentialis813V.P28-8(a)TheworkdoneisW=Fz=Eqz=qz=20.(b)SinceW=q
V,
V=z=20,soV=V0(=20)z:54khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP28-9(a)Thepotentialatanypointwillbethesumofthecontributionfromeachcharge,1q11q2V=+;40r140r2wherer1isthedistancethepointinquestionfromq1andr2isthedistancethepointinquestionfromq2.Pickapoint,callit(x;y).Sinceq1isattheorigin,pr1=x2+y2:Sinceq2isat(d;0),whered=9:60nm,pr2=(xd)2+y2:DenetheStanleyNumber"asS=40V.Equipotentialsurfacesarealsoequi-Stanleysurfaces.Inparticular,whenV=0,sodoesS.Wecanthenwritethepotentialexpressioninasightlysimpliedformq1q2S=+:khdaw.comr1r2IfS=0wecanrearrangeandsquarethisexpression.q1q2=;r1r2r2r212=;q2q212x2+y2(xd)2+y2=;q2q212Let=q2=q1,thenwecanwritewww.khdaw.com
2x2+y2=(xd)2+y2;2x2+2y2=x22xd+d2+y2;(21)x2+2xd+(21)y2=d2:Wecompletethesquareforthe(21)x2+2xdtermbyaddingd2=(21)tobothsidesoftheequation.Then"#22d221(1)x++y=d1+:2121Thecenterofthecircleisat课后答案网d(9:60nm)==5:4nm:21(10=6)21(b)Theradiusofthecircleisvuu1+1t21d2;21whichcanbesimpliedtoj(10=6)jd=(9:6nm)=9:00nm:21(10=6)21(c)No.khdaw.com55若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP28-10Anannulusiscomposedofdierentialringsofvaryingradiirandwidthdr;thechargeonanyringistheproductoftheareaofthering,dA=2rdr,andthesurfacechargedensity,ork2kdq=dA=2rdr=dr:r3r2Thepotentialatthecentercanbefoundbyaddingupthecontributionsfromeachring.Sinceweareatthecenter,thecontributionswilleachbedV=dq=40r.ThenZb22kdrk11kbaV==:=:a20r340a2b240b2a2ThetotalchargeontheannulusisZb2k11baQ=dr=2k=2k:ar2abbaCombining,khdaw.comQa+bV=:80abP28-11Addthethreecontributions,andthendoaseriesexpansionfordr.q111V=++;40r+drrdq11=+1+;40r1+d=r1d=rqdd1++1+1+;40rwww.khdaw.comrrq2d1+:40rrP28-12(a)Addthecontributionsfromeachdierentialcharge:dq=dy.ThenZy+Ly+LV=dy=ln:y40y40y(b)Takethederivative:课后答案网@VyLLEy===:@y40y+Ly240y(y+L)(c)Bysymmetryitmustbezero,sincethesystemisinvariantunderrotationsabouttheaxisoftherod.Notethatwecan"tdetermineE?fromderivativesbecausewedon"thavethefunctionalformofVforpointso-axis!P28-13(a)WefollowtheworkdoneinSection28-6forauniformlineofcharge,startingwithEq.28-26,1dxdV=p;40x2+y2ZL1kxdxdV=p;400x2+y2khdaw.com56若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comkpL=x2+y2;400pk=L2+y2y:40(b)Theycomponentoftheelectriceldcanbefoundfrom@VEy=;@ywhich(usingacomputer-aidedmathprogram)is!kyEy=1p:40L2+y2(c)WecouldndExifweknewthexvariationofV.Butwedon"t;weonlyfoundthevaluesofValongaxedvalueofx.khdaw.com(d)WewanttondysuchthattheratiopkkL2+y2y=(L)4040pisone-half.Simplifying,L2+y2y=L=2;whichcanbewrittenasL2+y2=L2=4+Ly+y2;or3L2=4=Ly;withsolutiony=3L=4.P28-14Thespheresaresmallcomparedtotheseparationdistance.Assumingonlywww.khdaw.comonesphereatapotentialof1500V,thechargewouldbeq=4rV=4(8:851012C2=N
m)(0:150m)(1500V)=2:50108C:0Thepotentialfromthesphereatadistanceof10.0mwouldbe(0:150m)V=(1500V)=22:5V:(10:0m)Thisissmallcomparedto1500V,sowewilltreatitasaperturbation.Thismeansthatwecanassumethatthesphereshavechargesofq=4rV课后答案网=4(8:851012C2=N
m)(0:150m)(1500V+22:5V)=2:54108C:0P28-15Calculatingthefractionofexcesselectronsisthesameascalculatingthefractionofexcesscharge,sowe"llskipcountingtheelectrons.ThisproblemiseectivelythesameasExercise28-47;wehaveatotalchargethatisdividedbetweentwounequalsizesphereswhichareatthesamepotentialonthesurface.UsingtheresultfromthatexercisewehaveQr1q1=;r2+r1whereQ=6:2nCisthetotalchargeavailable,andq1isthechargeleftonthesphere.r1istheradiusofthesmallball,r2istheradiusofEarth.Sincethefractionofchargeremainingisq1=Q,wecanwriteq1r1r18==2:010:Qr2+r1r2khdaw.com57若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP28-16Thepositivechargeonthespherewouldbeq=4rV=4(8:851012C2=N
m2)(1:08102m)(1000V)=1:20109C:0Thenumberofdecaysrequiredtobuildupthischargeisn=2(1:20109C)=(1:601019C)=1:501010:Theextrafactoroftwoisbecauseonlyhalfofthedecaysresultinanincreaseincharge.Thetimerequiredist=(1:501010)=(3:70108s1)=40:6s:P28-17(a)None.(b)None.(c)None.(d)None.khdaw.com(e)No.P28-18(a)OutsideofanisolatedchargedsphericalobjectE=q=4r2andV=q=4r.00ThenE=V=r.Consequently,thespheremusthavearadiuslargerthanr=(9:15106V)=(100106V=m)=9:15102m.(b)Thepowerrequiredis(320106C=s)(9:15106V)=2930W.(c)wv=(320106C=s),so(320106C=s)==2:00105C=m2:(0:485m)(33:0m=s)www.khdaw.com课后答案网58khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE29-1(a)Thechargewhich owsthroughacrosssectionalsurfaceareainatimetisgivenbyq=it;whereiisthecurrent.Forthisexercisewehaveq=(4:82A)(4:6060s)=1330Casthechargewhichpassesthroughacrosssectionofthisresistor.(b)Thenumberofelectronsisgivenby(1330C)=(1:601019C)=8:311021electrons.E29-2Q=t=(200106A=s)(60s=min)=(1:601019C)=7:51016electronsperminute.E29-3(a)j=nqv=(2:101014=m3)2(1:601019C)(1:40105m=s)=9:41A=m2:Sincetheionshavepositivechargethenthecurrentdensityisinthesamedirectionasthevelocity.(b)Weneedanareatocalculatethecurrent.E29-4(a)j=i=A=(1231012A)=(1:23103m)2=2:59105A=m2.(b)v=j=ne=(2:59105A=m2)=(8:491028=m3)(1:601019C)=1:911015m=s:khdaw.comdE29-5ThecurrentratingofafuseofcrosssectionalareaAwouldbei=(440A=cm2)A;maxandifthefusewireiscylindricalA=d2=4.Thens4(0:552A)2d==4:0010cm:(440A=m2)E29-6Currentdensityiscurrentdividedbycrosssectionofwire,sothegraphwouldlooklike:www.khdaw.com43课后答案网2I(A/mil^2x10^−3)150100150khdaw.com200d(mils)59若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE29-7Thecurrentisinthedirectionofthemotionofthepositivecharges.Themagnitudeofthecurrentisi=(3:11018=s+1:11018=s)(1:601019C)=0:672A:E29-8(a)Thetotalcurrentisi=(3:501015=s+2:251015=s)(1:601019C)=9:20104A:(b)Thecurrentdensityisj=(9:20104A)=(0:165103m)2=1:08104A=m2:E29-9(a)j=(8:70106=m3)(1:601019C)(470103m=s)=6:54107A=m2:(b)i=(6:54107A=m2)(6:37106m)2=8:34107A:E29-10i=wv,sokhdaw.com=(95:0106A)=(0:520m)(28:0m=s)=6:52106C=m2:E29-11ThedriftvelocityisgivenbyEq.29-6,ji(115A)4vd====2:7110m=s:neAne(31:2106m2)(8:491028=m3)(1:601019C)Thetimeittakesfortheelectronstogettothestartermotorisx(0:855m)3t===3:2610s:v(2:7110www.khdaw.com4m=s)That"sabout54minutes.E29-12
V=iR=(50103A)(1800)=90V:E29-13TheresistanceofanobjectwithconstantcrosssectionisgivenbyEq.29-13,L7(11;000m)R==(3:010
m)=0:59:A(0:0056m2)E29-14Theslopeisapproximately[(8课后答案网:21:7)=1000]
cm=C,so133=6:510
cm=C410=C1:7
cmE29-15(a)i=
V=R=(23V)=(15103)=1500A:(b)j=i=A=(1500A)=(3:0103m)2=5:3107A=m2:(c)=RA=L=(15103)(3:0103m)2=(4:0m)=1:1107
m.Thematerialispossiblyplatinum.E29-16UsetheequationfromExercise29-17.
R=8;then
T=(8)=(50)(4:3103=C)=37C:Thenaltemperatureisthen57C.khdaw.com60若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE29-17StartwithEq.29-16,0=0av(TT0);andmultiplythroughbyL=A,LL(0)=0av(TT0);AAtogetRR0=R0av(TT0):E29-18ThewirehasalengthL=(250)2(0:122m)=192m.Thediameteris0.129inches;thecrosssectionalareaisthenA=(0:1290:0254m)2=4=8:43106m2:Theresistanceiskhdaw.comR=L=A=(1:69108
m)(192m)=(8:43106m2)=0:385:E29-19IfthelengthofeachconductorisLandhasresistivity,thenL4LRA==D2=4D2andL4LRB==:(4D2=4www.khdaw.comD2=4)3D2TheratiooftheresistancesisthenRA=3:RBE29-20R=R,soL=(d=2)2=L=(d=2)2.Simplifying,=d2==d2.Then1112221122pd=(1:19103m)(9:68108
m)=(1:69108
m)=2:85103m:2E29-21(a)(750103A)=(125)=6:00103A.(b)
V=iR=(6课后答案网:00103A)(2:65106)=1:59108V:(c)R=
V=i=(1:59108V)=(750103A)=2:12108:E29-22Since
V=iR,thenif
Vandiarethesame,thenRmustbethesame.(a)SinceR=R,L=r2=L=r2,or=r2==r2.Then1112221122priron=rcopper=(9:68108
m)(1:69108
m)=2:39:(b)Startwiththedenitionofcurrentdensity:i
V
Vj===:ARALSince
VandListhesame,butisdierent,thenthecurrentdensitieswillbedierent.61khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE29-23ConductivityisgivenbyEq.29-8,~j=E~.Ifthewireislongandthin,thenthemagnitudeoftheelectriceldinthewirewillbegivenbyE
V=L=(115V)=(9:66m)=11:9V=m:Wecannowndtheconductivity,j(1:42104A=m2)===1:19103(
m)1:E(11:9V=m)E29-24(a)vd=j=en=E=en:Thenv=(2:701014=
m)(120V=m)=(1:601019C)(620106=m3+550106=m3)=1:73102m=s:d(b)j=E=(2:701014=
m)(120V=m)=3:241014A=m2:E29-25khdaw.com(a)R=L==A,soj=i=A=(R=L)i=.Forcopper,j=(0:152103=m)(62:3A)=(1:69108
m)=5:60105A=m2;foraluminum,j=(0:152103=m)(62:3A)=(2:75108
m)=3:44105A=m2:(b)A=L=R;ifisdensity,thenm=lA=l=(R=L).Forcopper,m=(1:0m)(8960kg=m3)(1:69108
m)=(0:152103=m)=0:996kg;foraluminum,www.khdaw.comm=(1:0m)(2700kg=m3)(2:75108
m)=(0:152103=m)=0:488kg:E29-26Theresistanceforpotentialdierenceslessthan1.5Varebeyondthescale.108课后答案网6R(Kilo−ohms)421234V(Volts)62khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE29-27(a)Theresistanceisdenedas
V(3:55106V=A2)i2R===(3:55106V=A2)i:iiWheni=2:40mAtheresistancewouldbeR=(3:55106V=A2)(2:40103A)=8:52k:(b)Inverttheaboveexpression,andi=R=(3:55106V=A2)=(16:0)=(3:55106V=A2)=4:51A:E29-28First,n=3(6:021023)(2700kg=m3)(27:0103kg)=1:811029=m3.Thenm(9:111031kg)===7:151015s:ne2(1:811029=m3)(1:601019C)2(2:75108
m)E29-29khdaw.com(a)E=E==q=4R2,so0e0e(1:00106C)E==4(8:851012C2=N
m2)(4:7)(0:10m)2(b)E=E=q=4R2,so00(1:00106C)E==4(8:851012C2=N
m2)(0:10m)2(c)=(EE)=q(11=)=4R2.Thenind00ewww.khdaw.com(1:00106C)1=1=6:23106C=m2:ind4(0:10m)2(4:7)E29-30MidwaybetweenthechargesE=q=0d,soq=(8:851012C2=N
m2)(0:10m)(3106V=m)=8:3106C:E29-31(a)AtthesurfaceofaconductorofradiusRwithchargeQthemagnitudeoftheelectriceldisgivenby12课后答案网E=QR;40whilethepotential(assumingV=0atinnity)isgivenby1V=QR:40TheratioisV=E=R.Thepotentialonthespherethatwouldresultinsparking"isV=ER=(3106N=C)R:(b)Itiseasier"togetasparkoofaspherewithasmallerradius,becauseanypotentialonthespherewillresultinalargerelectriceld.(c)Thepointsofalightingrodarelikesmallhemispheres;theelectriceldwillbelargenearthesepointssothatthiswillbethelikelyplaceforsparkstoformandlightningboltstostrike.khdaw.com63若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP29-1Ifthereismorecurrent owingintothespherethanis owingoutthentheremustbeachangeinthenetchargeonthesphere.Thenetcurrentisthedierence,or2A.Thepotentialonthesurfaceofthespherewillbegivenbythepoint-chargeexpression,1qV=;40randthechargewillberelatedtothecurrentbyq=it.Combining,1itV=;40ror4Vr4(8:851012C2=N
m2)(980V)(0:13m)0t===7:1ms:i(2A)P29-2Thenetcurrentdensityisinthedirectionofthepositivecharges,whichistotheeast.Therearetwoelectronsforeveryalphaparticle,andeachalphaparticlehasachargeequalinmagnitudekhdaw.comtotwoelectrons.Thecurrentdensityisthenj=qeneve+q+nv;=(1:61019C)(5:61021=m3)(88m=s)+(3:21019C)(2:81021=m3)(25m=s);=1:0105C=m2:P29-3(a)TheresistanceofthesegmentofthewireisR=L=A=(1:69108
m)(4:0102m)=(2:6103m)2=3:18105:Thepotentialdierenceacrossthesegmentiswww.khdaw.com
V=iR=(12A)(3:18105)=3:8104V:(b)Thetailisnegative.(c)Thedriftspeedisv=j=en=i=Aen,sov=(12A)=(2:6103m)2(1:61019C)(8:491028=m3)=4:16105m=s:Theelectronswillmove1cmin(1:0102m)=(4:16105m=s)=240s.P29-4(a)N=it=q课后答案网=(250109A)(2:9s)=(3:21019C)=2:271012.p(b)Thespeedoftheparticlesinthebeamisgivenbyv=2K=m,sopv=2(22:4MeV)=4(932MeV=c2)=0:110c:Ittakes(0:180m)=(0:110)(3:00108m=s)=5:45109sforthebeamtotravel18.0cm.ThenumberofchargesisthenN=it=q=(250109A)(5:45109s)=(3:21019C)=4260:(c)W=q
V,so
V=(22:4MeV)=2e=11:2MV:64khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP29-5(a)Thetimeittakestocompleteoneturnist=(250m)=c.Thetotalchargeisq=it=(30:0A)(950m)=(3:00108m=s)=9:50105C:(b)ThenumberofchargesisN=q=e,thetotalenergyabsorbedbytheblockisthen
U=(28:0109eV)(9:50105C)=e=2:66106J:Thiswillraisethetemperatureoftheblockby
T=
U=mC=(2:66106J)=(43:5kg)(385J=kgC)=159C:RRP29-6(a)i=jdA=2jrdr;Zi=20Rj(1r=R)rdr=2j(R2=2R3=3R)=jR2=6:000khdaw.com(b)Integrate,again:Zi=20Rj(r=R)rdr=2j(R3=3R)=jR2=3:000P29-7(a)Solve2=[1+(T20C)],or00T=20C+1=(4:3103=C)=250C:(b)Yes,ignoringchangesinthephysicaldimensionsoftheresistor.P29-8Theresistancewhenonis(2:90V)=(0:310A)=9www.khdaw.com:35.ThetemperatureisgivenbyT=20C+(9:351:12)=(1:12)(4:5103=C)=1650C:P29-9OriginallywehavearesistanceR1madeoutofawireoflengthl1andcrosssectionalareaA1.ThevolumeofthiswireisV1=A1l1.Whenthewireisdrawnouttothenewlengthwehavel2=3l1,butthevolumeofthewireshouldbeconstantsoA2l2=A1l1;A2(3l1)=A1l1;课后答案网A2=A1=3:Theoriginalresistanceisl1R1=:A1Thenewresistanceisl23l1R2===9R1;A2A1=3orR2=54.P29-10(a)i=(35:8V)=(935)=3:83102A:(b)j=i=A=(3:83102A)=(3:50104m2)=109A=m2.(c)v=(109A=m2)=(1:61019C)(5:331022=m3)=1:28102m=s.(d)E=(35:8V)=(0:158m)=227V=m:khdaw.com65若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP29-11(a)=(1:09103)(5:5103m)2=4(1:6m)=1:62108
m.Thisispossiblysilver.(b)R=(1:62108
m)(1:35103m)4=(2:14102m)2=6:08108:P29-12(a)
L=L=1:7105foratemperaturechangeof1:0C.Areachangesaretwicethis,or
A=A=3:4105.TakethedierentialofRA=L:RdA+AdR=dL+Ld,ordR=dL=A+Ld=ARdA=A.Fornitechangesthiscanbewrittenas
R
L

A=+:RLA
==4:3103.Sincethistermissomuchlargerthantheothertwoitistheonlysignicanteect.P29-13WewillusetheresultsofExercise29-17,khdaw.comRR0=R0av(TT0):Tosaveonsubscriptswewilldroptheav"notation,andjustspecifywhetheritiscarbonc"orironi".Thediskswillbeeectivelyinseries,sowewilladdtheresistancestogetthetotal.Lookingonlyatonediskpair,wehaveRc+Ri=R0;c(c(TT0)+1)+R0;i(i(TT0)+1);=R0;c+R0;i+(R0;cc+R0;ii)(TT0):Thislastequationwillonlybeconstantifthecoecientfortheterm(TT0)vanishes.Thenwww.khdaw.comR0;cc+R0;ii=0;butR=L=A,andthediskshavethesamecrosssectionalarea,soLccc+Liii=0;orL(9:68108
m)(6:5103=C)ciiL==(3500108
m)(0:50103=C)=0:036:iccP29-14Thecurrententeringtheconeis课后答案网i.Thecurrentdensityasafunctionofdistancexfromtheleftendisthenij=:[a+x(ba)=L]2TheelectriceldisgivenbyE=j.ThepotentialdierencebetweentheendsisthenZLZLiiL
V=Edx=dx=00[a+x(ba)=L]2abTheresistanceisR=
V=i=L=ab.66khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP29-15ThecurrentisfoundfromEq.29-5,Zi=~j
dA~;wheretheregionofintegrationisoverasphericalshellconcentricwiththetwoconductingshellsbutbetweenthem.ThecurrentdensityisgivenbyEq.29-10,~j=E~=;andwewillhaveanelectriceldwhichisperpendiculartothesphericalshell.Consequently,ZZ11i=E~
dA~=EdABysymmetryweexpecttheelectriceldtohavethesamemagnitudeanywhereonasphericalshellwhichisconcentricwiththetwoconductingshells,sowecanbringitoutoftheintegralsign,andthenZkhdaw.com14r2Ei=EdA=;whereEisthemagnitudeoftheelectriceldontheshell,whichhasradiusrsuchthatb>r>a.Theaboveexpressioncanbeinvertedtogivetheelectriceldasafunctionofradialdistance,sincethecurrentisaconstantintheaboveexpression.ThenE=i=4r2ThepotentialisgivenbyZa
V=E~
d~s;bwewillintegratealongaradialline,whichisparalleltotheelectriceld,sowww.khdaw.comZa
V=Edr;bZai=dr;4r2bZaidr=;4bri11=:4abWedividethisexpressionbythecurrenttogettheresistance.Then课后答案网11R=4abP29-16Since=p=vd,p/vd.Foranidealgasthekineticenergyisproportionaltothetemperature,so/K/T.67khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE30-1WeapplyEq.30-1,q=C
V=(501012F)(0:15V)=7:51012C;E30-2(a)C=
V=q=(73:01012C)=(19:2V)=3:801012F:(b)Thecapacitancedoesn"tchange!(c)
V=q=C=(2101012C)=(3:801012F)=55:3V:E30-3q=C
V=(26:0106F)(125V)=3:25103C:E30-4(a)C=A=d=(8:851012F=m)(8:22102m)2=(1:31103m)=1:431010F:0(b)q=C
V=(1:431010F)(116V)=1:66108C.E30-5Eq.30-11givesthecapacitanceofacylinder,L12(0:0238m)13khdaw.comC=20=2(8:8510F=m)=5:4610F:ln(b=a)ln((9:15mm)=(0:81mm))E30-6(a)A=Cd==(9:701012F)(1:20103m)=(8:851012F=m)=1:32103m2:0(b)C=Cd=d=(9:701012F)(1:20103m)=(1:10103m)=1:061011F.00(c)
V=q0=C=[
V]0C0=C=[
V]0d=d0.Usingthisformula,thenewpotentialdierencewouldbe[
V]=(13:0V)(1:10103m)=(1:20103m)=11:9V:Thepotentialenergyhaschanged0by(11:9V)(30:0V)=1:1V.E30-7(a)FromEq.30-8,12(0:040m)(0www.khdaw.com:038m)11C=4(8:8510F=m)=8:4510F:(0:040m)(0:038m)(b)A=Cd==(8:451011F)(2:00103m)=(8:851012F=m)=1:91102m2:0E30-8Leta=b+d,wheredisthesmallseparationbetweentheshells.Thenab(b+d)bC=40=40;abdb240=0A=d:课后答案网dE30-9Thepotentialdierenceacrosseachcapacitorinparallelisthesame;itisequalto110V.Thechargeoneachofthecapacitorsisthenq=C
V=(1:00106F)(110V)=1:10104C:IfthereareNcapacitors,thenthetotalchargewillbeNq,andwewantthistotalchargetobe1:00C.Then(1:00C)(1:00C)N===9090:q(1:10104C)68khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE30-10Firstndtheequivalentcapacitanceoftheparallelpart:C=C+C=(10:3106F)+(4:80106F)=15:1106F:eq12Thenndtheequivalentcapacitanceoftheseriespart:11151=+=3:2310F:Ceq(15:1106F)(3:90106F)Thentheequivalentcapacitanceoftheentirearrangementis3:10106F.E30-11Firstndtheequivalentcapacitanceoftheseriespart:11151=+=3:0510F:Ceq(10:3106F)(4:80106F)Theequivalentcapacitanceis3:28106F.Thenndtheequivalentcapacitanceoftheparallelpart:C=C+C=(3:28106F)+(3:90106F)=7:18106F:eq12Thisistheequivalentcapacitancefortheentirearrangement.khdaw.comE30-12Foronecapacitorq=C
V=(25:0106F)(4200V)=0:105C.Therearethreecapaci-tors,sothetotalchargetopassthroughtheammeteris0:315C.E30-13(a)TheequivalentcapacitanceisgivenbyEq.30-21,111115=+=+=CeqC1C2(4:0F)(6:0F)(12:0F)orCeq=2:40F.(b)Thechargeontheequivalentcapacitoriswww.khdaw.comq=C
V=(2:40F)(200V)=0:480mC.Forseriescapacitors,thechargeontheequivalentcapacitoristhesameasthechargeoneachofthecapacitors.ThisstatementiswrongintheStudentSolutions!(c)Thepotentialdierenceacrosstheequivalentcapacitorisnotthesameasthepotentialdierenceacrosseachoftheindividualcapacitors.Weneedtoapplyq=C
Vtoeachcapacitorusingthechargefrompart(b).Thenforthe4:0Fcapacitor,q(0:480mC)
V===120V;C(4:0F)andforthe6:0Fcapacitor,q(0:480mC)
V===80V:课后答案网C(6:0F)Notethatthesumofthepotentialdierencesacrosseachofthecapacitorsisequaltothepotentialdierenceacrosstheequivalentcapacitor.E30-14(a)TheequivalentcapacitanceisCeq=C1+C2=(4:0F)+(6:0F)=(10:0F):(c)Forparallelcapacitors,thepotentialdierenceacrosstheequivalentcapacitoristhesameasthepotentialdierenceacrosseitherofthecapacitors.(b)Forthe4:0Fcapacitor,q=C
V=(4:0F)(200V)=8:0104C;andforthe6:0Fcapacitor,q=C
V=(6:0F)(200V)=12:0104C:69khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE30-15(a)Ceq=C+C+C=3C;0A0Addeq===:Ceq3C3(b)1=Ceq=1=C+1=C+1=C=3=C;0A0Adeq===3d:CeqC=3E30-16(a)Themaximumpotentialacrossanyindividualcapacitoris200V;sotheremustbeatleast(1000V)=(200V)=5seriescapacitorsinanyparallelbranch.ThisbranchwouldhaveanequivalentcapacitanceofC=C=5=(2:0106F)=5=0:40106F:eq(b)Forparallelbranchesweadd,whichmeansweneed(1:2106F)=(0:40106F)=3parallelbranchesofthecombinationfoundinpart(a).E30-17khdaw.comLookbackatthesolutiontoEx.30-10.IfC3breaksdownelectricallythenthecircuitiseectivelytwocapacitorsinparallel.(b)
V=115Vafterthebreakdown.(a)q=(10:3106F)(115V)=1:18103C:1E30-18The108Fcapacitororiginallyhasachargeofq=(108106F)(52:4V)=5:66103C.Afteritisconnectedtothesecondcapacitorthe108Fcapacitorhasachargeofq=(108106F)(35:8V)=3:87103C.Thedierenceinchargemustresideonthesecondcapacitor,sothecapacitanceisC=(1:79103C)=(35:8V)=5:00105F:E30-19ConsideranyjunctionotherthanAorwww.khdaw.comB.Callthisjunctionpoint0;labelthefournearestjunctionstothisaspoints1,2,3,and4.Thechargeonthecapacitorthatlinkspoint0topoint1isq1=C
V01;where
V01isthepotentialdierenceacrossthecapacitor,so
V01=V0V1;whereV0isthepotentialatthejunction0,andV1isthepotentialatthejunction1.Similarexpressionsexistfortheotherthreecapacitors.Forthejunction0thenetchargemustbezero;thereisnowayforchargetocrosstheplatesofthecapacitors.Thenq1+q2+q3+q4=0;andthismeansC
V01+C
V02+C
V03+C
V04=0or课后答案网
V01+
V02+
V03+
V04=0:Let
V0i=V0Vi,andthenrearrange,4V0=V1+V2+V3+V4;or1V0=(V1+V2+V3+V4):4E30-20U=uV=E2V=2,whereVisthevolume.Then0112237U=(8:8510F=m)(150V=m)(2:0m)=1:9910J:270khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE30-21Thetotalcapacitanceis(2100)(5:0106F)=1:05102F:Thetotalenergystoredis1212327U=C(
V)=(1:0510F)(5510V)=1:5910J:22Thecostis7$0:03(1:5910J)=$0:133:3600103JE30-22(a)U=1C(
V)2=1(0:061F)(1:0104V)2=3:05106J.22(b)(3:05106J)=(3600103J=kW
h)=0:847kW
h:E30-23(a)Thecapacitanceofanairlledparallel-platecapacitorisgivenbyEq.30-5,A(8:851012F=m)(42:0104m2)C=0==2:861011F:khdaw.comd(1:30103m)(b)Themagnitudeofthechargeoneachplateisgivenbyq=C
V=(2:861011F)(625V)=1:79108C:(c)ThestoredenergyinacapacitorisgivenbyEq.30-25,regardlessofthetypeorshapeofthecapacitor,so121112U=C(
V)=(2:8610F)(625V)=5:59J:22(d)Assumingaparallelplatearrangementwithnofringingeects,themagnitudeoftheelectriceldbetweentheplatesisgivenbyEd=
V,wheredistheseparationbetweentheplates.Thenwww.khdaw.com5E=
V=d=(625V)=(0:00130m)=4:8110V=m:(e)TheenergydensityisEq.30-28,12112523u=0E=((8:8510F=m))(4:8110V=m)=1:02J=m:22E30-24Theequivalentcapacitanceisgivenby1=C=1=(2:12106F)+1=(3:88106F)=1=(1:37106F):eqTheenergystoredis课后答案网U=1(1:37106F)(328V)2=7:37102J:2E30-25V=r=q=4r2=E,sothatifVisthepotentialofthespherethenE=V=risthe0electriceldonthesurface.Thentheenergydensityoftheelectriceldnearthesurfaceis12212(8:8510F=m)(8150V)23u=0E==7:4110J=m:22(0:063m)E30-26ThechargeonC3canbefoundfromconsideringtheequivalentcapacitance.q3=(3:10106F)(112V)=3:47104C:ThepotentialacrossCisgivenby[
V]=(3:47104C)=(3:9033106F)=89:0V.Thepotentialacrosstheparallelsegmentisthen(112V)(89:0V)=23:0V.So[
V]1=[
V]2=23:0V.Thenq=(10:3106F)(23:0V)=2:37104Candq=(4:80106F)(23:0V)=1:10104C:.12khdaw.com71若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE30-27Thereisenoughworkonthisproblemwithoutderivingonceagaintheelectriceldbetweenchargedcylinders.IwillinsteadreferyoubacktoSection26-4,andstate1qE=;20LrwhereqisthemagnitudeofthechargeonacylinderandListhelengthofthecylinders.TheenergydensityasafunctionofradialdistanceisfoundfromEq.30-28,11q2u=E2=2082L2r20ThetotalenergystoredintheelectriceldisgivenbyEq.30-24,1q2q2ln(b=a)U==;2C220LwherewesubstitutedintothelastpartEq.30-11,thecapacitanceofacylindricalcapacitor.pkhdaw.comWewanttoshowthatintegratingavolumeintegralfromr=ator=abovertheenergydensityfunctionwillyieldU=2.Sincewewanttodothisproblemthehardway,wewillpretendwedon"tknowtheanswer,andintegratefromr=ator=c,andthenndoutwhatcis.ThenZ1U=udV;2ZcZ2ZL21q=rdrddz;a00820L2r22ZcZ2ZLqdr=ddz;820L2awww.khdaw.com00r2Zcqdr=;40Larq2c=ln:40LaNowweequatethistothevalueforUthatwefoundabove,andwesolveforc.1q2ln(b=a)q2c=ln;2220L40La课后答案网ln(b=a)=2ln(c=a);(b=a)=(c=a)2;pab=c:E30-28(a)d=0A=C,ord=(8:851012F=m)(0:350m2)=(51:31012F)=6:04103m:(b)C=(5:60)(51:31012F)=2:871010F.E30-29Originally,C1=0A=d1.Afterthechanges,C2=0A=d2.DividingC2byC1yieldsC2=C1=d1=d2,so=dC=dC=(2)(2:571012F)=(1:321012F)=3:89:2211khdaw.com72若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE30-30TherequiredcapacitanceisfoundfromU=1C(
V)2,or2C=2(6:61106J)=(630V)2=3:331011F:Thedielectricconstantrequiredis=(3:331011F)=(7:401012F)=4:50.Trytransformeroil.E30-31CapacitancewithdielectricmediaisgivenbyEq.30-31,e0AC=:dThevarioussheetshavedierentdielectricconstantsanddierentthicknesses,andwewanttomaximizeC,whichmeansmaximizing=d.Formicathisratiois54mm1,forglassthisratioise35mm1,andforparanthisratiois0.20mm1.Micawins.E30-32Theminimumplateseparationisgivenbykhdaw.comd=(4:13103V)=(18:2106V=m)=2:27104m:TheminimumplateareaisthendC(2:27104m)(68:4109F)A===0:627m2:0(2:80)(8:851012F=m)E30-33ThecapacitanceofacylindricalcapacitorisgivenbyEq.30-11,1:0103mC=2(8:851012F=m)(2:6)=8:63108F:ln(0:588=0:11)E30-34(a)U=C0(
V)2=2,C0=A=d,and
V=dislessthanorequaltothedielectrice0www.khdaw.comstrength(whichwewillcallS).Then
V=Sdand12U=e0AdS;2sothevolumeisgivenbyV=2U=S2:e0Thisquantityisaminimumformica,soV=2(250103J)=(5:4)(8:851012F=m)(160106V=m)2=0:41m3:(b)=2U=VS2,soe0课后答案网=2(250103J)=(0:087m3)(8:851012F=m)(160106V=m)2=25:eE30-35(a)ThecapacitanceofacylindricalcapacitorisgivenbyEq.30-11,LC=20e:ln(b=a)Thefactorofeisintroducedbecausethereisnowadielectric(thePyrexdrinkingglass)betweentheplates.WecanlookbacktoTable29-2togetthedielectricpropertiesofPyrex.Thecapacitanceofourglass"isthen12(0:15m)10C=2(8:8510F=m)(4:7)=7:310F:ln((3:8cm)=(3:6cm)(b)Thebreakdownpotentialis(14kV/mm)(2mm)=28kV.khdaw.com73若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE30-36(a)C0=C=(6:5)(13:51012F)=8:81011F.e(b)Q=C0
V=(8:81011F)(12:5V)=1:1109C.(c)E=
V=d,butwedon"tknowd.(d)E0=E=,butwecouldn"tndE.eE30-37(a)Inserttheslabsothatitisadistanceaabovethelowerplate.Thenthedistancebetweentheslabandtheupperplateisdab.Insertingtheslabhasthesameeectashavingtwocapacitorswiredinseries;theseparationofthebottomcapacitorisa,whilethatofthetopcapacitorisdab.ThebottomcapacitorhasacapacitanceofC1=0A=a;whilethetopcapacitorhasacapacitanceofC2=0A=(dab):Addingtheseinseries,111=+;CeqC1C2adab=+;khdaw.com0A0Adb=:0ASothecapacitanceofthesystemafterputtingthecopperslabinisC=0A=(db):(b)Theenergystoredinthesystembeforetheslabisinsertedisq2q2dUi==2Ci20Awhiletheenergystoredaftertheslabisinsertedisq2www.khdaw.comq2dbUf==2Cf20ATheratioisUi=Uf=d=(db):(c)Sincetherewasmoreenergybeforetheslabwasinserted,thentheslabmusthavegoneinwillingly,itwaspulledin!.Togettheslabbackoutwewillneedtodoworkontheslabequaltotheenergydierence.q2dq2dbq2bUiUf==:20A20A20AE30-38(a)Inserttheslabsothatitisadistance课后答案网aabovethelowerplate.Thenthedistancebetweentheslabandtheupperplateisdab.Insertingtheslabhasthesameeectashavingtwocapacitorswiredinseries;theseparationofthebottomcapacitorisa,whilethatofthetopcapacitorisdab.ThebottomcapacitorhasacapacitanceofC1=0A=a;whilethetopcapacitorhasacapacitanceofC2=0A=(dab):Addingtheseinseries,111=+;CeqC1C2adab=+;0A0Adb=:0ASothecapacitanceofthesystemafterputtingthecopperslabiniskhdaw.comC=0A=(db):74若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com(b)TheenergystoredinthesystembeforetheslabisinsertedisC(
V)2(
V)2Ai0Ui==22dwhiletheenergystoredaftertheslabisinsertedisC(
V)2(
V)2Af0Uf==22dbTheratioisUi=Uf=(db)=d:(c)Sincetherewasmoreenergyaftertheslabwasinserted,thentheslabmustnothavegoneinwillingly,itwasbeingrepelled!.Togettheslabinwewillneedtodoworkontheslabequaltotheenergydierence.(
V)2A(
V)2A(
V)2Ab000UfUi==:khdaw.com2db2d2d(db)E30-39C=e0A=d,sod=e0A=C.(a)E=
V=d=C
V=e0A,or(1121012F)(55:0V)E==13400V=m:(5:4)(8:851012F=m)(96:5104m2)(b)Q=C
V=(1121012F)(55:0V)=6:16109C:.(c)Q0=Q(11=)=(6:16109C)(11=(5:4))=5:02109C.eE30-40(a)E=q=e0A,sowww.khdaw.com9(89010C)e==6:53(1:40106V=m)(8:851012F=m)(110104m2)(b)q0=q(11=)=(890109C)(11=(6:53))=7:54107C:eP30-1ThecapacitanceofthecylindricalcapacitorisfromEq.30-11,20LC=:ln(b=a)Ifthecylindersareveryclosetogetherwecanwrite课后答案网b=a+d,whered,theseparationbetweenthecylinders,isasmallnumber,so20L20LC==:ln((a+d)=a)ln(1+d=a)Expandingaccordingtothehint,20L2a0LC=d=adNow2aisthecircumferenceofthecylinder,andListhelength,so2aListheareaofacylindricalplate.Hence,forsmallseparationbetweenthecylinderswehave0AC;dwhichistheexpressionfortheparallelplates.khdaw.com75若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP30-2(a)C=0A=x;takethederivativeanddC0dA0Adx=;dTxdTx2dT1dA1dx=C:AdTxdT(b)Since(1=A)dA=dT=2aand(1=x)dx=dT=s,weneed=2=2(23106=C)=46106=C:saP30-3Inserttheslabsothatitisadistancedabovethelowerplate.Thenthedistancebetweentheslabandtheupperplateisabd.Insertingtheslabhasthesameeectashavingtwocapacitorswiredinseries;theseparationofthebottomcapacitorisd,whilethatofthetopcapacitorisabd.ThebottomcapacitorhasacapacitanceofC1=0A=d;whilethetopcapacitorhasacapacitanceofkhdaw.comC2=0A=(abd):Addingtheseinseries,111=+;CeqC1C2dabd=+;0A0Aab=:0ASothecapacitanceofthesystemafterputtingtheslabinisC=0A=(ab):P30-4Thepotentialdierencebetweenanytwoadjacentplatesis
www.khdaw.comV.Eachinteriorplatehasachargeqoneachsurface;theexteriorplate(onepink,onegray)hasachargeofqontheinteriorsurfaceonly.Thecapacitanceofonepink/grayplatepairisC=0A=d.Therearenplates,butonlyn1platepairs,sothetotalchargeis(n1)q.ThismeansthetotalcapacitanceisC=0(n1)A=d.P30-5Let
V0=96:6V.Asfaraspointeisconcernedpointalookslikeitisoriginallypositivelycharged,andpointdisoriginallynegativelycharged.Itisthenconvenienttodenethechargesonthecapacitorsintermsofthechargesonthetopsides,sotheoriginalchargeonC1isq1;i=C1
V0whiletheoriginalchargeonC2isq2;i=C2
V0.Notethenegativesignre ectingtheoppositepolarityofC2.(a)Conservationofchargerequires课后答案网q1;i+q2;i=q1;f+q2;f;butsinceq=C
VandthetwocapacitorswillbeatthesamepotentialaftertheswitchesareclosedwecanwriteC1
V0C2
V0=C1
V+C2
V;(C1C2)
V0=(C1+C2)
V;C1C2
V0=
V:C1+C2Withnumbers,(1:16F)(3:22F)
V=(96:6V)=45:4V:(1:16F)+(3:22F)khdaw.com76若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comThenegativesignmeansthatthetopsidesofbothcapacitorwillbenegativelychargedaftertheswitchesareclosed.(b)ThechargeonC1isC1
V=(1:16F)(45:4V)=52:7C:(c)ThechargeonC2isC2
V=(3:22F)(45:4V)=146C:P30-6C2andC3formaneectivecapacitorwithequivalentcapacitanceCa=C2C3=(C2+C3).ThechargeonC1isoriginallyq0=C1
V0.AfterthrowingtheswitchthepotentialacrossC1isgivenbyq1=C1
V1.ThesamepotentialisacrossCa;q2=q3,soq2=Ca
V1.Chargeisconserved,soq1+q2=q0.Combiningsomeoftheabove,q0C1
V1==
V0;C1+CaC1+CaandthenC2C2(C+C)q=1
V=123
V:100C1+CaC1C2+C1C3+C2C3Similarly,khdaw.com1CaC1111q2=
V0=++
V0:C1+CaC1C2C3q3=q2becausetheyareinseries.P30-7(a)Ifterminalaismorepositivethanterminalbthencurrentcan owthatwillchargethecapacitorontheleft,thecurrentcan owthroughthediodeonthetop,andthecurrentcanchargethecapacitorontheright.Currentwillnot owthroughthediodeontheleft.Thecapacitorsareeectivelyinseries.Sincethecapacitorsareidenticalandseriescapacitorshavethesamecharge,weexpectthewww.khdaw.comcapacitorstohavethesamepotentialdierenceacrossthem.Butthetotalpotentialdierenceacrossbothcapacitorsisequalto100V,sothepotentialdierenceacrosseithercapacitoris50V.Theoutputpinsareconnectedtothecapacitorontheright,sothepotentialdierenceacrosstheoutputis50V.(b)Ifterminalbismorepositivethanterminalathecurrentcan owthroughthediodeontheleft.Ifweassumethediodeisresistancelessinthiscongurationthenthepotentialdierenceacrossitwillbezero.Thenetresultisthatthepotentialdierenceacrosstheoutputpinsis0V.Inreallifethepotentialdierenceacrossthediodewouldnotbezero,evenifforwardbiased.Itwillbesomewherearound0.5Volts.P30-8Dividethestripofwidth课后答案网aintoNsegments,eachofwidth
x=a=N.Thecapacitanceofeachstripis
C=0a
x=y.Ifissmallthen111d=1x=d):yd+xsind+x(Sinceparallelcapacitancesadd,XZa20a0aaC=
C=(1x=d)dx=1:d0d2d77khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP30-9(a)WhenS2isopenthecircuitactsastwoparallelcapacitors.Thebranchonthelefthasaneectivecapacitancegivenby1111=+=;Cl(1:0106F)(3:0106F)7:5107Fwhilethebranchontherighthasaneectivecapacitancegivenby1111=+=:Cl(2:0106F)(4:0106F)1:33106FThechargeoneithercapacitorinthebranchontheleftisq=(7:5107F)(12V)=9:0106C;whilethechargeoneithercapacitorinthebranchontherightiskhdaw.comq=(1:33106F)(12V)=1:6105C:(b)AfterclosingS2thecircuitiseectivelytwocapacitorsinseries.ThetopparthasaneectivecapacitanceofC=(1:0106F)+(2:0106F)=(3:0106F);twhiletheeectivecapacitanceofthebottompartisC=(3:0106F)+(4:0106F)=(7:0106F):bTheeectivecapacitanceoftheseriescombinationisgivenby11www.khdaw.com11=+=:Ceq(3:0106F)(7:0106F)2:1106FThechargeoneachpartisq=(2:1106F)(12V)=2:52105C.Thepotentialdierenceacrossthetoppartis
V=(2:52105C)=(3:0106F)=8:4V;tandthenthechargeonthetoptwocapacitorsisq=(1:0106F)(8:4V)=8:4106Cand1q=(2:0106F)(8:4V)=1:68105C.Thepotentialdierenceacrossthebottompartis2
V=(2:52105C)=(7:0106F)=3:6V;tandthenthechargeonthetoptwocapacitorsis课后答案网q=(3:0106F)(3:6V)=1:08105Cand1q=(4:0106F)(3:6V)=1:44105C.2P30-10Let
V=
Vxy.Bysymmetry
V2=0and
V1=
V4=
V5=
V3=
V=2.Suddenlytheproblemisveryeasy.Thechargesoneachcapacitorisq1,exceptforq2=0.Thentheequivalentcapacitanceofthecircuitisqq1+q46Ceq===C1=4:010F:
V2
V178khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP30-11(a)ThechargeonthecapacitorwithstoredenergyU0=4:0Jisq0,whereq20U0=:2CWhenthiscapacitorisconnectedtoanidenticalunchargedcapacitorthechargeissharedequally,sothatthechargeoneithercapacitorisnowq=q0=2.Thestoredenergyinonecapacitoristhenq2q2=410U===U0:2C2C4Buttherearetwocapacitors,sothetotalenergystoredis2U=U0=2=2:0J.(b)Goodquestion.Currenthadto owthroughtheconnectingwirestogetthechargefromonecapacitortotheother.Originallythesecondcapacitorwasuncharged,sothepotentialdierenceacrossthatcapacitorwouldhavebeenzero,whichmeansthepotentialdierenceacrossthecon-nectingwireswouldhavebeenequaltothatoftherstcapacitor,andtherewouldthenhavebeenenergydissipationinthewiresaccordingtokhdaw.comP=i2R:That"swherethemissingenergywent.P30-12R=L=AandC=0A=L.Combining,R=0=C,orR=(9:40
m)(8:851012F=m)=(1101012F)=0:756:P30-13(a)u=1E2=e2=322r4.R200(b)U=udVwheredV=4r2dr.Thenwww.khdaw.comZ122e2e1U=4pirdr=:R3220r480R(c)R=e2=8mc2,or0(1:601019C)2R==1:401015m:8(8:851012F=m)(9:111031kg)(3:00108m=s)2P30-14U=1q2=C=q2x=2A.F=dU=dx=q2=2A.200课后答案网P30-15AccordingtoProblem14,theforceonaplateofaparallelplatecapacitorisq2F=:20ATheforceperunitareaisthenFq22==;A20A220where=q=Aisthesurfacechargedensity.ButweknowthattheelectriceldnearthesurfaceofaconductorisgivenbyE==0,soF12=0E:A279khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP30-16AsmallsurfaceareaelementdAcarriesachargedq=qdA=4R2.Therearethreeforcesontheelementswhichbalance,sop(V=V)dA+qdq=4R2=pdA;00orpR3+q2=162R=pR3:00Thiscanberearrangedasq2=162pR(R3R3):00P30-17ThemagnitudeoftheelectriceldinthecylindricalregionisgivenbyE==20r;whereisthelinearchargedensityontheanode.Thepotentialdierenceisgivenby
V=(=20)ln(b=a);whereaistheradiusoftheanodebtheradiusofthecathode.Combining,E=
V=rln(b=a),thiswillbeamaximumwhenr=a,sokhdaw.com
V=(0:180103m)ln[(11:0103m)=(0:180103m)](2:20106V=m)=1630V:P30-18Thisiseectivelytwocapacitorsinparallel,eachwithanareaofA=2.Then0A=20A=20Ae1+e2Ceq=e1+e2=:ddd2P30-19Wewilltreatthesystemastwocapacitorsinseriesbypretendingthereisaninnitesi-mallythinconductorbetweenthem.Theslabsare(Iassume)thesamethickness.ThecapacitanceofoneoftheslabsisthengivenbyEq.30-31,e10AC1=;www.khdaw.comd=2whered=2isthethicknessoftheslab.Therewouldbeasimilarexpressionfortheotherslab.TheequivalentseriescapacitancewouldbegivenbyEq.30-21,111=+;CeqC1C2d=2d=2=+;e10Ae20Ade2+e1=;20Ae1e2课后答案网20Ae1e2Ceq=:de2+e1P30-20Treatthisasthreecapacitors.Findtheequivalentcapacitanceoftheseriescombinationontheright,andthenaddontheparallelpartontheleft.Therighthandsideis1dd2de2+e3=+=:Ceqe20A=2e30A=20Ae2e3Addthistothelefthandside,ande10A=20Ae2e3Ceq=+;2d2de2+e30Ae1e2e3=+:2d2e2+e3khdaw.com80若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP30-21(a)qdoesn"tchange,butC0=C=2.Then
V0=q=C=2
V.(b)U=C(
V)2=2=A(
V)2=2d.U0=C0(
V0)2=2=A(2
V)2=4d=2U.00(c)W=U0U=2UU=U=A(
V)2=2d:0P30-22ThetotalenergyisU=qV=2=(7:021010C)(52:3V)=2=1:84108J.(a)IntheairgapwehaveE2V(8:851012F=m)(6:9103V=m)2(1:15102m2)(4:6103m)U=00==1:11108J:a22Thatis(1:11=1:85)=60%ofthetotal.(b)Theremaining40%isintheslab.P30-23(a)C=A=d=(8:851012F=m)(0:118m2)=(1:22102m)=8:561011F.0(b)UsetheresultsofProblem30-24.(4:8)(8:851012F=m)(0:118m2)khdaw.comC0==1:191010F(4:8)(1:22102m)(4:3103m)(4:81)(c)q=C
V=(8:561011F)(120V)=1:03108C;sincethebatteryisdisconnectedq0=q.(d)E=q=A=(1:03108C)=(8:851012F=m)(0:118m2)=9860V=minthespacebetween0theplates.(e)E0=E==(9860V=m)=(4:8)=2050V=minthedielectric.e(f)
V0=q=C0=(1:03108C)=(1:191010F)=86:6V.(g)W=U0U=q2(1=C1=C0)=2,or(1:03108C)2W=[1=(8:561011F)www.khdaw.com1=(1:191010F)]=1:73107J:2P30-24Theresultiseectivelythreecapacitorsinseries.Twoareairlledwiththicknessesofxanddbx,thethirdisdielectriclledwiththicknessb.AllhaveanareaA.Theeectivecapacitanceisgivenby1xdbxb=++;C0A0Ae0A1b=(db)+;0Ae0A课后答案网C=;db+b=ee0A=:eb(e1)81khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE31-1(5:12A)(6:00V)(5:75min)(60s/min)=1:06104J:E31-2(a)(12:0V)(1:601019C)=1:921018J.(b)(1:921018J)(3:401018=s)=6:53W.E31-3Iftheenergyisdeliveredatarateof110W,thenthecurrentthroughthebatteryisP(110W)i===9:17A:
V(12V)Currentisthe owofchargeinsomeperiodoftime,so
q(125A
h)
t===13:6h;i(9:2A)whichisthesameas13hoursand36minutes.E31-4khdaw.com(100W)(8h)=800W
h.(a)(800W
h)=(2:0W
h)=400batteries,atacostof(400)($0:80)=$320.(b)(800W
h)($0:12103W
h)=$0:096:E31-5Goallofthewayaroundthecircuit.Itisasimpleoneloopcircuit,andalthoughitdoesnotmatterwhichwaywegoaround,wewillfollowthedirectionofthelargeremf.Then(150V)i(2:0)(50V)i(3:0)=0;whereiispositiveifitiscounterclockwise.Rearranging,100V=www.khdaw.comi(5:0);ori=20A.AssumingthepotentialatPisVP=100V,thenthepotentialatQwillbegivenbyVQ=VP(50V)i(3:0)=(100V)(50V)(20A)(3:0)=10V:E31-6(a)Req=(10)+(140)=150.i=(12:0V)=(150)=0:080A.(b)Req=(10)+(80)=90.i=(12:0V)=(90)=0:133A.(c)Req=(10)+(20)=30.i=(12:0V)=(30)=0:400A.E31-7(a)Req=(3课后答案网:0V2:0V)=(0:050A)=20.ThenR=(20)(3:0)(3:0)=14.(b)P=i
V=i2R=(0:050A)2(14)=3:5102W.E31-8(5:0A)R1=
V.(4:0A)(R1+2:0)=
V.Combining,5R1=4R1+8:0,orR1=8:0.E31-9(a)(53:0W)=(1:20A)=44:2V.(b)(1:20A)(19:0)=22:8VisthepotentialdierenceacrossR.Thenanadditionalpotentialdierenceof(44:2V)(22:8V)=21:4VmustexistacrossC.(c)Theleftsideispositive;itisareverseemf.pE31-10(a)Thecurrentintheresistoris(9:88W)=(0:108)=9:56A.Thetotalresistanceofthecircuitis(1:50V)=(9:56A)=0:157.Theinternalresistanceofthebatteryisthen(0:157)(0:108)=0:049.(b)(9:88W)=(9:56A)=1:03V.khdaw.com82若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE31-11Weassigndirectionstothecurrentsthroughthefourresistorsasshowninthegure.12ab34Sincetheammeterhasnoresistancethepotentialataisthesameasthepotentialatb.Con-sequentlythepotentialdierence(
Vb)acrossbothofthebottomresistorsisthesame,andthepotentialdierence(
Vt)acrossthetwotopresistorsisalsothesame(butdierentfromthebottom).Wethenhavethefollowingrelationships:khdaw.com
Vt+
Vb=E;i1+i2=i3+i4;
Vj=ijRj;wherethejsubscriptinthelastlinereferstoresistor1,2,3,or4.Forthetopresistors,
V1=
V2implies2i1=i2;whileforthebottomresistors,
V3=
V4impliesi3=i4:Thenthejunctionrulerequiresi4=3i1=2,andthelooprulerequireswww.khdaw.com(i1)(2R)+(3i1=2)(R)=Eori1=2E=(7R):Thecurrentthat owsthroughtheammeteristhedierencebetweeni2andi4,or4E=(7R)3E=(7R)=E=(7R).E31-12(a)Denethecurrenti1asmovingtotheleftthroughr1andthecurrenti2asmovingtotheleftthroughr2.i3=i1+i2ismovingtotherightthroughR.Thentherearetwoloopequations:E1=i1r1+i3R;课后答案网E2=(i3i1)r2+i3R:Multiplythetopequationbyr2andthebottombyr1andthenadd:r2E1+r1E2=i3r1r2+i3R(r1+r2);whichcanberearrangedasr2E1+r1E2i3=:r1r2+Rr1+Rr2(b)Thereisonlyonecurrent,soE1+E2=i(r1+r2+R);orE1+E2i=:r1+r2+Rkhdaw.com83若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE31-13(a)Assumethatthecurrent owsthrougheachsourceofemfinthesamedirectionastheemf.ThethelooprulewillgiveusthreeequationsE1i1R1+i2R2E2i1R1=0;E2i2R2+i3R1E3+i3R1=0;E1i1R1+i3R1E3+i3R1i1R1=0:Thejunctionrule(looksatpointa)givesusi1+i2+i3=0.Usethistoeliminatei2fromthesecondloopequation,E2+i1R2+i3R2+2i3R1E3=0;andthencombinethiswiththethethirdequationtoeliminatei3,ERER+2iRR+2ER+2iRR+4iR22ER=0;1232312213123131or2E3R1+E3R2E1R22E2R1i3=2=0:582A:khdaw.com4R1R2+4R1Thenwecanndi1fromE3E2i3R22i3R1i1==0:668A;R2wherethenegativesignindicatesthecurrentisdown.Finally,wecanndi2=(i1+i3)=0:0854A:(b)Startataandgotob(nalminusinitial!),+i2R2E2=3:60V:E31-14(a)Thecurrentthroughthecircuitiswww.khdaw.comi=E=(r+R).ThepowerdeliveredtoRisthenP=i
V=i2R=E2R=(r+R)2:EvaluatedP=dRandsetitequaltozerotondthemaximum.ThendP2rR0==ER;dR(r+R)3whichhasthesolutionr=R.(b)Whenr=Rthepoweris1E2P=E2R=:(R+R)24r课后答案网E31-15(a)WerstuseP=Fvtondthepoweroutputbytheelectricmotor.ThenP=(2:0N)(0:50m=s)=1:0W.Thepotentialdierenceacrossthemotoris
Vm=Eir:Thepoweroutputfromthemotoristherateofenergydissipation,soPm=
Vmi:Combiningthesetwoexpressions,Pm=(Eir)i;=Eii2r;0=i2r+EiP;m0=(0:50)i2(2:0V)i+(1:0W):Rearrangeandsolvefori,p(2:0V)(2:0V)24(0:50)(1:0W)i=;2(0:50)khdaw.com84若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comwhichhassolutionsi=3:4Aandi=0:59A.(b)Thepotentialdierenceacrosstheterminalsofthemotoris
Vm=Eirwhichifi=3:4Ayields
Vm=0:3V,butifi=0:59Ayields
Vm=1:7V.Thebatteryprovidesanemfof2.0V;itisn"tpossibleforthepotentialdierenceacrossthemotortobelargerthanthis,butbothsolutionsseemtosatisfythisconstraint,sowewillmovetothenextpartandseewhathappens.(c)Sowhatisthesignicanceofthetwopossiblesolutions?Itisaconsequenceofthefactthatpowerisrelatedtothecurrentsquared,andwithanyquadraticsweexpecttwosolutions.Botharepossible,butitmightbethatonlyoneisstable,oreventhatneitherisstable,andasmallperturbationtothefrictioninvolvedinturningthemotorwillcausethesystemtobreakdown.Wewilllearninalaterchapterthattheeectiveresistanceofanelectricmotordependsonthespeedatwhichitisspinning,andalthoughthatwon"taecttheproblemhereasworded,itwillaectthephysicalproblemthatprovidedthenumbersinthisproblem!E31-16req=4r=4(18)=72.Thecurrentisi=(27V)=(72)=0:375A.E31-17khdaw.comInparallelconnectionsoftworesistorstheeectiveresistanceislessthanthesmallerresistancebutlargerthanhalfthesmallerresistance.Inseriesconnectionsoftworesistorstheeectiveresistanceisgreaterthanthelargerresistancebutlessthantwicethelargerresistance.Sincetheeectiveresistanceoftheparallelcombinationislessthaneithersingleresistanceandtheeectiveresistanceoftheseriescombinationsislargerthaneithersingleresistancewecanconcludethat3:0musthavebeentheparallelcombinationand16musthavebeentheseriescombination.Theresistorsarethen4:0and12resistors.E31-18PointsBandCareeectivelythesamepoint!(a)Thethreeresistorsareinparallel.Thenrwww.khdaw.comeq=R=3.(b)See(a).(c)0,sincethereisnoresistancebetweenBandC.E31-19Focusontheloopthroughthebattery,the3:0,andthe5:0resistors.Theloopruleyields(12:0V)=i[(3:0)+(5:0)]=i(8:0):Thepotentialdierenceacrossthe5:0resistoristhen
V=i(5:0)=(5:0)(12:0V)=(8:0)=7:5V:E31-20Eachlampdrawsacurrentof(500W)课后答案网=(120V)=4:17A.Furthermore,thefusecansupport(15A)=(4:17A)=3:60lamps.Thatisamaximumof3.E31-21Thecurrentintheseriescombinationisis=E=(R1+R2).ThepowerdissipatedisP=iE=E2=(R+R):s12InaparallelarrangementRdissipatesP=iE=E2=R.AsimilarexpressionexistsforR,11112sothetotalpowerdissipatedisP=E2(1=R+1=R):p12Theratiois5,so5=P=P=(1=R+1=R)(R+R),or5RR=(R+R)2.Solvingforps12121212R2yields2:618R1or0:382R1.ThenR2=262orR2=38:2.85khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE31-22Combiningnidenticalresistorsinseriesresultsinanequivalentresistanceofreq=nR.Combiningnidenticalresistorsinparallelresultsinanequivalentresistanceofreq=R=n.Iftheresistorsarearrangedinasquarearrayconsistingofnparallelbranchesofnseriesresistors,thentheeectiveresistanceisR.EachwilldissipateapowerP,togethertheywilldissipaten2P.Sowewantnineresistors,sincefourwouldbetoosmall.E31-23(a)Workthroughthecircuitonestepatatime.Werstadd"R2,R3,andR4inparallel:11111=++=Re42:061:675:018:7Wethenadd"thisresistanceinserieswithR1,Re=(112)+(18:7)=131:(b)Thecurrentthroughthebatteryisi=E=R=(6:22V)=(131)=47:5mA.ThisisalsothecurrentthroughR1,sinceallthecurrentthroughthebatterymustalsogothroughR1.khdaw.comThepotentialdierenceacrossR1is
V1=(47:5mA)(112)=5:32V.Thepotentialdierenceacrosseachofthethreeremainingresistorsis6:22V5:32V=0:90V.Thecurrentthrougheachresistoristheni2=(0:90V)=(42:0)=21:4mA;i3=(0:90V)=(61:6)=14:6mA;i4=(0:90V)=(75:0)=12:0mA:E31-24Theequivalentresistanceoftheparallelpartisr0=RR=(R+R):Theequivalent22resistanceforthecircuitisr=R+RR=(R+R):Thecurrentthroughthecircuitisi0=E=r.122ThepotentialdierenceacrossRis
V=Ei0R,orwww.khdaw.com1
V=E(1R1=r);R2+R=E1R1;R1R2+R1R+RR2RR2=E:R1R2+R1R+RR2SinceP=i
V=(
V)2=R,RR2P=E22:(R1R2+R1R+RR2)2SetdP=dR=0,thesolutionis课后答案网R=R1R2=(R1+R2).E31-25(a)Firstadd"thelefttworesistorsinseries;theeectiveresistanceofthatbranchis2R.Thenadd"therighttworesistorsinseries;theeectiveresistanceofthatbranchisalso2R.Nowwecombinethethreeparallelbranchesandndtheeectiveresistancetobe11114=++=;Re2RR2R2RorRe=R=2.(b)Firstweadd"therighttworesistorsinseries;theeectiveresistanceofthatbranchis2R.WethencombinethisbranchwiththeresistorwhichconnectspointsFandH.Thisisaparallelconnection,sotheeectiveresistanceis1113=+=;Re2RR2Rkhdaw.com86若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comor2R=3.ThisvalueiseectivelyinserieswiththeresistorwhichconnectsGandH,sothetotal"is5R=3.Finally,wecancombinethisvalueinparallelwiththeresistorthatdirectlyconnectsFandGaccordingto1138=+=;ReR5R5RorRe=5R=8.E31-26Theresistanceofthesecondresistorisr2=(2:4V)=(0:001A)=2400.Thepotentialdierenceacrosstherstresistoris(12V)(2:4V)=9:6V.Theresistanceoftherstresistoris(9:6V)=(0:001A)=9600.E31-27SeeExercise31-26.Theresistanceratioisr1(0:950:1V)=;khdaw.comr1+r2(1:50V)orr2(1:50V)=1:r1(0:950:1V)Theallowedrangefortheratior2=r1isbetween0:5625and0:5957.Wecanchooseanystandardresistorswewant,andwecoulduseanytolerance,butthenwewillneedtocheckourresults.22and39wouldwork;aswould27and47.Thereareotherchoices.E31-28ConsideranyjunctionotherthanAwww.khdaw.comorB.Callthisjunctionpoint0;labelthefournearestjunctionstothisaspoints1,2,3,and4.Thecurrentthroughtheresistorthatlinkspoint0topoint1isi1=
V01=R;where
V01isthepotentialdierenceacrosstheresistor,so
V01=V0V1;whereV0isthepotentialatthejunction0,andV1isthepotentialatthejunction1.Similarexpressionsexistfortheotherthreeresistor.Forthejunction0thenetcurrentmustbezero;thereisnowayforchargetoaccumulateonthejunction.Theni1+i2+i3+i4=0;andthismeans
V01=R+
V02=R+
V03=R+
V04=R=0or课后答案网
V01+
V02+
V03+
V04=0:Let
V0i=V0Vi,andthenrearrange,4V0=V1+V2+V3+V4;or1V0=(V1+V2+V3+V4):4E31-29Thecurrentthroughtheradioisi=P=
V=(7:5W)=(9:0V)=0:83A.Theradiowasleftonefor6hours,or2:16104s.Thetotalchargeto owthroughtheradiointhattimeis(0:83A)(2:16104s)=1:8104C.E31-30Thepowerdissipatedbytheheadlightsis(9:7A)(12:0V)=116W.Thepowerrequiredbytheengineis(116W)=(0:82)=142W,whichisequivalentto0:khdaw.com190hp.87若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE31-31(a)P=(120V)(120V)=(14:0)=1030W:(b)W=(1030W)(6:42h)=6:61kW
h:Thecostis$0:345.E31-32E31-33WewanttoapplyeitherEq.31-21,P=i2R;RorEq.31-22,P=(
V)2=R;RRdependingonwhetherweareinseries(thecurrentisthesamethrougheachbulb),orinparallel(thepotentialdierenceacrosseachbulbisthesame.ThebrightnessofabulbwillbemeasuredbyP,eventhoughPisnotnecessarilyameasureoftherateradiantenergyisemittedfromthebulb.(b)IfthebulbsareinparallelthenP=(
V)2=Rishowwewanttocomparethebrightness.RRThepotentialdierenceacrosseachbulbisthesame,sothebulbwiththesmallerresistanceiskhdaw.combrighter.(b)IfthebulbsareinseriesthenP=i2Rishowwewanttocomparethebrightness.BothRbulbshavethesamecurrent,sothelargervalueofRresultsinthebrighterbulb.Onedirectconsequenceofthiscanbetriedathome.Wireupa60W,120Vbulbanda100W,120Vbulbinseries.Whichisbrighter?Youshouldobservethatthe60Wbulbwillbebrighter.E31-34(a)j=i=A=(25A)=(0:05in)=3180A/in2=4:93106A=m2.(b)E=j=(1:69108
m)(4:93106A=m2)=8:33102V=m.(c)
V=Ed=(8:33102V=m)(305m)=25V.(d)P=i
V=(25A)(25V)=625W.www.khdaw.comE31-35(a)Thebulbisonfor744hours.Theenergyconsumedis(100W)(744h)=74:4kW
h,atacostof(74:4)(0:06)=$4:46.(b)r=V2=P=(120V)2=(100W)=144.(c)i=P=V=(100W)=(120V)=0:83A.E31-36P=(
V)2=randr=r(1+
T).Then0P0(500W)P===660W1+
T1+(4:0104=C)(600C)课后答案网E31-37(a)n=q=e=it=e,son=(485103A)(95109s)=(1:61019C)=2:881011:(b)i=(520=s)(485103A)(95109s)=2:4105A.av(c)P=i
V=(485103A)(47:7106V)=2:3106W;whileP=i
V=(2:4105A)(47:7ppaa106V)=1:14103W.E31-38r=L=A=(3:5105
m)(1:96102m)=(5:12103m)2=8:33103.pp(a)i=P=r=(1:55W)=(8:33103)=13:6A,soj=i=A=(13:6A)=(5:12103m)2=1:66105A=m2:pp(b)
V=Pr=(1:55W)(8:33103)=0:114V:khdaw.com88若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE31-39(a)Thecurrentthroughthewireisi=P=
V=(4800W)=(75V)=64A;TheresistanceofthewireisR=
V=i=(75V)=(64A)=1:17:ThelengthofthewireisthenfoundfromRA(1:17)(2:6106m2)L===6:1m:(5:0107m)Onecouldeasilywindthismuchnichrometomakeatoasteroven.Ofcourseallowing64Ampstobedrawnthroughhouseholdwiringwilllikelyblowafuse.(b)Wewanttocombinetheabovecalculationsintooneformula,soRAA
V=iA(
V)2khdaw.comL===;Pthen(2:6106m2)(110V)2L==13m:(4800W)(5:0107m)Hmm.Weneedmorewireifthepotentialdierenceisincreased?Doesthismakesense?Yes,itdoes.Weneedmorewirebecauseweneedmoreresistancetodecreasethecurrentsothatthesamepoweroutputoccurs.E31-40(a)Theenergyrequiredtobringthewatertoboilingiswww.khdaw.comQ=mC
T.ThetimerequiredisQ(2:1kg)(4200J=kg)(100C18:5C)t===2:22103s0:77P0:77(420W)(b)TheadditionaltimerequiredtoboilhalfofthewaterawayismL=2(2:1kg)(2:26106J=kg)=2t===7340s:0:77P0:77(420W)E31-41(a)IntegratebothsidesofEq.31-26;课后答案网ZqZtdqdt=;0qEC0RCtqtln(qEC)j=;0RC0qECtln=;ECRCqECt=RC=e;ECq=EC1et=RC:Thatwasn"tsobad,wasit?89khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com(b)RearrangeEq.31-26inordertogetqtermsontheleftandttermsontheright,thenintegrate;ZqZtdqdt=;q0q0RCtqtlnqj=;q0RC0qtln=;q0RCqt=RC=e;q0q=qet=RC:0Thatwasn"tsobadeither,wasit?E31-42(a)=RC=(1:42106)(1:80106F)=2:56s.Ckhdaw.com(b)q=C
V=(1:80106F)(11:0V)=1:98105C.0(c)t=Cln(1q=q0),sot=(2:56s)ln(115:5106C=1:98105C)=3:91s:E31-43Solven=t=C=ln(10:99)=4:61.E31-44(a)
V=E(1et=C),so=(1:28106s)=ln(15:00V=13:0V)=2:64106sC(b)C==R=(2:64106s)=(15:2103)=1:731010FCwww.khdaw.comE31-45(a)
V=Eet=C,soC=(10:0s)=ln(1:06V=100V)=2:20s(b)
V=(100V)e17s=2:20s=4:4102V.E31-46
V=Eet=Cand=RC,soCtttR===:Cln(
V=
V0)(220109F)ln(0:8V=5V)4:03107FIftisbetween10:0课后答案网sand6:0ms,thenRisbetweenR=(10106s)=(4:03107F)=24:8;andR=(6103s)=(4:03107F)=14:9103:E31-47ThechargeonthecapacitorneedstobuilduptoapointwherethepotentialacrossthecapacitorisVL=72V,andthisneedstohappenwithin0.5seconds.ThismeansthatwewanttosolveC
V=CE1eT=RCLforRknowingthatT=0:5s.ThisexpressioncanbewrittenasT(0:5s)6R===2:3510:Cln(1VL=E)(0:15C)ln(1(72V)=(95V))khdaw.com90若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comppE31-48(a)q=2UC=2(0:50J)(1:0106F)=1103C.0(b)i=
V=R=q=RC=(1103C)=(1:0106)(1:0106F)=1103A.000(c)
V=
Vet=C,soC03(110C)t=(1:0106)(1:0106F)t=(1:0s)
VC=e=(1000V)e(1:0106F)Notethat
VR=
VC.(d)P=(
V)2=R,soRRP=(1000V)2e2t=(1:0s)=(1106!)=(1W)e2t=(1:0s):RE31-49(a)i=dq=dt=Eet=C=R,so(4:0V)(1:0s)=(3:0106)(1:0106F)7i=e=9:5510A:(3:0106)(b)P=i
V=(E2=R)et=C(1et=C),sokhdaw.comC2(4:0V)(1:0s)=(3:0106)(1:0106F)(1:0s)=(3:0106)(1:0106F)6PC=e1e=1:0810W:(3:0106)(c)P=i2R=(E2=R)e2t=C,soR2(4:0V)2(1:0s)=(3:0106)(1:0106F)6PR=e=2:7410W:(3:0106)(d)P=PR+PC,orP=2:74106W+1:08www.khdaw.com106W=3:82106WE31-50TherateofenergydissipationintheresistorisP=i2R=(E2=R)e2t=C:REvaluatingZ12Z12E2t=RCEPRdt=edt=C;0R02butthatistheoriginalenergystoredinthecapacitor.P31-1TheterminalvoltageofthebatteryisgivenbyV=Eir;sotheinternalresistanceis课后答案网EV(12:0V)(11:4V)r===0:012;i(50A)sothebatteryappearswithinspecs.Theresistanceofthewireisgivenby
V(3:0V)R===0:06;i(50A)sothecableappearstobebad.Whataboutthemotor?Tryingit,
V(11:4V)(3:0V)R===0:168;i(50A)soitappearstobewithinspec.khdaw.com91若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP31-2TraversingthecircuitwehaveEir1+Eir2iR=0;soi=2E=(r1+r2+R):Thepotentialdierenceacrosstherstbatteryisthen2r1r2r1+R
V1=Eir1=E1=Er1+r2+Rr1+r2+RThisquantitywillonlyvanishifr2r1+R=0,orr1=R+r2.Sincer1>r2thisisactuallypossible;R=r1r2.P31-3
V=Eiriandi=E=(ri+R),soR
V=E;ri+RTherearethentwosimultaneousequations:(0:10V)(500)+(0:10V)ri=E(500)andkhdaw.com(0:16V)(1000)+(0:16V)ri=E(1000);withsolution(a)r=1:5103andi(b)E=0:400V.(c)Thecellreceivesenergyfromthesunatarate(2:0mW/cm2)(5:0cm2)=0:010W.ThecellconvertsenergyatarateofV2=R=(0:16V)2=(1000)=0:26%P31-4(a)TheemfofthebatterycanbefoundfromE=iri+
Vl=(10A)(0www.khdaw.com:05)+(12V)=12:5V(b)Assumethatresistanceisnotafunctionoftemperature.Theresistanceoftheheadlightsisthenrl=(12:0V)=(10:0A)=1:2:Thepotentialdierenceacrossthelightswhenthestartermotorisonis
Vl=(8:0A)(1:2)=9:6V;andthisisalsothepotentialdierenceacrosstheterminalsofthebattery.ThecurrentthroughthebatteryisthenE
V(12:5V)(9:6V)i===58A;课后答案网ri(0:05)sothecurrentthroughthemotoris50Amps.P31-5(a)Theresistivitiesare=rA=L=(76:2106)(91:0104m2)=(42:6m)=1:63108
m;AAand=rA=L=(35:0106)(91:0104m2)=(42:6m)=7:48109
m:BB(b)Thecurrentisi=
V=(r+r)=(630V)=(111:2)=5:67106A.ThecurrentdensityABisthenj=(5:67106A)=(91:0104m2)=6:23108A=m2:(c)E=j=(1:63108
m)(6:23108A=m2)=10:2V=mandE=j=(7:48109
AABBm)(6:23108A=m2)=4:66V=m.(d)
VA=EAL=(10:2V=m)(42:6m)=435Vand
VB=EBkhdaw.comL=(4:66V=m)(42:6m)=198V.92若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP31-6SetuptheproblemwiththetraditionalpresentationoftheWheatstonebridgeproblem.Thenthesymmetryoftheproblem( ipitoveronthelinebetweenxandy)impliesthatthereisnocurrentthroughr.Assuch,theproblemisequivalenttotwoidenticalparallelbrancheseachwithtwoidenticalseriesresistances.EachbranchhasresistanceR+R=2R,sotheoverallcircuithasresistance1111=+=;Req2R2RRsoReq=R.P31-7P31-8(a)TheloopthroughR1istrivial:i1=E2=R1=(5:0V)=(100)=0:05A.TheloopthroughR2isonlyslightlyharder:i2=(E2+E3E1)=R2=0:06A.khdaw.com(b)
Vab=E3+E2=(5:0V)+(4:0V)=9:0V.P31-9(a)Thethreewaylight-bulbhastwolaments(orsowearetoldinthequestion).Therearefourwaysforthesetwolamentstobewired:eitheronealone,bothinseries,orbothinparallel.Wiringthelamentsinserieswillhavethelargesttotalresistance,andsinceP=V2=Rthisarrangementwouldresultinthedimmestlight.Butwearetoldthelightstilloperatesatthelowestsetting,andifalamentburnedoutinaseriesarrangementthelightwouldgoout.Wethenconcludethatthelowestsettingisonelament,themiddlesettingisanotherlament,andthebrightestsettingisbothlamentsinparallel.(b)Thebeautyofparallelsettingsisthatthenpowerisadditive(itisalsoaddictive,butthat"sadierenteld.)Onelamentdissipates100Wat120V;theotherlament(theonethatburnsout)dissipates200Wat120V,andbothtogetherdissipate300Wat120V.www.khdaw.comTheresistanceofonelamentisthen(
V)2(120V)2R===144:P(100W)Theresistanceoftheotherlamentis(
V)2(120V)2R===72:P(200W)P31-10Wecanassumethat课后答案网Rcontains"alloftheresistanceoftheresistor,thebatteryandtheammeter,thenR=(1:50V)=(1:0m=A)=1500:ForeachofthefollowingpartsweapplyR+r=
V=i,so(a)r=(1:5V)=(0:1mA)(1500)=1:35104,(b)r=(1:5V)=(0:5mA)(1500)=1:5103,(c)r=(1:5V)=(0:9mA)(1500)=167.(d)R=(1500)(18:5)=1482P31-11(a)TheeectiveresistanceoftheparallelbranchesonthemiddleandtherightisR2R3:R2+R393khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comTheeectiveresistanceofthecircuitasseenbythebatteryisthenR2R3R1R2+R1R3+R2R3R1+=;R2+R3R2+R3ThecurrentthroughthebatteryisR2+R3i=E;R1R2+R1R3+R2R3ThepotentialdierenceacrossR1isthenR2+R3
V1=ER1;R1R2+R1R3+R2R3while
V3=E
V1,orR2R3
V3=E;khdaw.comR1R2+R1R3+R2R3sothecurrentthroughtheammeteris
V3R2i3==E;R3R1R2+R1R3+R2R3or(4)i3=(5:0V)=0:45A:(2)(4)+(2)(6)+(4)(6)(b)ChangingthelocationsofthebatteryandtheammeterisequivalenttoswappingR1andR3.Butsincetheexpressionforthecurrentdoesn"tchange,thenthecurrentisthesame.www.khdaw.comP31-12
V1+
V2=
VS+
VX;ifVa=Vb,then
V1=
VS.Usingtherstexpression,ia(R1+R2)=ib(RS+RX);usingthesecond,iaR1=ibR2:Dividingtherstbythesecond,1+R2=R1=1+RX=RS;orRX=RS(R2=R1课后答案网).P31-13P31-14Lv=
Q=
mand
Q=
t=P=i
V,soi
V(5:2A)(12V)6Lv===2:9710J=kg:
m=
t(21106kg=s)P31-15P=i2R.W=p
V,whereVisvolume.p=mg=AandV=Ay,whereyistheheightofthepiston.ThenP=dW=dt=mgv.Combiningallofthis,i2R(0:240A)2(550)v===0:274m=s:mg(11:8kg)(9:8m=s2)khdaw.com94若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP31-16(a)Sinceq=CV,thenq=(32106F)(6V)+(4V=s)(0:5s)(2V=s2)(0:5s)2=2:4104C:(b)Sincei=dq=dt=CdV=dt,theni=(32106F)(4V=s)2(2V=s2)(0:5s)=6:4105A:(c)SinceP=iV,P=[(4V=s)2(2V=s2)(0:5s)(6V)+(4V=s)(0:5s)(2V=s2)(0:5s)2=4:8104W:P31-17(a)WehaveP=30P0andi=4i0.ThenP30P030R===R0:i2(4i0)216Wedon"treallycarewhathappenedwiththepotentialdierence,sinceknowingthechangeinkhdaw.comresistanceofthewireshouldgivealltheinformationweneed.Thevolumeofthewireisaconstant,evenupondrawingthewireout,soLA=L0A0;theproductofthelengthandthecrosssectionalareamustbeaconstant.ResistanceisgivenbyR=L=A,butA=L0A0=L,sothelengthofthewireisssA0L0R30A0L0R0L===1:37L0:16(b)WeknowthatA=L0A0=L,sowww.khdaw.comLA0A=A0==0:73A0:L01:37P31-18(a)ThecapacitorchargeasafunctionoftimeisgivenbyEq.31-27,q=CE1et=RC;whilethecurrentthroughthecircuit(andtheresistor)isgivenbyEq.31-28,Et=RCi=e:课后答案网RTheenergysuppliedbytheemfisZZU=Eidt=Edq=Eq;buttheenergyinthecapacitorisUC=q
V=2=Eq=2.(b)Integrating,ZZE2E2EqU=i2Rdt=e2t=RCdt==:RR2C295khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP31-19ThecapacitorchargeasafunctionoftimeisgivenbyEq.31-27,q=CE1et=RC;whilethecurrentthroughthecircuit(andtheresistor)isgivenbyEq.31-28,Et=RCi=e:RTheenergystoredinthecapacitorisgivenbyq2U=;2CsotheratethatenergyisbeingstoredinthecapacitorisdUqdqqPC===i:khdaw.comdtCdtCTherateofenergydissipationintheresistorisP=i2R;RsothetimeatwhichtherateofenergydissipationintheresistorisequaltotherateofenergystorageinthecapacitorcanbefoundbysolvingPC=PR;2qiR=i;Cwww.khdaw.comiRC=q;ECet=RC=CE1et=RC;et=RC=1=2;t=RCln2:课后答案网96khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE32-1ApplyEq.32-3,F~=q~vB~.Allofthepathswhichinvolvelefthandturnsarepositiveparticles(path1);thosepathswhichinvolverighthandturnsarenegativeparticle(path2andpath4);andthosepathswhichdon"tturninvolveneutralparticles(path3).E32-2(a)ThegreatestmagnitudeofforceisF=qvB=(1:61019C)(7:2106m=s)(83103T)=9:61014N.Theleastmagnitudeofforceis0.(b)TheforceontheelectronisF=ma;theanglebetweenthevelocityandthemagneticeldis,givenbyma=qvBsin.Then(9:11031kg)(4:91016m=s2)=arcsin=28:(1:61019C)(7:2106m=s)(83103T)E32-3(a)v=E=B=(1:5103V=m)=(0:44T)=3:4103m=s.E32-4(a)v=F=qBsin=(6:481017N=(1:601019C)(2:63103T)sin(23:0)=3:94105m=s:khdaw.com(b)K=mv2=2=(938MeV=c2)(3:94105m=s)2=2=809eV:E32-5ThemagneticforceontheprotonisF=qvB=(1:61019C)(2:8107m=s)(30eex6T)=1:31016N:BThegravitationalforceontheprotonismg=(1:71027kg)(9:8m=s2)=1:71026N:Theratioisthen7:6109.If,however,youcarrythenumberofsignicantdigitsfortheintermediateanswersfartheryouwillgettheanswerwhichisinthebackofthebook.www.khdaw.compE32-6Thespeedoftheelectronisgivenbyv=2q
V=m,orpv=2(1000eV)=(5:1105eV=c2)=0:063c:TheelectriceldbetweentheplatesisE=(100V)=(0:020m)=5000V=m.TherequiredmagneticeldisthenB=E=v=(5000V=m)=(0:063c)=2:6104T:E32-7Bothhavethesamevelocity.ThenK=K=mv2=mv2=m=m=.pepepe课后答案网pE32-8Thespeedoftheionisgivenbyv=2q
V=m,orpv=2(10:8keV)=(6:01)(932MeV=c2)=1:96103c:TherequiredelectriceldisE=vB=(1:96103c)(1:22T)=7:17105V=m.E32-9(a)ForachargedparticlemovinginacircleinamagneticeldweapplyEq.32-10;mv(9:111031kg)(0:1)(3:00108m=s)r===3:4104m:jqjB(1:61019C)(0:50T)(b)The(non-relativistic)kineticenergyoftheelectronis12123K=mv=(0:511MeV)(0:10c)=2:610MeV:22khdaw.com97若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comppE32-10(a)v=2K=m=2(1:22keV)=(511keV=c2)=0:0691c.(b)B=mv=qr=(9:111031kg)(0:0691c)=(1:601019C)(0:247m)=4:78104T:(c)f=qB=2m=(1:601019C)(4:78104T)=2(9:111031kg)=1:33107Hz:(d)T=1=f=1=(1:33107Hz)=7:48108s:ppE32-11(a)v=2K=m=2(350eV)=(511keV=c2)=0:037c.(b)r=mv=qB=(9:111031kg)(0:037c)=(1:601019C)(0:20T)=3:16104m:E32-12Thefrequencyisf=(7:00)=(1:29103s)=5:43103Hz.Themassisgivenbym=qB=2f,or(1:601019C)(45:0103T)m==2:111025kg=127u:2(5:43103Hz)E32-13(a)ApplyEq.32-10,butrearrangeitasjqjrB2(1:61019C)(0:045m)(1:2T)khdaw.comv===2:6106m=s:m4:0(1:661027kg)(b)Thespeedisequaltothecircumferencedividedbytheperiod,so2r2m24:0(1:661027kg)T====1:1107s:vjqjB2(1:61019C)(1:2T)(c)The(non-relativistic)kineticenergyisjqj2r2B(21:61019C)2(0:045m)2(1:2T)2K===2:241014J:2m2(4:01:66www.khdaw.com1027kg))Tochangetoelectronvoltsweneedmerelydividethisanswerbythechargeononeelectron,so(2:241014J)K==140keV:(1:61019C)(d)
V=K=(140keV)=(2e)=70V:qE32-14(a)R=mv=qB=(938MeV=c2)(0:100c)=e(1:40T)=0:223m:(b)f=qB=2m=e(1:40T)=2(938MeV=c2)=2:13107Hz:课后答案网E32-15(a)K=K=(q2=m)=(q2=m)=22=4=1:ppp(b)K=K=(q2=m)=(q2=m)=12=2=1=2:dpddppE32-16(a)K=q
V.ThenKp=e
Vp,Kd=e
V,andKp=2e
V.(b)r=sqrt2mK=qB.Thenrd=rp=(2=1)(1=1)=(1=1)=2:pp(c)r=sqrt2mK=qB.Thenr=rp=(4=1)(2=1)=(2=1)=2:pppE32-17r=2mK=jqjB=(m=jqj)(2K=B):Allthreeparticlesaretravelingwiththesamekineticenergyinthesamemagneticeld.Therelevantfactorsareinfront;wejustneedtocomparethemassandchargeofeachofthethreeparticles.p(a)Theradiusofthedeuteronpathis2r:1pp(b)Theradiusofthealphaparticlepathis4r=r.2ppkhdaw.com98若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE32-18Theneutron,beingneutral,isunaectedbythemagneticeldandmovesoinalinetangenttotheoriginalpath.Theprotonmovesatthesameoriginalspeedasthedeuteronandhasthesamecharge,butsinceithashalfthemassitmovesinacirclewithhalftheradius.E32-19(a)Theprotonmomentumwouldbepc=qcBR=e(3:0108m=s)(41106T)(6:4106m)=7:9104MeV:Since79000MeVismuch,muchgreaterthan938MeVtheprotonisultra-relativistic.ThenEpc,andsince=E=mc2wehave=p=mc.Inverting,rsv1m2c2m2c2=1=110:99993:c2p22p2pE32-20(a)Classically,R=2mK=qB,orpR=2(0:511MeV=c2)(10:0MeV)=e(2:20T)=4:84103m:khdaw.com(b)Thiswouldbeanultra-relativisticelectron,soKEpc,thenR=p=qB=K=qBc,orR=(10:0MeV)=e(2:2T)(3:00108m=s)=1:52102m:(c)Theelectroniseectivelytravelingatthespeedoflight,soT=2R=c,orT=2(1:52102m)=(3:00108m=s)=3:181010s:Thisresultdoesdependonthespeed!E32-21UseEq.32-10,exceptwerearrangeforthemass,19www.khdaw.comjqjrB2(1:6010C)(4:72m)(1:33T)27m===9:4310kgv0:710(3:00108m=s)However,ifitismovingatthisvelocitythenthemass"whichwehavehereisnotthetruemass,butarelativisticcorrection.Foraparticlemovingat0:710cwehave11=p=p=1:42;1v2=c21(0:710)2sothetruemassoftheparticleis(9:431027kg)=(1:42)=6:641027kg:Thenumberofnucleonspresentinthisparticleisthen(6课后答案网:641027kg)=(1:671027kg)=3:974:Thechargewas+2,whichimpliestwoprotons,theothertwonucleonswouldbeneutrons,sothismustbeanalphaparticle.E32-22(a)Since950GeVismuch,muchgreaterthan938MeVtheprotonisultra-relativistic.=E=mc2,sorrv1m2c4m2c4=1=110:9999995:c2E22E2(b)Ultra-relativisticmotionrequirespcE,soB=pc=qRc=(950GeV)=e(750m)(3:00108m=s)=4:44T:99khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE32-23Firstuse2f=qB=m.TheuseK=q2B2R2=2m=mR2(2f)2=2.Thenumberofpturnsisn=K=2q
V,onaveragetheparticleislocatedatadistanceR=2fromthecenter,sotheppdistancetraveledisx=n2R=2=n2R.Combining,pp23R3mf223(0:53m)3(2932103keV=c2)(12106=s)2x===240m:q
Ve(80kV)E32-24Theparticlemovesinacircle.x=Rsin!tandy=Rcos!t.E32-25WewilluseEq.32-20,EH=vdB,exceptwewillnottakethederivationthroughtoEq.32-21.Instead,wewillsetthedriftvelocityequaltothespeedofthestrip.Wewill,however,setEH=
VH=w.ThenE
V=w(3:9106V)=(0:88102m)v=H=H==3:7101m=s:BB(1:2103T)E32-26khdaw.com(a)v=E=B=(40106V)=(1:2102m)=(1:4T)=2:4103m=s:(b)n=(3:2A)(1:4T)=(1:61019C)(9:5106m)(40106V)=7:41028=m3:;Silver.E32-27EH=vdBandvd=j=ne.Combineandrearrange.E32-28(a)UsetheresultofthepreviousexerciseandEc=j.(b)(0:65T)=(8:491028=m3)(1:601019C)(1:69108
m)=0:0028:E32-29SinceL~isperpendiculartoB~canusewww.khdaw.comFB=iLB:Equatingthetwoforces,iLB=mg;mg(0:0130kg)(9:81m=s2)i===0:467A:LB(0:620m)(0:440T)Useofanappropriaterighthandrulewillindicatethatthecurrentmustbedirectedtotherightinordertohaveamagneticforcedirectedupward.课后答案网36E32-30F=iLBsin=(5:1210A)(100m)(5810T)sin(70)=27:9N.Thedirectionishorizontallywest.E32-31(a)WeuseEq.32-26again,andsincethe(horizontal)axleisperpendiculartotheverticalcomponentofthemagneticeld,F(10;000N)8i===3:310A:BL(10T)(3:0m)(b)ThepowerlostperohmofresistanceintherailsisgivenbyP=r=i2=(3:3108A)2=1:11017W:(c)Ifsuchatrainweretobedevelopedtherailswouldmeltwellbeforethetrainleftthestation.khdaw.com100若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE32-32F=idB,soa=F=m=idB=m.Sinceaisconstant,v=at=idBt=m.Thedirectionistotheleft.RRE32-33Onlythe^jcomponentofB~isofinterest.ThenF=dF=iBydx,orZ3:2F=(5:0A)(8103T=m2)x2dx=0:414N:1:2Thedirectionisk^.E32-34Themagneticforcewillhavetwocomponents:onewillliftvertically(Fy=Fsin),theotherpushhorizontally(Fx=Fcos).TherodwillmovewhenFx>(WFy).WeareinterestedintheminimumvalueforFasafunctionof.ThisoccurswhendFdW==0:ddcos+sinThishappenswhenkhdaw.com=tan.Then=arctan(0:58)=30,and(0:58)(1:15kg)(9:81m=s2)F==5:66Ncos(30)+(0:58)sin(30)istheminimumforce.ThenB=(5:66N)=(53:2A)(0:95m)=0:112T.E32-35Wechoosethattheeldpointsfromtheshortersidetothelongerside.(a)Themagneticeldisparalleltothe130cmsidesothereisnomagneticforceonthatside.Themagneticforceonthe50cmsidehasmagnitudeFB=iLBwww.khdaw.comsin;whereistheanglebetweenthe50cmsideandthemagneticeld.Thisangleislargerthan90,butthesinecanbefounddirectlyfromthetriangle,(120cm)sin==0:923;(130cm)andthentheforceonthe50cmsidecanbefoundby3(120cm)FB=(4:00A)(0:50m)(75:010T)=0:138N;(130cm)andisdirectedoutoftheplaneofthetriangle.Themagneticforceonthe120cmsidehasmagnitude课后答案网FB=iLBsin;whereistheanglebetweenthe1200cmsideandthemagneticeld.Thisangleislargerthan180,butthesinecanbefounddirectlyfromthetriangle,(50cm)sin==0:385;(130cm)andthentheforceonthe50cmsidecanbefoundby3(50cm)FB=(4:00A)(1:20m)(75:010T)=0:138N;(130cm)andisdirectedintotheplaneofthetriangle.(b)Lookatthethreenumbersabove.khdaw.com101若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE32-36=NiABsin,so=(20)(0:1A)(0:12m)(0:05m)(0:5T)sin(9033)=5:0103N
m:E32-37Theexternalmagneticeldmustbeintheplaneoftheclock/wireloop.Theclockwisecurrentproducesamagneticdipolemomentdirectedintotheplaneoftheclock.(a)Sincethemagneticeldpointsalongthe1pmlineandthetorqueisperpendiculartoboththeexternaleldandthedipole,thenthetorquemustpointalongeitherthe4pmorthe10pmline.ApplyingEq.32-35,thedirectionisalongthe4pmline.Itwilltaketheminutehand20minutestogetthere.(b)=(6)(2:0A)(0:15m)2(0:07T)=0:059N
m:pP32-1SinceF~mustbeperpendiculartoB~thenB~mustbealongpk^.Themagnitudeofvis(40)2+(35)2km/s=53:1km/s;themagnitudeofFis(4:2)2+(4:8)2fN=6:38fN:ThenB=F=qv=(6:381015N)=(1:61019C)(53:1103m=s)=0:75T:orkhdaw.comB~=0:75Tk^:P32-2~a=(q=m)(E~+~vB~).Fortheinitialvelocitygiven,~vB~=(15:0103m=s)(400106T)^j(12:0103m=s)(400106T)k^:Butsincethereisnoaccelerationinthe^jork^directionthismustbeosetbytheelectriceld.Consequently,twooftheelectriceldcomponentsareEy=6:00V=mandEz=4:80V=m.Thethirdcomponentoftheelectriceldisthesourceoftheacceleration,soE=ma=q=(9:111031kg)(2:001012m=s2)=(1:601019C)=11:4V=m:xxwww.khdaw.comP32-3(a)Considerrstthecrossproduct,~vB~.Theelectronmoveshorizontally,thereisacomponentoftheB~whichisdown,sothecrossproductresultsinavectorwhichpointstotheleftoftheelectron"spath.ButtheforceontheelectronisgivenbyF~=q~vB~,andsincetheelectronhasanegativechargetheforceontheelectronwouldbedirectedtotherightoftheelectron"spath.(b)Thekineticenergyoftheelectronsismuchlessthantherestmassenergy,sothisisnon-prelativisticmotion.Thespeedoftheelectronisthenv=2K=m,andthemagneticforceontheelectronisFB=qvB,whereweareassumingsin=1becausetheelectronmoveshorizontallythroughamagneticeldwithaverticalcomponent.Wecanignoretheeectofthemagneticeld"shorizontalcomponentbecausetheelectronismovingparalleltothiscomponent.Theaccelerationoftheelectronbecauseofthemagneticforceisthen课后答案网rqvBqB2Ka==;mmms(1:601019C)(55:0106T)2(1:921015J)==6:271014m=s2:(9:111031kg)(9:111031kg)(c)Theelectrontravelsahorizontaldistanceof20.0cminatimeof(20:0cm)(20:0cm)9t=p=p=3:0810s:2K=m2(1:921015J)=(9:111031kg)Inthistimetheelectronisacceleratedtothesidethroughadistanceof12114292d=at=(6:2710m=s)(3:0810s)=2:98mm:22khdaw.com102若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP32-4(a)dneedstobelargerthantheturnradius,soRd;but2mK=q2B2=R2d2,orpB2mK=q2d2:(b)Outofthepage.P32-5Onlyunde ectedionsemergefromthevelocityselector,sov=E=B.Theionsarethende ectedbyB0witharadiusofcurvatureofr=mv=qB;combiningandrearranging,q=m=E=rBB0.P32-6TheionsaregivenakineticenergyK=q
V;theyarethende ectedwitharadiusofcurvaturegivenbyR2=2mK=q2B2.Butx=2R.Combinealloftheabove,andm=B2qx2=8
V:P32-7(a)StartwiththeequationinProblem6,andtakethesquarerootofbothsidestoget1pB2q2m=x;khdaw.com8
Vandthentakethederivativeofxwithrespecttom,11dmB2q2p=dx;2m8
Vandthenconsidernitedierencesinsteadofdierentialquantities,1mB2q2
m=
x;2
V(b)Inverttheaboveexpression,www.khdaw.com12
V2
x=
m;mB2qandthenputinthegivenvalues,12(7:33103V)2
x=(2:0)(1:661027kg);(35:0)(1:661027kg)(0:520T)2(1:601019C)=8:02mm:Notethatweused35.0uforthemass;ifwehadused37.0utheresultwouldhavebeencloserto课后答案网theanswerinthebackofthebook.ppP32-8(a)B=2
Vm=qr2=2(0:105MV)(238)(932MeV=c2)=2e(0:973m)2=5:23107T:(b)Thenumberofatomsinagramis6:021023=238=2:531021.Thecurrentisthen(0:090)(2:531021)(2)(1:61019C)=(3600s)=20:2mA:P32-9(a)q.(b)Regardlessofspeed,theorbitalperiodisT=2m=qB.Buttheycollidehalfwayaroundacompleteorbit,sot=m=qB.P32-10103khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP32-11(a)TheperiodofmotioncanbefoundfromthereciprocalofEq.32-12,2m2(9:111031kg)T===7:86108s:jqjB(1:601019C)(455106T)(b)Weneedtondthevelocityoftheelectronfromthekineticenergy,ppv=2K=m=2(22:5eV)(1:601019J/eV)=(9:111031kg)=2:81106m=s:Thevelocitycanwrittenintermsofcomponentswhichareparallelandperpendiculartothemagneticeld.Thenvjj=vcosandv?=vsin:Thepitchistheparalleldistancetraveledbytheelectroninonerevolution,sop=vT=(2:81106m=s)cos(65:5)(7:86108s)=9:16cm:jj(c)TheradiusofthehelicalpathisgivenbyEq.32-10,exceptthatweusetheperpendicularvelocitycomponent,sokhdaw.commv(9:111031kg)(2:81106m=s)sin(65:5)?R===3:20cmjqjB(1:601019C)(455106T)RbP32-12F~=id~lB~.d~lhastwocomponents,thoseparalleltothepath,sayd~xandthoseaperpendicular,sayd~y.ThentheintegralcanbewrittenasZbZbF~=d~xB~+d~yB~:aawww.khdaw.comRbButB~isconstant,andcanberemovedfromtheintegral.d~x=~l,avectorthatpointsfromatoRabb.d~y=0,becausethereisnonetmotionperpendicularto~l.aP32-13qvyB=Fx=mdvx=dt;qvxB=Fy=mdvy=dt.Takingthetimederivativeofthesecondexpressionandinsertingintotherstwegetmd2vyqvyB=m;qBdt2whichhassolutionvy=vsin(mt=qB),wherevisaconstant.Usingthesecondequationwendthatthereisasimilarsolutionfor课后答案网vx,exceptthatitisoutofphase,andsovx=vcos(mt=qB):Integrating,ZZqBvx=vxdt=vcos(mt=qB)=sin(mt=qB):mSimilarly,ZZqBvy=vydt=vsin(mt=qB)=cos(mt=qB):mThisistheequationofacircle.P32-14dL~=^idx+^jdy+k^dz.B~isuniform,sothattheintegralcanbewrittenasIIIIF~=i(^idx+^jdy+k^dz)B~=i^iB~dx+i^jB~dy+ik^B~dz;HHHbutsincedx=dy=dz=0,theentireexpressionvanishes.khdaw.com104若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP32-15ThecurrentpulseprovidesanimpulsewhichisequaltoZZZFdt=BiLdt=BLidt=BLq:Thisgivesaninitialvelocityofv0=BLq=m,whichwillcausetherodtohoptoaheightofh=v2=2g=B2L2q2=2m2g:0Solvingforq,mp(0:013kg)pq=2gh=2(9:8m=s2)(3:1m)=4:2C:BL(0:12T)(0:20m)P32-16P32-17Thetorqueonacurrentcarryingloopdependsontheorientationoftheloop;themaximumtorqueoccurswhentheplaneoftheloopisparalleltothemagneticeld.Inthiscasethekhdaw.commagnitudeofthetorqueisfromEq.32-34withsin=1|=NiAB:TheareaofacircularloopisA=r2whereristheradius,butsincethecircumferenceisC=2r,wecanwriteC2A=:4Thecircumferenceisnotthelengthofthewire,becausetheremaybemorethanoneturn.Instead,C=L=N,whereNisthenumberofturns.Finally,wecanwritethetorqueaswww.khdaw.comL2iL2B=NiB=;4N24NwhichisamaximumwhenNisaminimum,orN=1.P32-18dF~=idL~B~;thedirectionofdF~willbeupwardandsomewhattowardthecenter.L~andB~arearightangles,butonlytheupwardcomponentofdF~willsurvivetheintegrationasthecentralcomponentswillcanceloutbysymmetry.HenceZ课后答案网F=iBsindL=2riBsin:P32-19Thetorqueonthecylinderfromgravityisg=mgrsin;whereristheradiusofthecylinder.Thetorquefrommagnetismneedstobalancethis,somgrsin=NiABsin=Ni2rLBsin;ormg(0:262kg)(9:8m=s2)i===1:63A:2NLB2(13)(0:127m)(0:477T)105khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE33-1(a)ThemagneticeldfromamovingchargeisgivenbyEq.33-5.Iftheprotonsaremovingsidebysidethentheangleis==2,so0qvB=4r2andweareinterestedisadistancer=d.Theelectriceldatthatdistanceis1qE=;40r2whereinbothoftheaboveexpressionsqisthechargeofthesourceproton.Onthereceivingendistheotherproton,andtheforceonthatprotonisgivenbyF~=q(E~+~vB~):Thevelocityisthesameasthatoftherstproton(otherwisetheywouldn"tbemovingsidebyside.)Thisvelocityisthenperpendiculartothemagneticeld,andtheresultingdirectionforthecrossproductwillbeoppositetothedirectionofkhdaw.comE~.Thenforbalance,E=vB;1q0qv=v;40r24r212=v:00Wecansolvethiseasilyenough,andwendv3108m=s.(b)Thisisclearlyarelativisticspeed!E33-2B=i=2d=(4107T
m=A)(120A)=2(6:3m)=3:8106T:Thiswillde ectthe0www.khdaw.comcompassneedlebyasmuchasonedegree.However,thereisunlikelytobeaplaceontheEarth"ssurfacewherethemagneticeldis210T.Thiswaslikelyatypo,andshouldprobablyhavebeen21.0T.Thede ectionwouldthenbesometendegrees,andthatissignicant.E33-3B=i=2d=(4107T
m=A)(50A)=2(1:3103m)=37:7103T:0E33-4(a)i=2dB==2(8:13102m)(39:0106T)=(4107T
m=A)=15:9A:0(b)DueEast.E33-5UseB=0课后答案网i=(4107N=A2)(1:61019C)(5:61014s1)=1:2108T:2d2(0:0015m)E33-6Zero,bysymmetry.Anycontributionsfromthetopwireareexactlycanceledbycontribu-tionsfromthebottomwire.E33-7B=i=2d=(4107T
m=A)(48:8A)=2(5:2102m)=1:88104T:0F~=q~vB~.Allcasesareeitherparallelorperpendicular,soeitherF=0orF=qvB.(a)F=qvB=(1:601019C)(1:08107m=s)(1:88104T)=3:241016N.ThedirectionofF~isparalleltothecurrent.(b)F=qvB=(1:601019C)(1:08107m=s)(1:88104T)=3:241016N.ThedirectionofF~isradiallyoutwardfromthecurrent.(c)F=0.khdaw.com106若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE33-8WewantB1=B2,butwithoppositedirections.Theni1=d1=i2=d2,sinceallconstantscancelout.Theni2=(6:6A)(1:5cm)=(2:25cm)=4:4A,directedoutofthepage.E33-9Forasinglelongstraightwire,B=0i=2dbutweneedafactorof2"sincetherearetwowires,theni=dB=0:FinallydB(0:0405m)(296;T)i===30A0(4107N=A2)E33-10(a)Thesemi-circlepartcontributeshalfofEq.33-21,or0i=4R.EachlongstraightwirecontributeshalfofEq.33-13,or0i=4R.Addthethreecontributionsandgeti2(4107N=A2)(11:5A)2B=0+1=+1=1:14103T:a4R4(5:20103m)Thedirectionisoutofthepage.(b)EachlongstraightwirecontributesEq.33-13,or0i=2R.Addthetwocontributionsandgetkhdaw.comi(4107N=A2)(11:5A)B=0==8:85104T:aR(5:20103m)Thedirectionisoutofthepage.E33-11z3=iR2=2B=(4107N=A2)(320)(4:20A)(2:40102m)2=2(5:0106T)=9:730102m3.Thenz=0:46m.E33-12ThecircularpartcontributesafractionofEq.33-21,or0i=4R.EachlongstraightwirecontributeshalfofEq.33-13,or0i=4R.Addthethreecontributionsandgetwww.khdaw.com0iB=(2):4RThegoalistogetB=0thatwillhappenif=2radians.E33-13Therearefourcurrentsegmentsthatcouldcontributetothemagneticeld.Thestraightsegments,however,contributenothingbecausethestraightsegmentscarrycurrentseitherdirectlytowardordirectlyawayfromthepointP.Thatleavesthetworoundedsegments.EachcontributiontoB~canbefoundbystartingwithEq.33-21,or0i=4b.Thedirectionisoutofthepage.Thereisalsoacontributionfromthetoparc;thecalculationsarealmostidenticalexceptthat课后答案网thisispointingintothepageandr=a,so0i=4a.ThenetmagneticeldatPisthen0i11B=B1+B2=:4baE33-14ForeachstraightwiresegmentuseEq.33-12.WhenthelengthofwireisL,thedistancetothecenterisW=2;whenthelengthofwireisWthedistancetothecenterisL=2.Therearefourterms,buttheyareequalinpairs,so!0i4L4WB=p+p;4WL2=4+W2=4LL2=4+W2=4p2iL2W22iL2+W200=p+=:L2+W2WLWLkhdaw.comWL107若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE33-15Weimaginetheribbonconductortobeacollectionofthinwires,eachofthicknessdxandcarryingacurrentdi.dianddxarerelatedbydi=dx=i=w.ThecontributionofoneofthesethinwirestothemagneticeldatPisdB=0di=2x;wherexisthedistancefromthisthinwiretothepointP.Wewanttochangevariablestoxandintegrate,soZZZ0idx0idxB=dB==:2wx2wxThelimitsofintegrationarefromdtod+w,so0id+wB=ln:2wdE33-16TheeldsfromeachwireareperpendicularatP.EachcontributesanamountB0=pp0i=2d,butsincetheyareperpendicularthereisaneteldofmagnitudeB=2B02=20i=2d:pNotethata=2d,soB=0i=a.(a)B=(4107T
m=A)(115A)=(0:122m)=3:77104T:Thedirectionistotheleft.khdaw.com(b)Samenumericalresult,exceptthedirectionisup.E33-17FollowalongwithSampleProblem33-4.Reversingthedirectionofthesecondwire(sothatnowbothcurrentsaredirectedoutofthepage)willalsoreversethedirectionofB2.Then0i11B=B1B2=;2b+xbx0i(bx)(b+x)=;2b2x20ix=:x2b2www.khdaw.comE33-18(b)Bysymmetry,onlythehorizontalcomponentofB~survives,andmustpointtotheright.(a)Thehorizontalcomponentoftheeldcontributedbythetopwireisgivenby0i0ib=20ibB=sin==;2h2hh(4R2+b2)psincehisthehypotenuse,orh=R2+b2=4.Buttherearetwosuchcomponents,onefromthetopwire,andanidenticalcomponentfromthebottomwire.课后答案网E33-19(a)WecanuseEq.33-21tondthemagneticeldstrengthatthecenterofthelargeloop,i(4107T
m=A)(13A)B=0==6:8105T:2R2(0:12m)(b)ThetorqueonthesmallerloopinthecenterisgivenbyEq.32-34,~=NiA~B~;butsincethemagneticeldfromthelargeloopisperpendiculartotheplaneofthelargeloop,andtheplaneofthesmallloopisalsoperpendiculartotheplaneofthelargeloop,themagneticeldisintheplaneofthesmallloop.ThismeansthatjA~B~j=AB.Consequently,themagnitudeofthetorqueonthesmallloopis=NiAB=(50)(1:3A)()(8:2103m)2(6:8105T)=9:3107N
m:108khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE33-20(a)Therearetwocontributionstotheeld.Oneisfromthecircularloop,andisgivenby0i=2R.Theotherisfromthelongstraightwire,andisgivenby0i=2R.Thetwoeldsareoutofthepageandparallel,so0iB=(1+1=):2R(b)Thetwocomponentsarenowatrightangles,so0ipB=1+1=2:2RThedirectionisgivenbytan=1=or=18.E33-21TheforcepermeterforanypairofparallelcurrentsisgivenbyEq.33-25,F=L=i2=2d,0wheredistheseparation.Thedirectionoftheforceisalongthelineconnectingtheintersectionofthecurrentswiththeperpendicularplane.Eachcurrentexperiencesthreeforces;twoareatrightppanglesandequalinmagnitude,sojF~+F~j=L=F2+F2=L=2i2=2a.Thethirdforcep121412140pointsparalleltothissum,butkhdaw.comd=a,sotheresultantforceispF2i2i24107N=A2(18:7A)2pp=0+p0=(2+1=2)=6:06104N=m:L2a22a2(0:245m)Itisdirectedtowardthecenterofthesquare.E33-22Bysymmetryweexpectthemiddlewiretohaveanetforceofzero;thetwoontheoutsidewilleachbeattractedtowardthecenter,buttheanswerswillbesymmetricallydistributed.Forthewirewhichisthefarthestleft,Fi211114107N=A2(3:22A)2111=0+++=www.khdaw.com1+++=5:21105N=m:L2a2a3a4a2(0:083m)234Forthesecondwireover,thecontributionsfromthetwoadjacentwiresshouldcancel.ThisleavesFi2114107N=A2(3:22A)211=0++=+=2:08105N=m:L22a3a2(0:083m)23E33-23(a)Theforceontheprojectileisgivenbytheintegralof课后答案网dF~=id~lB~overthelengthoftheprojectile(whichisw).Themagneticeldstrengthcanbefoundfromaddingtogetherthecontributionsfromeachrail.IftherailsarecircularandthedistancebetweenthemissmallcomparedtothelengthofthewirewecanuseEq.33-13,0iB=;2xwherexisthedistancefromthecenteroftherail.Thereisoneproblem,however,becausethesearenotwiresofinnitelength.Sincethecurrentstopstravelingalongtherailwhenitreachestheprojectilewehavearodthatisonlyhalfofaninniterod,soweneedtomultiplybyafactorof1/2.Buttherearetworails,andeachwillcontributetotheeld,sothenetmagneticeldstrengthbetweentherailsis0i0iB=+:4x4(2r+wx)khdaw.com109若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comInthatlasttermwehaveanexpressionthatisameasureofthedistancefromthecenterofthelowerrailintermsofthedistancexfromthecenteroftheupperrail.ThemagnitudeoftheforceontheprojectileisthenZr+wF=iBdx;r2Zr+w0i11=+dx;4rx2r+wxi2r+w0=2ln4rThecurrentthroughtheprojectileisdownthepage;themagneticeldthroughtheprojectileisintothepage;sotheforceontheprojectile,accordingtoF~=i~lB~;istotheright.(b)Numericallythemagnitudeoftheforceontherailis(450103A)2(4107N=A2)(0:067m)+(0:012m)khdaw.comF=ln=6:65103N2(0:067m)Thespeedoftherailcanbefoundfromeitherenergyconservationsowerstndtheworkdoneontheprojectile,W=Fd=(6:65103N)(4:0m)=2:66104J:Thisworkresultsinachangeinthekineticenergy,sothenalspeedisppv=2K=m=2(2:66104J)=(0:010kg)=2:31103m=s:E33-24Thecontributionsfromtheleftendandtherightendofthesquarecancelout.Thisleaveswww.khdaw.comthetopandthebottom.Thenetforceisthedierence,or(4107N=A2)(28:6A)(21:8A)(0:323m)11F=;2(1:10102m)(10:30102m)=3:27103N:E33-25Themagneticforceontheupperwirenearthepointdis0iaibL0iaibL0iaibLFB=x;2(d+x)2d2d2wherexisthedistancefromtheequilibriumpoint课后答案网d.Theequilibriummagneticforceisequaltotheforceofgravitymg,soneartheequilibriumpointwecanwritexFB=mgmg:dThereisthenarestoringforceagainstsmallperturbationsofmagnitudemgx=dwhichcorrespondstoaspringconstantofk=mg=d.Thiswouldgiveafrequencyofoscillationof1p1pf=k=m=g=d;22whichisidenticaltothependulum.E33-26B=(4107N=A2)(3:58A)(1230)=(0:956m)=5:79103T:110khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE33-27ThemagneticeldinsideanidealsolenoidisgivenbyEq.33-28B=0in;wherenistheturnsperunitlength.Solvingforn,B(0:0224T)31n===1:0010=m:0i(4107N=A2)(17:8A)Thesolenoidhasalengthof1.33m,sothetotalnumberofturnsisN=nL=(1:00103=m1)(1:33m)=1330;andsinceeachturnhasalengthofonecircumference,thenthetotallengthofthewirewhichmakesupthesolenoidis(1330)(0:026m)=109m:E33-28FromthesolenoidwehaveBs=0nis=0(11500=m)(1:94mA)=0(22:3A=m):Fromthewirewehavekhdaw.com0iw0(6:3A)0Bw===(1:002A)2r2rrTheseeldsareatrightangles,soweareinterestedinwhentan(40)=B=B,orws(1:002A)2r==5:3510m:tan(40)(22:3A=m)E33-29Letu=zd.Then2Zd+L=20niRduB=;2dL=2[R2+www.khdaw.comu2]3=22d+L=20niRu=p;2R2R2+u2dL=2!0nid+L=2dL=2=pp:2R2+(d+L=2)2R2+(dL=2)2IfLismuch,muchgreaterthanRanddthenjL=2dj>>R,andRcanbeignoredinthedenominatoroftheaboveexpressions,whichthensimplifyto!课后答案网0nid+L=2dL=2B=pp:2R2+(d+L=2)2R2+(dL=2)2!0nid+L=2dL=2=pp:2(d+L=2)2(dL=2)2=0in:ItisimportantthatweconsidertherelativesizeofL=2andd!HE33-30Thenetcurrentintheloopis1i0+3i0+7i06i0=5i0.ThenB~
d~s=50i0:E33-31(a)Thepathisclockwise,soapositivecurrentisintopage.Thenetcurrentis2.0Aout,HsoB~
d~s=i=2:5106T
m.00H(b)Thenetcurrentiszero,soB~
d~s=0.khdaw.com111若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE33-32LetR0betheradiusofthewire.OnthesurfaceofthewireB0=0i=2R0.OutsidethewirewehaveB=0i=2R,thisishalfB0whenR=2R0.InsidethewirewehaveB=iR=2R2,thisishalfBwhenR=R=2.0000E33-33(a)Wedon"twanttoreinventthewheel.TheanswerisfoundfromEq.33-34,exceptitlookslike0irB=:2c2(b)IntheregionbetweenthewiresthemagneticeldlookslikeEq.33-13,0iB=:2rThisisderivedontherighthandsideofpage761.H(c)Ampere"slaw(Eq.33-29)isB~
d~s=0i;whereiisthecurrentenclosed.OurAmperianloopwillstillbeacirclecenteredontheaxisoftheproblem,sothelefthandsideoftheaboveequationwillreduceto2khdaw.comrB,justlikeinEq.33-32.Therighthandside,however,dependsonthenetcurrentenclosedwhichisthecurrentiinthecenterwireminusthefractionofthecurrentenclosedintheouterconductor.Thecrosssectionalareaoftheouterconductoris(a2b2);sothefractionoftheoutercurrentenclosedintheAmperianloopis(r2b2)r2b2i=i:(a2b2)a2b2Thenetcurrentintheloopisthenr2b2a2r2ii=i;a2b2www.khdaw.coma2b2sothemagneticeldinthisregionisia2r20B=:2ra2b2(d)Thispartiseasysincethenetcurrentiszero;consequentlyB=0.HE33-34(a)Ampere"slaw(Eq.33-29)isB~
d~s=0i;whereiisthecurrentenclosed.OurAmperianloopwillstillbeacirclecenteredontheaxisoftheproblem,sothelefthandsideoftheaboveequationwillreduceto2课后答案网rB,justlikeinEq.33-32.Therighthandside,however,dependsonthenetcurrentenclosedwhichisthefractionofthecurrentenclosedintheconductor.Thecrosssectionalareaoftheconductoris(a2b2);sothefractionofthecurrentenclosedintheAmperianloopis(r2b2)r2b2i=i:(a2b2)a2b2Themagneticeldinthisregionisir2b20B=:2ra2b2(b)Ifr=a,thenia2b2i00B==;2aa2b22awhichiswhatweexpect.112khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comIfr=b,thenib2b20B==0;2ba2b2whichiswhatweexpect.Ifb=0,thenir202ir00B==2ra2022a2whichiswhatIexpected.E33-35Themagnitudeofthemagneticeldduetothecylinderwillbezeroatthecenterofthecylinderand0i0=2(2R)atpointP.Themagnitudeofthemagneticeldeldduetothewirewillbe0i=2(3R)atthecenterofthecylinderbut0i=2RatP.Inorderfortheneteldtohavedierentdirectionsinthetwolocationsthecurrentsinthewireandpipemustbeindierentdirection.Theneteldatthecenterofthepipeis0i=2(3R),whilethatatPisthen0i0=2(2R)0i=2R.Settheseequalandsolvefori;khdaw.comi=3=i0=2i;ori=3i0=8.E33-36(a)B=(4107N=A2)(0:813A)(535)=2(0:162m)=5:37104T.(b)B=(4107N=A2)(0:813A)(535)=2(0:162m+0:052m)=4:07104T.E33-37(a)Apositiveparticlewouldexperienceamagneticforcedirectedtotherightforamagneticeldoutofthepage.Thisparticleisgoingtheotherway,soitmustbenegative.(b)ThemagneticeldofatoroidisgivenbyEq.33-36,www.khdaw.com0iNB=;2rwhiletheradiusofcurvatureofachargedparticleinamagneticeldisgivenbyEq.32-10mvR=:jqjBWeusetheRtodistinguishitfromr.Combining,2mvR=r;0iNjqjsothetworadiiaredirectlyproportional.Thismeans课后答案网R=(11cm)=(110cm)=(125cm);soR=9:7cm.P33-1TheeldfromonecoilisgivenbyEq.33-19iR20B=:2(R2+z2)3=2ThereareNturnsinthecoil,soweneedafactorofN.TherearetwocoilsandweareinterestedinthemagneticeldatP,adistanceR=2fromeachcoil.Themagneticeldstrengthwillbetwicetheaboveexpressionbutwithz=R=2,so2NiR28Ni00B==:2(R2+(R=2)2)3=2(5)3=2Rkhdaw.com113若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP33-2(a)ChangethelimitsofintegrationthatleadtoEq.33-12:ZL0iddzB=;40(z2+d2)3=2L0idz=4(z2+d2)1=2;00idL=:4(L2+d2)1=2(b)TheangleinEq.33-11wouldalwaysbe0,sosin=0,andthereforeB=0.P33-3ThisproblemistheallimportantderivationoftheHelmholtzcoilproperties.(a)ThemagneticeldfromonecoilisNiR20B1=:khdaw.com2(R2+z2)3=2Themagneticeldfromtheothercoil,locatedadistancesaway,butforpointsmeasuredfromtherstcoil,isNiR20B2=:2(R2+(zs)2)3=2Themagneticeldontheaxisbetweenthecoilsisthesum,NiR2NiR200B=+:2(R2+z2)3=22(R2+(zs)2)3=2Takethederivativewithrespecttozandgetwww.khdaw.comdB3NiR23NiR200=z(zs):dz2(R2+z2)5=22(R2+(zs)2)5=2Atz=s=2thisexpressionvanishes!Weexpectthisbysymmetry,becausethemagneticeldwillbestrongestintheplaneofeithercoil,sothemid-pointshouldbealocalminimum.(b)Takethederivativeagainandd2B3NiR215NiR2002=+zdz22(R2+z2)5=22(R2+z2)5=2课后答案网3NiR215NiR2002+(zs):2(R2+(zs)2)5=22(R2+(zs)2)5=2Wecouldtryandsimplifythis,butwedon"treallywantto;weinsteadwanttosetitequaltozero,thenletz=s=2,andthensolvefors.Thesecondderivativewillequalzerowhen3(R2+z2)+15z23(R2+(zs)2)+15(zs)2=0;andisz=s=2thisexpressionwillsimplifyto30(s=2)2=6(R2+(s=2)2);4(s=2)2=R2;s=R:114khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP33-4(a)Eachofthesideofthesquareisastraightwiresegmentoflengthawhichcontributesaeldstrengthof0iaB=p;4ra2=4+r2whereristhedistancetothepointontheaxisoftheloop,sopr=a2=4+z2:Thiseldisnotparalleltothezaxis;thezcomponentisBz=B(a=2)=r.Therearefourofthesecontributions.Theoaxiscomponentscancel.Consequently,theeldforthesquareis0iaa=2B=4p;4ra2=4+r2ria20=p;2r2a2=4+r2ia2khdaw.com0=p;2(a2=4+z2)a2=2+z24ia20=p:(a2+4z2)2a2+4z2(b)Whenz=0thisreducesto4ia24i00B=p=p:(a2)2a22aP33-5(a)Thepolygonhasnsides.Aperpendicularbisectorofeachsidecanbedrawntothewww.khdaw.comcenterandhaslengthxwherex=a=cos(=n):EachsidehasalengthL=2asin(=n).Eachofthesideofthepolygonisastraightwiresegmentwhichcontributesaeldstrengthof0iLB=p;4xL2=4+x2Thiseldisparalleltothezaxis.Therearenofthesecontributions.Theoaxiscomponentscancel.Consequently,theeldforthepolygon0iLB=np;课后答案网4xL2=4+x20i2=nptan(=n);4L2=4+x20i1=ntan(=n);2asince(L=2)2+x2=a2.(b)Evaluate:limntan(=n)=limnsin(=n)n=n=:n!1n!1Thentheanswertopart(a)simpliesto0iB=:2a115khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP33-6ForasquareloopofwirewehavefournitelengthsegmentseachcontributingatermwhichlookslikeEq.33-12,exceptthatLisreplacedbyL=4anddisreplacedbyL=8.Thenatthecenter,0iL=4160iB=4p=p:4L=8L2=64+L2=642LForacircularloopR=L=2so0i0B==:2RLpSince16=2>,thesquarewins.Butonlybysome7%!P33-7WewanttousethedierentialexpressioninEq.33-11,exceptthatthelimitsofintegra-tionaregoingtobedierent.Wehavefourwiresegments.Fromthetopsegment,3L=40idB1=p;4z2+d2L=4khdaw.com!0i3L=4L=4=pp:4d(3L=4)2+d2(L=4)2+d2Forthetopsegmentd=L=4,sothissimpliesevenfurtherto0ippB1=2(35+5):10LThebottomsegmenthasthesameintegral,butd=3L=4,so0ipwww.khdaw.compB3=2(5+5):30LBysymmetry,thecontributionfromtherighthandsideisthesameasthebottom,soB2=B3,andthecontributionfromthelefthandsideisthesameasthatfromthetop,soB4=B1.Addingallfourterms,20ippppB=32(35+5)+2(5+5);30L20ipp=(22+10):3LP33-8Assumeacurrentringhasaradius课后答案网randawidthdr,thechargeontheringisdq=2rdr,where=q=R2.Thecurrentintheringisdi=!dq=2=!rdr.TheringcontributesaelddB=0di=2r.Integrateoveralltherings:ZRB=0!rdr=2r=0!R=2=!q=2R:0P33-9B=0inandmv=qBr.Combine,andmv(5:11105eV=c2)(0:046c)i===0:271A:0qrn(4107N=A2)e(0:023m)(10000=m)116khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP33-10ThisshapeisatrianglewithareaA=(4d)(3d)=2=6d2.Theenclosedcurrentistheni=jA=(15A=m2)6(0:23m)2=4:76AThelineintegralistheni=6:0106T
m:0P33-11AssumethatBdoesvaryasthepictureimplies.Thenthelineintegralalongthepathshownmustbenonzero,sinceB~
~lontherightisnotzero,whileitisalongthethreeothersides.HHenceB~
d~lisnonzero,implyingsomecurrentpassesthroughthedottedpath.Butitdoesn"t,soB~cannothaveanabruptchange.P33-12(a)SketchanAmperianloopwhichisarectanglewhichenclosedNwires,hasaverticalHsideswithheighth,andhorizontalsideswithlengthL.ThenB~
d~l=0Ni:Evaluatetheintegralalongthefoursides.Theverticalsidecontributenothing,sinceB~isperpendicularto~h,andthenB~
~h=0.Iftheintegralisperformedinacounterclockwisedirection(itmust,sincethesenseofintegrationwasdeterminedbyassumingthecurrentispositive),wegetkhdaw.comBLforeachhorizontalsection.Then0iN1B==0in:2L2(b)Asa!1thentan1(a=2R)!=2.ThenB!i=2a.Ifweassumethatiismadeupof0severalwires,eachwithcurrenti0,theni=a=i0n.P33-13ApplyAmpere"slawwithanAmperianloopthatisacirclecenteredonthecenterofthewire.ThenIIIB~
d~s=Bds=Bds=2rB;becauseB~istangenttothepathandBisuniformalongthepathbysymmetry.Thecurrentwww.khdaw.comenclosedisZienc=jdA:Thisintegralisbestdoneinpolarcoordinates,sodA=(dr)(rd),andthenZrZ2ienc=(j0r=a)rdrd;00Zr=2j=ar2dr;00课后答案网2j03=r:3aWhenr=athecurrentenclosedisi,so2ja23i0i=orj0=:32a2ThemagneticeldstrengthinsidethewireisfoundbygluingtogetherthetwopartsofAmpere"slaw,2j032rB=0r;3ajr200B=;3air20=:2a3khdaw.com117若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP33-14(a)AccordingtoEq.33-34,themagneticeldinsidethewirewithoutaholehasmagnitudeB=ir=2R2=jr=2andisdirectedradially.Ifwesuperimposeasecondcurrenttocreatethe00hole,theadditionaleldatthecenteroftheholeiszero,soB=0jb=2.Butthecurrentintheremainingwireisi=jA=j(R2a2);so0ibB=:2(R2a2)khdaw.comwww.khdaw.com课后答案网118khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE34-1=B~
A~=(42106T)(2:5m2)cos(57)=5:7105Wb:BE34-2jEj=jd=dtj=AdB=dt=(=4)(0:112m)2(0:157T=s)=1:55mV:BE34-3(a)ThemagnitudeoftheemfinducedinaloopisgivenbyEq.34-4,dBjEj=N;dt=N(12mWb=s2)t+(7mWb=s)Thereisonlyoneloop,andwewanttoevaluatethisexpressionfort=2:0s,sojEj=(1)(12mWb=s2)(2:0s)+(7mWb=s)=31mV:(b)Thispartisn"tharder.Themagnetic uxthroughtheloopisincreasingwhent=2:0s.Theinducedcurrentneedsto owinsuchadirectiontocreateasecondmagneticeldtoopposethisincrease.Theoriginalmagneticeldisoutofthepageandweopposetheincreasebypointingtheotherway,sothekhdaw.comsecondeldwillpointintothepage(insidetheloop).Bytherighthandrulethismeanstheinducedcurrentisclockwisethroughtheloop,ortotheleftthroughtheresistor.E34-4E=dB=dt=AdB=dt.(a)E=(0:16m)2(0:5T)=(2s)=2:0102V:(b)E=(0:16m)2(0:0T)=(2s)=0:0102V:(c)E=(0:16m)2(0:5T)=(4s)=1:0102V:E34-5(a)R=L=A=(1:69108
m)[()(0:104m)]=[(=4)(2:50103m)2]=1:12103:(b)E=iR=(9:66A)(1:12103)=1:0810www.khdaw.com2V.TherequireddB=dtisthengivenbydBE22==(1:0810V)=(=4)(0:104m)=1:27T=s:dtAE34-6E=A
B=
t=AB=
t.ThepowerisP=iE=E2=R.TheenergydissipatedisE2
tA2B2E=P
t==:RR
tE34-7(a)Wecouldre-derivethestepsinthesampleproblem,orwecouldstartwiththeendresult.We"llstartwiththeresult,课后答案网diE=NA0n;dtexceptthatwehavegoneaheadandusedthederivativeinsteadofthe
.Therateofchangeinthecurrentisdi2=(3:0A=s)+(1:0A=s)t;dtsotheinducedemfis
E=(130)(3:46104m2)(4107Tm=A)(2:2104=m)(3:0A=s)+(2:0A=s2)t;=(3:73103V)+(2:48103V=s)t:(b)Whent=2:0stheinducedemfis8:69103V,sotheinducedcurrentisi=(8:69103V)=(0:15)=5:8102A:119khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE34-8(a)i=E=R=NAdB=dt.NotethatAreferstotheareaenclosedbytheoutersolenoidwhereBisnon-zero.ThisAisthenthecrosssectionalareaoftheinnersolenoid!Then1di(120)(=4)(0:032m)2(4107N=A2)(220102=m)(1:5A)i=NAn==4:7103A:0Rdt(5:3)(0:16s)E34-9P=Ei=E2=R=(AdB=dt)2=(L=a),whereAistheareaoftheloopandaisthecrosssectionalareaofthewire.Buta=d2=4andA=L2=4,soL3d2dB2(0:525m)3(1:1103m)2P==(9:82103T=s)2=4:97106W:64dt64(1:69108
m)E34-10=BA=B(2:3m)2=2.E=d=dt=AdB=dt,orBBB(2:3m)2EB=[(0:87T=s)]=2:30V;khdaw.com2soE=(2:0V)+(2:3V)=4:3V.E34-11(a)Theinducedemf,asafunctionoftime,isgivenbyEq.34-5,E(t)=dB(t)=dtThisemfdrivesacurrentthroughtheloopwhichobeysE(t)=i(t)RCombining,1dB(t)i(t)=:RdtSincethecurrentisdenedbyi=dq=dtwecanwritedq(t)www.khdaw.com1dB(t)=:dtRdtFactoroutthedtfrombothsides,andthenintegrate:1dq(t)=dB(t);RZZ1dq(t)=dB(t);R1q(t)q(0)=(B(0)B(t))R(b)No.Theinducedcurrentcouldhaveincreasedfromzerotosomepositivevalue,thendecreased课后答案网tozeroandbecamenegative,sothatthenetchargeto owthroughtheresistorwaszero.Thiswouldbelikesloshingthechargebackandforththroughtheloop.E34-12
PhiB=2B=2NBA.Thenthechargeto owthroughisq=2(125)(1:57T)(12:2104m2)=(13:3)=3:60102C:E34-13Thepartabovethelongstraightwire(adistancebaaboveit)cancelsoutcontributionsbelowthewire(adistancebabeneathit).The uxthroughtheloopisthenZa0i0ibaB=bdr=ln:2ab2r22ab120khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comTheemfintheloopisthendB0ba2E==ln[2(4:5A=s)t(10A=s)]:dt22abEvaluating,4107N=A2(0:16m)(0:12m)E=ln[2(4:5A=s2)(3:0s)(10A=s)]=2:20107V:22(0:12m)(0:16m)E34-14UseEq.34-6:E=BDv=(55106T)(1:10m)(25m=s)=1:5103V:E34-15Iftheangledoesn"tvarythenthe ux,givenby=B~
A~isconstant,sothereisnoemf.khdaw.comE34-16(a)UseEq.34-6:E=BDv=(1:18T)(0:108m)(4:86m=s)=0:619V:(b)i=(0:619V)=(0:415)=1:49A:(c)P=(0:619V)(1:49A)=0:922W.(d)F=iLB=(1:49A)(0:108m)(1:18T)=0:190N.(e)P=Fv=(0:190V)(4:86m=s)=0:923W.E34-17Themagneticeldisoutofthepage,andthecurrentthroughtherodisdown.ThenEq.32-26F~=iL~B~showsthatthedirectionofthemagneticforceistotheright;furthermore,sinceeverythingisperpendiculartoeverythingelse,wecangetridofthevectornatureoftheproblemwww.khdaw.comandwriteF=iLB.Newton"ssecondlawgivesF=ma,andtheaccelerationofanobjectfromrestresultsinavelocitygivenbyv=at.Combining,iLBv(t)=t:mE34-18(b)Therodwillaccelerateaslongasthereisanetforceonit.ThisnetforcecomesfromF=iLB.ThecurrentisgivenbyiR=EBLv,soasvincreasesidecreases.Wheni=0therodstopsacceleratingandassumesaterminalvelocity.(a)E=BLvwillgivetheterminalvelocity.Inthiscase,v=E=BL.课后答案网E34-19E34-20Theaccelerationisa=R!2;sinceE=B!R2=2,wecannda=4E2=B2R3=4(1:4V)2=(1:2T)2(5:3102m)3=3:7104m=s2:E34-21WewillusetheresultsofExercise11thatwereworkedoutabove.Allweneedtodoisndtheinitial ux; ippingthecoilup-side-downwillsimplychangethesignofthe ux.So(0)=B~
A~=(59T)()(0:13m)2sin(20)=1:1106Wb:B121khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comThenusingtheresultsofExercise11wehaveNq=(B(0)B(t));R95066=((1:110Wb)(1:110Wb));85=2:5105C:E34-22(a)The uxthroughtheloopisZvtZa+L0i0ivta+LB=dxdr=ln:0a2r2aTheemfisthendB0iva+LE==ln:dt2aPuttinginthenumbers,khdaw.com(4107N=A2)(110A)(4:86m=s)(0:0102m)+(0:0983m)E=ln=2:53104V:2(0:0102m)(b)i=E=R=(2:53104V)=(0:415)=6:10104A.(c)P=i2R=(6:10104A)2(0:415)=1:54107W.R(d)F=Bildl,orZa+L0i0ia+LF=ildr=illn:a2r2aPuttinginthenumbers,(4107N=A2)(110A)www.khdaw.com(0:0102m)+(0:0983m)F=(6:10104A)ln=3:17108N:2(0:0102m)(e)P=Fv=(3:17108N)(4:86m=s)=1:54107W.E34-23(a)Startingfromthebeginning,Eq.33-13gives0iB=:2yThe uxthroughtheloopisgivenby课后答案网ZB=B~
dA~;butsincethemagneticeldfromthelongstraightwiregoesthroughtheloopperpendiculartotheplaneoftheloopthisexpressionsimpliestoascalarintegral.Theloopisarectangular,sousedA=dxdy,andletxbeparalleltothelongstraightwire.Combiningtheabove,ZD+bZa0iB=dxdy;D02yZD+b0idy=a;2Dy0iD+b=aln2Dkhdaw.com122若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com(b)The uxthroughtheloopisafunctionofthedistanceDfromthewire.Iftheloopmovesawayfromthewireataconstantspeedv,thenthedistanceDvariesasvt.TheinducedemfisthendBE=;dt0ib=a:2t(vt+b)ThecurrentwillbethisemfdividedbytheresistanceR.Theback-of-the-book"answerissomewhatdierent;theanswerisexpressedintermsofDinsteadift.Thetwoanswersareotherwiseidentical.E34-24(a)TheareaofthetriangleisA=x2tan=2.Inthiscasex=vt,so=B(vt)2tan=2;BandthenE=2Bv2ttan=2;khdaw.com(b)t=E=2Bv2tan=2;so(56:8V)t==2:08s:2(0:352T)(5:21m=s)2tan(55)E34-25E=NBA!,so(24V)!==39:4rad=s:(97)(0:33T)(0:0190m2)That"s6.3rev/second.www.khdaw.comE34-26(a)Thefrequencyoftheemfisthesameasthefrequencyofrotation,f.(b)The uxchangesbyBA=Ba2duringahalfarevolution.Thisisasinusoidalchange,sotheamplitudeofthesinusoidalvariationintheemfisE=!=2.ThenE=B2a2f.BE34-27WecanuseEq.34-10;theemfisE=BA!sin!t;Thiswillbeamaximumwhensin!t=1.Theangularfrequency,!isequalto!=(1000)(2)=(60)rad/s=105rad/sThemaximumemfisthenE=(3:5T)[(100)(0:5m)(0:3m)](105rad/s)=5:5kV:E34-28(a)Theamplitudeoftheemfis课后答案网E=BA!,soA=E=2fB=(150V)=2(60=s)(0:50T)=0:798m2:(b)Dividethepreviousresultby100.A=79:8cm2.E34-29dB=dt=AdB=dt=A(8:50mT=s).(a)ForthispathIE~
d~s=d=dt=(0:212m)2(8:50mT=s)=1:20mV:B(b)ForthispathIE~
d~s=d=dt=(0:323m)2(8:50mT=s)=2:79mV:Bkhdaw.com123若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com(c)ForthispathIE~
d~s=d=dt=(0:323m)2(8:50mT=s)(0:323m)2(8:50mT=s)=1:59mV:BHE34-30dB=dt=AdB=dt=A(6:51mT=s),whileE~
d~s=2rE.(a)Thepathofintegrationisinsidethesolenoid,sor2(6:51mT=s)(0:022m)(6:51mT=s)E===7:16105V=m:2r2(b)Thepathofintegrationisoutsidethesolenoid,sor2(6:51mT=s)(0:063m)2(6:51mT=s)E===1:58104V=m2R2(0:082m)E34-31khdaw.comTheinducedelectriceldcanbefoundfromapplyingEq.34-13,IE~
d~s=dB:dtWestartwiththelefthandsideofthisexpression.Theproblemhascylindricalsymmetry,sotheinducedelectriceldlinesshouldbecirclescenteredontheaxisofthecylindricalvolume.Ifwechoosethepathofintegrationtoliealonganelectriceldline,thentheelectriceldE~willbeparalleltod~s,andEwillbeuniformalongthispath,soIIIE~
d~s=Eds=Eds=2rE;www.khdaw.comwhereristheradiusofthecircularpath.Nowfortherighthandside.The uxiscontainedinthepathofintegration,so=Br2.BAllofthetimedependenceofthe uxiscontainedinB,sowecanimmediatelywrite2dBrdB2rE=rorE=:dt2dtWhatdoesthenegativesignmean?Thepathofintegrationischosensothatifourrighthandngerscurlaroundthepathourthumbgivesthedirectionofthemagneticeldwhichcutsthroughthepath.Sincetheeldpointsintothepageapositiveelectriceldwouldhaveaclockwiseorientation.SinceBisdecreasingthederivativeisnegative,butwegetanothernegativefromtheequation课后答案网above,sotheelectriceldhasapositivedirection.Nowforthemagnitude.E=(4:82102m)(10:7103T=s)=2=2:58104N=C:Theaccelerationoftheelectronateitheraorcthenhasmagnitudea=Eq=m=(2:58104N=C)(1:601019C)=(9:111031kg)=4:53107m=s2:P34-1Theinducedcurrentisgivenbyi=E=R.TheresistanceoftheloopisgivenbyR=L=A,whereAisthecrosssectionalarea.Combining,andwritingintermsoftheradiusofthewire,wehaver2Ei=:Lkhdaw.com124若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comThelengthofthewireisrelatedtotheradiusofthewirebecausewehaveaxedmass.Thetotalvolumeofthewireisr2L,andthisisrelatedtothemassanddensitybym=r2L.EliminatingrwehavemEi=:L2Thelengthofthewireloopisthesameasthecircumference,whichisrelatedtotheradiusRoftheloopbyL=2R.Theemfisrelatedtothechanging uxbyE=dB=dt,butiftheshapeoftheloopisxedthisbecomesE=AdB=dt.Combiningallofthis,mAdBi=:(2R)2dtWedroppedthenegativesignbecauseweareonlyinterestedinabsolutevalueshere.NowA=R2,sothisexpressioncanalsobewrittenasmR2dBmdBi==:(2R)2dt4dtP34-2khdaw.comForthelowersurfaceB~
A~=(76103T)(=2)(0:037m)2cos(62)=7:67105Wb:FortheuppersurfaceB~
A~=(76103T)(=2)(0:037m)2cos(28)=1:44104Wb:.TheinducedemfisthenE=(7:67105Wb+1:44104Wb)=(4:5103s)=4:9102V:P34-3(a)Weareonlyinterestedintheportionoftheringintheyzplane.ThenE=(3:32103T=s)(=4)(0:104m)2=2:82105V.(b)Fromctob.PointyourrightthumbalongxtoopposetheincreasingB~eld.Yourrightngerswillcurlfromctob.2www.khdaw.comP34-4E/NA,butA=randN2r=L,soE/1=N.Thismeansuseonlyonelooptomaximizetheemf.P34-5Thisisaintegralbestperformedinrectangularcoordinates,thendA=(dx)(dy).Themagneticeldisperpendiculartothesurfacearea,soB~
dA~=BdA.The uxisthenZZB=B~
dA~=BdA;ZaZa=(4T=m
s2)t2ydydx;00课后答案网2212=(4T=m
s)taa;2=(2T=m
s2)a3t2:Buta=2:0cm,sothisbecomes=(2T=m
s2)(0:02m)3t2=(1:6105Wb=s2)t2:BTheemfaroundthesquareisgivenbydB52E==(3:210Wb=s)t;dtandatt=2:5sthisis8:0105V.Sincethemagneticeldisdirectedoutofthepage,apositiveemfwouldbecounterclockwise(holdyourrightthumbinthedirectionofthemagneticeldandyourngerswillgiveacounterclockwisesensearoundtheloop).Buttheanswerwasnegative,sotheemfmustbeclockwise.khdaw.com125若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP34-6(a)Farfromtheplaneofthelargeloopwecanapproximatethelargeloopasadipole,andtheniR20B=:2x3The uxthroughthesmallloopistheni2r2R220B=rB=:2x3(b)E=dB=dt,so3i2r2R20E=v:2x4(c)Anti-clockwisewhenviewedfromabove.P34-7Themagneticeldisperpendiculartothesurfacearea,soB~
dA~=BdA.The uxisthenZZkhdaw.comB=B~
dA~=BdA=BA;sincethemagneticeldisuniform.TheareaisA=r2,whereristheradiusoftheloop.TheinducedemfisdBdrE==2rB:dtdtItisgiventhatB=0:785T,r=1:23m,anddr=dt=7:50102m=s.Thenegativesignindicateadecreasingradius.ThenE=2(1:23m)(0:785T)(7:50102m=s)=0:455V:www.khdaw.comP34-8(a)dB=dt=BdA=dt,butdA=dtis
A=
t,where
Aistheareasweptoutduringonerotationand
t=1=f.ButtheareasweptoutisR2,sodB2jEj==fBR:dt(b)IftheoutputcurrentisithenthepowerisP=Ei.ButP=!=2f,soP2==iBR=2:2f课后答案网P34-9(a)E=dB=dt,andB=B~
A~,soE=BLvcos:ThecomponentoftheforceofgravityontherodwhichpullsitdowntheinclineisFG=mgsin.ThecomponentofthemagneticforceontherodwhichpullsituptheinclineisFB=BiLcos.Equating,BiLcos=mgsin;andsinceE=iR,EmgRsinv==:BLcosB2L2cos2(b)P=iE=E2=R=B2L2v2cos2=R=mgvsin.Thisisidenticaltotherateofchangeofgravitationalpotentialenergy.khdaw.com126若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP34-10Letthecrosssectionofthewirebea.(a)R=L=a=(r+2r)=a;withnumbers,R=(3:4103)(2+):(b)=Br2=2;withnumbers,B=(4:32103Wb):B(c)i=E=R=B!r2=2R=Ba!r=2(+2),orBatri=:(t2+4)Takethederivativeandsetitequaltozero,4at20=;khdaw.com(t2+4)2soat2=4,or=1at2=2rad.p2(d)!=2,soq(0:15T)(1:2106m2)2(12rad/s2)(2rad)(0:24m)i==2:2A:(1:7108
m)(6rad)P34-11Itdoessayapproximate,sowewillbemakingsomeratherboldassumptionshere.Firstwewillndanexpressionfortheemf.SinceBisconstant,theemfmustbecausedbyachangeinwww.khdaw.comthearea;inthiscaseashiftinposition.ThesmallsquarewhereB6=0hasawidthaandsweepsaroundthediskwithaspeedr!.AnapproximationfortheemfisthenE=Bar!:Thisemfcausesacurrent.Wedon"tknowexactlywherethecurrent ows,butwecanreasonablyassumethatitoccursnearthelocationofthemagneticeld.Letusassumethatitisconstrainedtothatregionofthedisk.Theresistanceofthisportionofthediskistheapproximately1L1a1R===;Aattwherewehaveassumedthatthecurrentis owingradiallywhendeningthecrosssectionalareaoftheresistor".Theinducedcurrentisthen(ontheorderof)课后答案网EBar!==Bar!t:R1=(t)ThiscurrentexperiencesabreakingforceaccordingtoF=BIl,soF=B2a2r!t;wherelisthelengththroughwhichthecurrent ows,whichisa.Finallywecanndthetorquefrom=rF,and=B2a2r2!t:127khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP34-12TheinducedelectriceldintheringisgivenbyEq.34-11:2RE=jdB=dtj.Thiselectriceldwillresultinaforceonthefreechargecarries(electrons?),givenbyF=Ee.Theaccelerationoftheelectronsisthena=Ee=me.ThenedBa=:2RmedtIntegratebothsideswithrespecttotimetondthespeedoftheelectrons.ZZedBadt=dt;2RmedtZedBv=2Rme;e=
B:2RmeThecurrentdensityisgivenbykhdaw.comj=nev,andthecurrentbyiA=ia2.Combining,ne2a2i=
PhiB:2RmeActually,itshouldbepointedoutthat
PhiBreferstothechangein uxfromexternalsources.Thecurrentinducedinthewirewillproducea uxwhichwillexactlyoset
PhiBsothatthenet uxthroughthesuperconductingringisxedatthevaluepresentwhentheringbecamesuperconducting.P34-13AssumethatEdoesvaryasthepictureimplies.Thenthelineintegralalongthepathshownmustbenonzero,sinceE~
~lontherightisnotzero,whileitisalongthethreeothersides.HHenceE~
d~lisnonzero,implyingachangeinthemagnetic uxthroughthedottedpath.Butitdoesn"t,soE~cannothaveanabruptchange.www.khdaw.comP34-14Theelectriceldadistancerfromthecenterisgivenbyr2dB=dTrdBE==:2r2dtThiseldisdirectedperpendiculartotheradiallines.Denehtobethedistancefromthecenterofthecircletothecenteroftherod,andevaluateRE=E~
d~s,Z课后答案网dBrhE=dx;dt2rdBL=h:dt2Buth2=R2(L=2)2,sodBLpE=R2(L=2)2:dt2P34-15(a)=r2B,soBavE(0:32m)E==2(0:28T)(120)=34V=m:2r2(b)a=F=m=Eq=m=(33:8V=m)(1:61019C)=(9:11031kg)=6:01012m=s2.128khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE35-1IftheEarth"smagneticdipolemomentwereproducedbyasinglecurrentaroundthecore,thenthatcurrentwouldbe(8:01022J=T)i===2:1109AA(3:5106m)2E35-2(a)i==A=(2:33A
m2)=(160)(0:0193m)2=12:4A.(b)=B=(2:33A
m2)(0:0346T)=8:06102N
m:E35-3(a)Usingtherighthandruleaclockwisecurrentwouldgenerateamagneticmomentwhichwouldbeintothepage.Bothcurrentsareclockwise,soaddthemoments:=(7:00A)(0:20m)2+(7:00A)(0:30m)2=2:86A
m2:(b)Reversingthecurrentreversesthemoment,sokhdaw.com=(7:00A)(0:20m)2(7:00A)(0:30m)2=1:10A
m2:E35-4(a)=iA=(2:58A)(0:16m)2=0:207A
m2.(b)=Bsin=(0:207A
m2)(1:20T)sin(41)=0:163N
m:E35-5(a)TheresultfromProblem33-4forasquareloopofwirewas4ia20B(z)=:(4z2+a2)(4z2+2a2)1=2Forzmuch,muchlargerthanawecanignoreanyatermswhichareaddedtoorsubtractedfromzterms.Thismeansthatwww.khdaw.com4z2+a2!4z2and(4z2+2a2)1=2!2z;butwecan"tignorethea2inthenumerator.TheexpressionforBthensimpliestoia20B(z)=;2z3whichcertainlylookslikeEq.35-4.(b)Wecanrearrangethisexpressionandget课后答案网02B(z)=ia;2z3whereitisratherevidentthatia2mustcorrespondto~,thedipolemoment,inEq.35-4.Sothatmustbetheanswer.E35-6=iA=(0:2A)(0:08m)2=4:02103A
m2;~=n^.(a)Forthetorque,~=~B~=(9:65104N
m)^i+(7:24104N
m)^j+(8:08104N
m)k^:(b)Forthemagneticpotentialenergy,U=~
B~=[(0:60)(0:25T)]=0:603103J:129khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE35-7=iA=i(a2+b2=2)=i(a2+b2)=2:E35-8IfthedistancetoPisverylargecomparedtoaorbwecanwritetheLawofBiotandSavartasB~=0i~s~r:4r3~sisperpendicularto~rfortheleftandrightsides,sotheleftsidecontributes0ibB1=;4(x+a=2)2andtherightsidecontributes0ibB3=:4(xa=2)2Thetopandbottomsideseachcontributeanequalamount0iasin0ia(b=2)khdaw.comB2=B4=4x2+b2=44x3:Addthefourterms,expandthedenominators,andkeeponlytermsinx3,0iab0B==:4x34x3Thenegativesignindicatesthatitisintothepage.E35-9(a)Theelectriceldatthisdistancefromtheprotoniswww.khdaw.com191(1:6010C)11E==5:1410N=C:4(8:851012C2=N
m2)(5:291011m)2(b)Themagneticeldatthisfromtheprotonisgivenbythedipoleapproximation,0B(z)=;2z3(4107N=A2)(1:411026A=m2)=;2(5:291011m)3课后答案网=1:90102TE35-101:50gofwaterhas(2)(6:021023)(1:5)=(18)=1:001023hydrogennuclei.Ifallarealignedthenetmagneticmomentwouldbe=(1:001023)(1:411026J=T)=1:41103J=T.Theeldstrengthisthen(1:41103J=T)B=0=(1:00107N=A2)=9:31013T:4x3(5:33m)3E35-11(a)Thereiseectivelyacurrentofi=fq=q!=2.Thedipolemomentisthen=iA=(q!=2)(r2)=1q!r2:2(b)Therotationalinertiaoftheringismr2soL=I!=mr2!.Then(1=2)q!r2q==:Lmr2!2mkhdaw.com130若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE35-12Themassofthebaris32m=V=(7:87g/cm)(4:86cm)(1:31cm)=50:1g:ThenumberofatomsinthebarisN=(6:021023)(50:1g)=(55:8g)=5:411023:Thedipolemomentofthebaristhen=(5:411023)(2:22)(9:271024J=T)=11:6J=T:(b)Thetorqueonthemagnetis=(11:6J=T)(1:53T)=17:7N
m:E35-13Themagneticdipolemomentisgivenby=MV,Eq.35-13.Thenkhdaw.com=(5;300A=m)(0:048m)(0:0055m)2=0:024A
m2:E35-14(a)TheoriginaleldisB0=0in.TheeldwillincreasetoB=mB0,sotheincreaseis
B=(1)in=(3:3104)(4107N=A2)(1:3A)(1600=m)=8:6107T:10(b)M=(1)B==(1)in=(3:3104)(1:3A)(1600=m)=0:69A=m.1001E35-15Theenergyto ipthedipolesisgivenbyU=2B.Thetemperatureisthen2B4(1:21023J=T)(0:5T)T===0:58K:3k=23(1:38www.khdaw.com1023J=K)E35-16TheCurietemperatureofironis770C,whichis750Chigherthanthesurfacetemper-ature.Thisoccursatadepthof(750C)=(30C=km)=25km.E35-17(a)Lookatthegure.At50%(whichis0.5ontheverticalaxis),thecurveisatB0=T0:55T=K.SinceT=300K,wehaveB0165T.(b)Samegure,butnowlookatthe90%mark.B0=T1:60T=K,soB0480T.(c)Goodquestion.Ithinkbotheldsarefarbeyondourcurrentabilities.E35-18(a)Lookatthegure.At50%(whichis0.5ontheverticalaxis),thecurveisat课后答案网B0=T0:55T=K.SinceB0=1:8T,wehaveT(1:8T)=(0:55T=K)=3:3K.(b)Samegure,butnowlookatthe90%mark.B0=T1:60T=K,soT(1:8T)=(1:60T=K)=1:1K.E35-19Since(0:5T)=(10K)=0:05T=K,andallhighertemperatureshavelowervaluesoftheratio,andthisputsallpointsintheregionnearwhereCurie"sLaw(thestraightline)isvalid,thentheanswerisyes.E35-20UsingEq.35-19,VMMM(108g/mol)(511103A=m)==r==8:741021A=m2nNA(10490kg=m3)(6:021023/mol)131khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE35-21(a)B==2z3,so0(4107N=A2)(1:51023J=T)B==9:4106T:2(10109m)3(b)U=2B=2(1:51023J=T)(9:4106T)=2:821028J.E35-22=(43106T)(295;000106m2)=1:3107Wb:BE35-23(a)We"llassume,however,thatalloftheironatomsareperfectlyaligned.ThenthedipolemomentoftheearthwillberelatedtothedipolemomentofoneatombyEarth=NFe;whereNisthenumberofironatomsinthemagnetizedsphere.IfmAistherelativeatomicmassofiron,thenthetotalmassisNmAmAEarthm==;khdaw.comAAFewhereAisAvogadro"snumber.Next,thevolumeofasphereofmassmismmAEarthV==;AFewhereisthedensity.Andnally,theradiusofaspherewiththisvolumewouldbe1=31=33V3EarthmAr==:44FeANowwendtheradiusbysubstitutingintheknownvalues,www.khdaw.com!1=33(8:01022J=T)(56g/mol)r==1:8105m:4(14106g/m3)(2:11023J=T)(6:01023/mol)(b)Thefractionalvolumeisthecubeofthefractionalradius,sotheansweris(1:8105m=6:4106)3=2:2105:E35-24(a)AtmagneticequatorLm=0,so(1:00107N=A2)(8:01022J=T)0B===31T:课后答案网4r3(6:37106m)3Thereisnoverticalcomponent,sotheinclinationiszero.(b)HereL=60,somq7222q02(1:0010N=A)(8:010J=T)2B=1+3sinLm=1+3sin(60)=56T:4r3(6:37106m)3Theinclinationisgivenbyarctan(B=B)=arctan(2tanL)=74:vhm(c)AtmagneticnorthpoleL=90,som2(1:00107N=A2)(8:01022J=T)0B===62T:2r3(6:37106m)3Thereisnohorizontalcomponent,sotheinclinationis90.khdaw.com132若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE35-25Thisproblemiseectivelysolving1=r3=1=2forrmeasuredinEarthradii.Thenr=1:26r,andthealtitudeabovetheEarthis(0:26)(6:37106m)=1:66106m:EE35-26Theradialdistancefromthecenterisr=(6:37106m)(2900103m)=3:47106m:Theeldstrengthis2(1:00107N=A2)(8:01022J=T)0B===380T:2r3(3:47106m)3E35-27HereL=9011:5=78:5,somq7222q02(1:0010N=A)(8:010J=T)2B=1+3sinLm=1+3sin(78:5)=61T:4r3(6:37106m)3Theinclinationisgivenbyarctan(B=B)=arctan(2tanL)=84:khdaw.comvhmE35-28The uxouttheother"endis(1:6103T)(0:13m)2=85Wb.Thenet uxthroughthesurfaceiszero,sothe uxthroughthecurvedsurfaceis0(85Wb)(25Wb)=60Wb:.Thenegativeindicatesinward.E35-29Thetotalmagnetic uxthroughaclosedsurfaceiszero.Thereisinward uxonfacesone,three,andveforatotalof-9Wb.Thereisoutward uxonfacestwoandfourforatotalof+6Wb.Thedierenceis+3Wb;consequentlytheoutward uxonthesixthfacemustbe+3Wb.E35-30Thestablearrangementsare(a)and(c).Thetorqueineachcaseiszero.www.khdaw.comE35-31Theeldonthexaxisbetweenthewiresis0i11B=+:22r+x2rxHSinceB~
dA~=0,wecanassumethe uxthroughthecurvedsurfaceisequaltothe uxthroughthexzplanewithinthecylinder.This uxisZr0i11B=L+dx;r22r+x2rx0i3rr课后答案网=Llnln;2r3r0i=Lln3:P35-1Wecanimaginetherotatingdiskasbeingcomposedofanumberofrotatingringsofradiusr,widthdr,andcircumference2r.Thesurfacechargedensityonthediskis=q=R2,andconsequentlythe(dierential)chargeonanyringis2qrdq=(2r)(dr)=drR2Theringsrotate"withangularfrequency!,orperiodT=2=!.Theeective(dierential)currentforeachringisthendqqr!di==dr:TR2khdaw.com133若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comEachringcontributestothemagneticmoment,andwecanglueallofthistogetherasZ=d;Z=r2di;ZR3qr!=dr;R20qR2!=:4P35-2(a)Thespherecanbeslicedintodisks.Thediskscanbeslicedintorings.Eachringhassomechargeqi,radiusri,andmassmi;theperiodofrotationforaringisT=2=!,sothecurrentintheringisqi=T=!qi=2.Themagneticmomentis=(!q=2)r2=!qr2=2:khdaw.comiiiiiNotethatthisiscloselyrelatedtotheexpressionforangularmomentumofaring:l=!mr2.iiiEquating,i=qili=2mi:Ifbothmassdensityandchargedensityareuniformthenwecanwriteqi=mi=q=m,ZZ=d=(q=2m)dl=qL=2mForasolidsphereL=!I=2!mR2=5,so=q!Rwww.khdaw.com2=5:(b)Seepart(a)P35-3(a)TheorbitalspeedisgivenbyK=mv2=2:Theorbitalradiusisgivenbymv=qBr,orr=mv=qB.Thefrequencyofrevolutionisf=v=2r.Theeectivecurrentisi=qf.Combiningalloftheabovetondthedipolemoment,vvrmv2K=iA=qr2=q=q=:2r22qBB(b)Sinceqandmcanceloutoftheaboveexpressiontheansweristhesame!(c)Workitout:课后答案网(5:281021=m3)(6:211020J)(5:281021=m3)(7:581021J)M==+=312A=m:V(1:18T)(1:18T)P35-4(b)Pointthethumboryourrighthandtotheright.Yourngerscurlinthedirectionofthecurrentinthewireloop.(c)InthevicinityofthewireoftheloopB~hasacomponentwhichisdirectedradiallyoutward.ThenB~d~shasacomponentdirectedtotheleft.Hence,thenetforceisdirectedtotheleft.P35-5(b)Pointthethumboryourrighthandtotheleft.Yourngerscurlinthedirectionofthecurrentinthewireloop.(c)InthevicinityofthewireoftheloopB~hasacomponentwhichisdirectedradiallyoutward.ThenB~d~shasacomponentdirectedtotheright.Hence,thenetforceisdirectedtotheright.khdaw.com134若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP35-6(a)Letx=B=kT.AdopttheconventionthatN+referstotheatomswhichhaveparallelalignmentandNthosewhichareanti-parallel.ThenN++N=N,soN=Nex=(ex+ex);+andN=Nex=(ex+ex);NotethatthedenominatorsarenecessarysothatN++N=N.Finally,exexM=(N+N)=N:ex+ex(b)IfBkTthenxisverysmallandex1x.Theaboveexpressionreducesto(1+x)(1x)2BM=N=Nx=:(1+x)+(1x)kTkhdaw.com(c)IfBkTthenxisverylargeandex!1whileex!0.TheaboveexpressionreducestoN=N:P35-7(a)Centripetalaccelerationisgivenbya=r!2.Thenaa=r(!+
!)2r!2;000=2r!
!+r(
!)2;002r!0
!:www.khdaw.com(b)Thechangeincentripetalaccelerationiscausedbytheadditionalmagneticforce,whichhasmagnitudeFB=qvB=er!B:Thenaa0eB
!==:2r!02mNotethatweboldlycanceled!against!0inthislastexpression;weareassumingthat
!issmall,andfortheseproblemsitis.P35-8(a)i==A=(8:01022J=T)=(6:37106m)2=6:3108A:(b)Farenoughawaybotheldsactlikeperfectdipoles,andcanthencancel.课后答案网(c)Closeenoughneithereldactslikeaperfectdipoleandtheeldswillnotcancel.pP35-9(a)B=Bh2+Bv2,soqq02202B=4r3cosLm+4sinLm=4r31+3sinLm:(b)tani=Bv=Bh=2sinLm=cosLm=2tanLm:135khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE36-1TheimportantrelationshipisEq.36-4,writtenasiL(5:0mA)(8:0mH)7B===1:010WbN(400)E36-2(a)=(34)(2:62103T)(0:103m)2=2:97103Wb:(b)L==i=(2:97103Wb)=(3:77A)=7:88104H:E36-3n=1=d,wheredisthediameterofthewire.ThenLA(4107H=m)(=4)(4:10102m)2=n2A=0==2:61104H=m:l0d2(2:52103m)2E36-4(a)Theemfsupportsthecurrent,sothecurrentmustbedecreasing.khdaw.com(b)L=E=(di=dt)=(17V)=(25103A=s)=6:8104H:E36-5(a)Eq.36-1canbeusedtondtheinductanceofthecoil.EL(3:0mV)4L===6:010H:di=dt(5:0A=s)(b)Eq.36-4canthenbeusedtondthenumberofturnsinthecoil.iL(8:0A)(6:0104H)N===120B(40Wb)E36-6UsetheequationbetweenEqs.36-9and36-10.www.khdaw.com(4107H=m)(0:81A)(536)(5:2102m)(5:2102m)+(15:3102m)B=ln;2(15:3102m)=1:32106Wb:E36-7L=n2Al=N2A=l,orm0m0L=(968)(4107H=m)(1870)2(=4)(5:45102m)2=(1:26m)=7:88H:E36-8IneachcaseapplyE=L
i=
t.(a)E=(4:6H)(7A)课后答案网=(2103s)=1:6104V:(b)E=(4:6H)(2A)=(3103s)=3:1103V:(c)E=(4:6H)(5A)=(1103s)=2:3104V:E36-9(a)Iftwoinductorsareconnectedinparallelthenthecurrentthrougheachinductorwilladdtothetotalcurrentthroughthecircuit,i=i1+i2;Takethederivativeofthecurrentwithrespecttotimeandthendi=dt=di1=dt+di2=dt;Thepotentialdierenceacrosseachinductoristhesame,soifwedividebyEandapplywegetdi=dtdi1=dtdi2=dt=+;EEEButdi=dt1=;ELkhdaw.com136若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comsothepreviousexpressioncanalsobewrittenas111=+:LeqL1L2(b)Iftheinductorsarecloseenoughtogetherthenthemagneticeldfromonecoilwillinducecurrentsintheothercoil.Thenwewillneedtoconsidermutualinductioneects,butthatisatopicnotcoveredinthistext.E36-10(a)Iftwoinductorsareconnectedinseriesthentheemfacrosseachinductorwilladdtothetotalemfacrossboth,E=E1+E2;Thenthecurrentthrougheachinductoristhesame,soifwedividebydi=dtandapplywegetEE1E2=+;di=dtdi=dtdi=dtButEkhdaw.com=L;di=dtsothepreviousexpressioncanalsobewrittenasLeq=L1+L2:(b)Iftheinductorsarecloseenoughtogetherthenthemagneticeldfromonecoilwillinducecurrentsintheothercoil.Thenwewillneedtoconsidermutualinductioneects,butthatisatopicnotcoveredinthistext.E36-11UseEq.36-17,butrearrange:www.khdaw.comt(1:50s)L===0:317s:ln[i0=i]ln[(1:16A)=(10:2103A)]ThenR=L=L=(9:44H)=(0:317s)=29:8:E36-12(a)Thereisnocurrentthroughtheresistor,soER=0andthenEL=E.(b)E=Ee2=(0:135)E.L(c)n=ln(EL=E)=ln(1=2)=0:693.E36-13(a)FromEq.36-4wendtheinductancetobe课后答案网3NB(26:210Wb)3L===4:7810H:i(5:48A)NotethatBisthe ux,whilethequantityNBisthenumberof uxlinkages.(b)WecanndthetimeconstantfromEq.36-14,=L=R=(4:78103H)=(0:745)=6:42103s:LThewecaninvertEq.36-13togetRi(t)t=Lln1;E3(0:745A)(2:53A)3=(6:4210s)ln1=2:4210s:(6:00V)khdaw.com137若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE36-14(a)Rearrange:diE=iR+L;dtELdii=;RRdtRdidt=:LE=Ri(b)Integrate:ZtZiRdidt=;0L0iE=RRi+E=Rt=ln;LE=REt=Le=i+E=R;Rkhdaw.comE1et=L=i:RE36-15di=dt=(5:0A=s).ThendiE=iR+L=(3:0A)(4:0)+(5:0A=s)t(4:0)+(6:0H)(5:0A=s)=42V+(20V=s)t:dtE36-16(1=3)=(1et=L),sot(5:22s)L==www.khdaw.com=12:9s:ln(2=3)ln(2=3)E36-17WewanttotakethederivativeofthecurrentinEq.36-13withrespecttotime,di=E1et=L=Eet=L:dtRLLThen=(5:0102H)=(180)=2:78104s.UsingthiswendtherateofchangeinthecurrentLwhent=1:2mstobedi(45V)(1:2103s)=(2:78104s)=e=12A=s:课后答案网dt((5:0102H)E36-18(b)Considersometimeti:E(t)=Eeti=L:LiTakingaratiofortwodierenttimes,EL(t1)=e(t2t1)=L;EL(t2)ort2t1(2ms)(1ms)L===3:58msln[EL(t1)=EL(t2)]ln[(18:24V)=(13:8V)](a)Chooseanytime,andE=Eet=L=(18:24V)e(1ms)=(3:58ms)=24V:Lkhdaw.com138若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE36-19(a)Whentheswitchisjustclosedthereisnocurrentthroughtheinductor.Soi1=i2isgivenbyE(100V)i1===3:33A:R1+R2(10)+(20)(b)Alongtimelaterthereiscurrentthroughtheinductor,butitisasiftheinductorhasnoeectonthecircuit.Thentheeectiveresistanceofthecircuitisfoundbyrstndingtheequivalentresistanceoftheparallelpart1=(30)+1=(20)=1=(12);andthenndingtheequivalentresistanceofthecircuit(10)+(12)=22:Finally,i1=(100V)=(22)=4:55Aandkhdaw.com
V2=(100V)(4:55A)(10)=54:5V;consequently,i2=(54:5V)=(20)=2:73A:Itdidn"task,buti2=(4:55A)(2:73A)=1:82A:(c)Aftertheswitchisjustopenedthecurrentthroughthebatterystops,whilethatthroughtheinductorcontinueson.Theni2=i3=1:82A.(d)Allgotozero.E36-20(a)FortoroidsL=N2hln(b=a)=2.Thenumberofturnsislimitedbytheinnerradius:0Nd=2a.Inthiscase,N=2(0:10m)=(0:00096m)=654:Theinductanceisthen(4107H=m)(654)2(0:02m)www.khdaw.com(0:12m)L=ln=3:1104H:2(0:10m)(b)Eachturnhasalengthof4(0:02m)=0:08m.TheresistanceisthenR=N(0:08m)(0:021=m)=1:10Thetimeconstantis=L=R=(3:1104H)=(1:10)=2:8104s:LE36-21(I)Whentheswitchisjustclosedthereis课后答案网nocurrentthroughtheinductororR2,sothepotentialdierenceacrosstheinductormustbe10V.ThepotentialdierenceacrossR1isalways10Vwhentheswitchisclosed,regardlessoftheamountoftimeelapsedsinceclosing.(a)i1=(10V)=(5:0)=2:0A.(b)Zero;readtheaboveparagraph.(c)Thecurrentthroughtheswitchisthesumoftheabovetwocurrents,or2:0A.(d)Zero,sincethecurrentthroughR2iszero.(e)10V,sincethepotentialacrossR2iszero.(f)LookattheresultsofExercise36-17.Whent=0therateofchangeofthecurrentisdi=dt=E=L.Thendi=dt=(10V)=(5:0H)=2:0A=s:(II)Aftertheswitchhasbeenclosedforalongperiodoftimethecurrentsarestableandtheinductornolongerhasaneectonthecircuit.Thenthecircuitisasimpletworesistorparallelnetwork,eachresistorhasapotentialdierenceof10Vacrossit.khdaw.com139若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com(a)Still2.0A;nothinghaschanged.(b)i2=(10V)=(10)=1:0A.(c)Addthetwocurrentsandthecurrentthroughtheswitchwillbe3.0A.(d)10V;seetheabovediscussion.(e)Zero,sincethecurrentisnolongerchanging.(f)Zero,sincethecurrentisnolongerchanging.E36-22U=(71J=m3)(0:022m3)=1:56J.ThenusingU=i2L=2wegetppi=2U=L=2(1:56J)=(0:092H)=5:8A:E36-23(a)L=2U=i2=2(0:0253J)=(0:062A)2=13:2H.(b)Sincethecurrentissquaredintheenergyexpression,doublingthecurrentwouldquadrupletheenergy.Theni0=2i=2(0:062A)=0:124A.0E36-24(a)B=inandu=B2=2,orkhdaw.com00u=i2n2=2=(4107N=A2)(6:57A)2(950=0:853m)2=2=33:6J=m3:0(b)U=uAL=(33:6J=m3)(17:2104m2)(0:853m)=4:93102J:E36-25u=B2=2,andfromSampleProblem33-2weknowB,henceB0(12:6T)2u==6:32107J=m3:B2(4107N=A2)E36-26(a)u=B2=2,sowww.khdaw.comB0(1001012T)21u==2:5102eV/cm3:B2(4107N=A2)(1:61019J/eV)(b)x=(10)(9:461015m)=9:461016m.Usingtheresultsfrompart(a)expressedinJ/m3wendtheenergycontainedisU=(3:981015J=m3)(9:461016m)3=3:41036JE36-27TheenergydensityofanelectriceldisgivenbyEq.36-23;thatofamagneticeldis课后答案网givenbyEq.36-22.Equating,0212E=B;220BE=p:00TheansweristhenqE=(0:50T)=(8:851012C2=N
m2)(4107N=A2)=1:5108V=m:140khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE36-28TherateofinternalenergyincreaseintheresistorisgivenbyP=i
VR.TherateofenergystorageintheinductorisdU=dt=Lidi=dt=i
VL.Sincethecurrentisthesamethroughbothwewanttondthetimewhen
VR=
VL.UsingEq.36-15wend1et=L=et=L;ln2=t=L;sot=(37:5ms)ln2=26:0ms:E36-29(a)StartwithEq.36-13:i=E(1et=L)=R;iRt=L1=e;EtL=;ln(1iR=E)khdaw.com(5:20103s)=;ln[1(1:96103A)(10:4103)=(55:0V)]=1:12102s:ThenL=R=(1:12102s)(10:4103)=116H:L(b)U=(1=2)(116H)(1:96103A)2=2:23104J:RE36-30(a)U=E
q;q=idt.ZEt=LU=Ewww.khdaw.com1edt;RE22=t+et=L;LR0E2E2L=t+(et=L1):RR2Usingthenumbersprovided,L=(5:48H)=(7:34)=0:7466s:Then2hi课后答案网(12:2V)(2s)=0:7466s)U=(2s)+(0:7466s)(e1)=26:4J(7:34)(b)TheenergystoredintheinductorisU=Li2=2,orLLE2Z2U=1et=Ldt;L2R2=6:57J:(c)UR=UUL=19:8J.E36-31Thisshellhasavolumeof4
V=(R+a)3R3:EE3khdaw.com141若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comSincea<XC,thishappenswhenimlagsEm.If,ontheotherhand,XLXCandthecircuitispredomi-nantlyinductive.Butwhatdoesthisreallymean?Itmeansthattheinductorplaysamajorroleinthecurrentthroughthecircuitwhilethecapacitorplaysaminorrole.Themoreinductiveacircuitis,thelesssignicantanycapacitanceisonthebehaviorofthecircuit.Forfrequenciesbelowtheresonantfrequencythereverseistrue.(b)Rightattheresonantfrequencytheinductiveeectsareexactlycanceledbythecapacitiveeects.Theimpedanceisequaltotheresistance,anditis(almost)asifneitherthecapacitoror课后答案网inductorareeveninthecircuit.E37-10ThenetycomponentisXCXL.ThenetxcomponentisR.ThemagnitudeoftheresultantispZ=R2+(XCXL)2;whilethephaseangleis(XCXL)tan=:RE37-11Yes.pAtresonance!=1=(1:2H)(1:3106F)=800rad/sandZ=R.Thenim=E=Z=(10V)=(9:6)=1:04A,so[
VL]m=imXL=(1:08A)(800rad/s)(1:2H)=1000Vkhdaw.com:152若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE37-12(a)LetO=XXandA=R,thenH2=A2+O2=Z2,soLCsin=(XLXC)=Zandcos=R=Z:E37-13(a)Thevoltageacrossthegeneratoristhegeneratoremf,sowhenitisamaximumfromSampleProblem37-3,itis36V.Thiscorrespondsto!t==2.(b)Thecurrentthroughthecircuitisgivenbyi=imsin(!t).WefoundinSampleProblem37-3thati=0:196Aand=29:4=0:513rad.mForaresistiveloadweapplyEq.37-3,
VR=imRsin(!t)=(0:196A)(160)sin((=2)(0:513))=27:3V:(c)ForacapacitiveloadweapplyEq.37-12,
VC=imXCsin(!t=2)=(0:196A)(177)sin((0:513))=17:0V:khdaw.com(d)ForaninductiveloadweapplyEq.37-7,
VL=imXLsin(!t+=2)=(0:196A)(87)sin((0:513))=8:4V:(e)(27:3V)+(17:0V)+(8:4V)=35:9V.E37-14Ifcircuit1and2havethesameresonantfrequencythenL1C1=L2C2.TheseriescombinationfortheinductorsisL=L1+L2;Theseriescombinationforthecapacitorsis1=C=1=Cwww.khdaw.com1+1=C2;soC1C2L1C1C2+L2C2C1LC=(L1+L2)==L1C1;C1+C2C1+C2whichisthesameasbothcircuit1and2.E37-15(a)Z=(125V)=(3:20A)=39:1.(b)LetO=XXandA=R,thenH2=A2+O2=Z2,soLCcos=R=Z:Usingthisrelation,课后答案网R=(39:1)cos(56:3)=21:7:(c)Ifthecurrentleadstheemfthenthecircuitiscapacitive.E37-16(a)Integratingoverasinglecycle,ZTZT1211sin!tdt=(1cos2!t)dt;T0T0211=T=:2T2(b)Integratingoverasinglecycle,ZTZT111sin!tcos!tdt=sin2!tdt;T0T02=0:khdaw.com153若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE37-17TheresistancewouldbegivenbyEq.37-32,Pav(0:10)(746W)R===177:irms2(0:650A)2Thiswouldnotbethesameasthedirectcurrentresistanceofthecoilsofastoppedmotor,becausetherewouldbenoinductiveeects.E37-18Sinceirms=Erms=Z,thenE2R2rmsPav=irmsR=:Z2pE37-19(a)Z=(160)2+(177)2=239;then1(36V)2(160)Pav==1:82W:khdaw.com2(239)2p(b)Z=(160)2+(87)2=182;then1(36V)2(160)Pav==3:13W:2(182)2pE37-20(a)Z=(12:2)2+(2:30)2=12:4(b)P=(120V)2(12:2)=(12:4)2=1140W:av(c)irms=(120V)=(12:4)=9:67A.www.khdaw.compE37-21Thermsvalueofanysinusoidalquantityisrelatedtothemaximumvalueby2vrms=pvmax.Sincethisfactorof2appearsinalloftheexpressions,wecanconcludethatifthermsvaluesareequalthensoarethemaximumvalues.Thismeansthat(
VR)max=(
VC)max=(
VL)maxorimR=imXC=imXLor,withonelastsimplication,R=XL=XC:Focusontherighthandsideofthelastequality.IfXC=XLthenwehavearesonancecondition,andtheimpedance(seeEq.37-20)isaminimum,andisequaltoR.Then,accordingtoEq.37-21,课后答案网Emim=;Rwhichhastheimmediateconsequencethatthermsvoltageacrosstheresistoristhesameasthermsvoltageacrossthegenerator.Soeverythingis100V.pE37-22(a)Theantennaisin-tune"whentheimpedanceisaminimum,or!=1=LC.Sopf=!=2=1=2(8:22106H)(0:2701012F)=1:07108Hz:(b)i=(9:13V)=(74:7)=1:22107A.rms(c)XC=1=2fC,soV=iX=(1:22107A)=2(1:07108Hz)(0:2701012F)=6:72104V:CC154khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE37-23Assumingnoinductorsorcapacitorsinthecircuit,thenthecircuiteectivelybehavesasaDCcircuit.Thecurrentthroughthecircuitisi=E=(r+R).ThepowerdeliveredtoRisthenP=i
V=i2R=E2R=(r+R)2:EvaluatedP=dRandsetitequaltozerotondthemaximum.ThendP2rR0==ER;dR(r+R)3whichhasthesolutionr=R.E37-24(a)SinceP=i2R=2=E2R=2Z2,thenPisamaximumwhenZisaminimum,andavmmavpvise-versa.Zisaminimumatresonance,whenZ=Randf=1=2LC.WhenZisaminimumC=1=42f2L=1=42(60Hz)2(60mH)=1:2107F:(b)ZisamaximumwhenXCisamaximum,whichoccurswhenCisverysmall,likezero.(c)WhenXCisamaximumP=0.WhenPisamaximumZ=RsoP=(30V)2=2(5:0)=90W:(d)Thephaseangleiszeroforresonance;itis90forinniteXorX.khdaw.comCL(e)Thepowerfactoriszeroforasystemwhichhasnopower.Thepowerfactorisoneforasysteminresonance.E37-25(a)TheresistanceisR=15:0.Theinductivereactanceis11XC===61:3:!C2(550s1)(4:72F)TheinductivereactanceisgivenbyX=!L=2(550s1)(25:3mH)=87:4:Lwww.khdaw.comTheimpedanceisthenqZ=(15:0)2+((87:4)(61:3))2=30:1:Finally,thermscurrentisErms(75:0V)irms===2:49A:Z(30:1)(b)Thermsvoltagesbetweenanytwopointsisgivenby(
V)rms=irmsZ;whereZisnottheimpedanceofthecircuitbutinsteadtheimpedancebetweenthetwopointsinquestion.Whenonlyonedeviceisbetweenthetwopointstheimpedanceisequaltothereactance课后答案网(orresistance)ofthatdevice.We"renotgoingtoshowalloftheworkhere,butwewillputtogetheranicetableforyouPointsImpedanceExpressionImpedanceValue(
V)rms,abZ=RZ=15:037.4V,bcZ=XCZ=61:3153V,cdZ=XLZ=87:4218V,bdZ=pjXLXCjZ=26:165V,acZ=R2+X2Z=63:1157V,CNotethatthislastonewas
Vac,andnot
Vad,becauseitismoreentertaining.Youprobablyshoulduse
Vadforyourhomework.(c)Theaveragepowerdissipatedfromacapacitororinductoriszero;thatoftheresistorisP=[(
V)]2=R=(37:4V)2=(15:0)=93:3W:RRrmskhdaw.com155若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE37-26(a)Theenergystoredinthecapacitorisafunctionofthechargeonthecapacitor;althoughthechargedoesvarywithtimeitvariesperiodicallyandattheendofthecyclehasreturnedtotheoriginalvalues.Assuch,theenergystoredinthecapacitordoesn"tchangefromoneperiodtothenext.(b)Theenergystoredintheinductorisafunctionofthecurrentintheinductor;althoughthecurrentdoesvarywithtimeitvariesperiodicallyandattheendofthecyclehasreturnedtotheoriginalvalues.Assuch,theenergystoredintheinductordoesn"tchangefromoneperiodtothenext.(c)P=Ei=Emimsin(!t)sin(!t),sotheenergygeneratedinonecycleisZTZTU=Pdt=Emimsin(!t)sin(!t)dt;00ZT=Emimsin(!t)sin(!t)dt;0T=Emimcos:khdaw.com2(d)P=i2Rsin2(!t),sotheenergydissipatedinonecycleismZTZTU=Pdt=i2Rsin2(!t)dt;m00ZT=i2Rsin2(!t)dt;m0T2=imR:2(e)Sincecos=R=ZandEm=Z=imwecanequatetheanswersfor(c)and(d).www.khdaw.comE37-27ApplyEq.37-41,Ns(780)3
Vs=
Vp=(150V)=1:810V:Np(65)E37-28(a)ApplyEq.37-41,Ns(10)
Vs=
Vp=(120V)=2:4V:课后答案网Np(500)(b)is=(2:4V)=(15)=0:16A;Ns(10)3ip=is=(0:16A)=3:210A:Np(500)E37-29TheautotransformercouldhaveaprimaryconnectedbetweentapsT1andT2(200turns),T1andT3(1000turns),andT2andT3(800turns).Thesamepossibilitiesaretrueforthesecondaryconnections.Ignoringtheone-to-oneconnectionsthereare6choices|threearestepup,andthreearestepdown.Thestepupratiosare1000=200=5,800=200=4,and1000=800=1:25.Thestepdownratiosarethereciprocalsofthesethreevalues.156khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE37-30=(1:69108
m)[1(4:3103=C)(14:6C)]=1:58108
m.TheresistanceofthetwowiresisL(1:58108
m)2(1:2103m)R===14:9:A(0:9103m)2P=i2R=(3:8A)2(14:9)=220W.E37-31Thesupplycurrentispi=(0:270A)(74103V=2)=(220V)=64:2A:pThepotentialdropacrossthesupplylinesis
V=(64:2A)(0:62)=40V:Thisistheamountbywhichthesupplyvoltagemustbeincreased.E37-32khdaw.comUseEq.37-46:pNp=Ns=(1000)=(10)=10:P37-1(a)Theemfisamaximumwhen!t=4==2,sot=3=4!=3=4(350rad/s)=6:73103s.(b)Thecurrentisamaximumwhen!t3=4==2,sot=5=4!=5=4(350rad/s)=1:12102s.(c)Thecurrentlagstheemf,sothecircuitcontainsaninductor.(d)XL=Em=imandXL=!L,soEm(31www.khdaw.com:4V)L===0:144H:im!(0:622A)(350rad/s)P37-2(a)Theemfisamaximumwhen!t=4==2,sot=3=4!=3=4(350rad/s)=6:73103s.(b)Thecurrentisamaximumwhen!t+=4==2,sot==4!==4(350rad/s)=2:24103s.(c)Thecurrentleadstheemf,sothecircuitcontainsancapacitor.(d)XC=Em=imandXC=1=!C,soim(0:622A)5C===5:6610F:Em!(31:4V)(350rad/s)课后答案网P37-3(a)Sincethemaximumvaluesforthevoltagesacrosstheindividualdevicesarepropor-tionaltothereactances(orresistances)fordevicesinseries(theconstantofproportionalityisthemaximumcurrent),wehaveXL=2RandXC=R.FromEq.37-18,XLXC2RRtan===1;RRor=45.(b)Theimpedanceofthecircuit,intermsoftheresistiveelement,ispppZ=R2+(XLXC)2=R2+(2RR)2=2R:ButEm=imZ,soZ=(34:4V)=(0:320A)=108.ThenwecanuseourpreviousworktondsthatR=76.khdaw.com157若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP37-4WhentheswitchisopenthecircuitisanLRCcircuit.Inposition1thecircuitisanRLCcircuit,butthecapacitanceisequaltothetwocapacitorsofCinparallel,or2C.Inposition2thecircuitisasimpleLCcircuitwithnoresistance.Theimpedancewhentheswitchisinposition2isZ2=j!L1=!Cj.ButZ2=(170V)=(2:82A)=60:3:Thephaseanglewhentheswitchisopenis=20.But0!L1=!CZ2tan0==;RRsoR=(60:3)=tan(20)=166.Thephaseanglewhentheswitchisinposition1is!L1=!2Ctan1=;Rsokhdaw.com!L1=!2C=(166)tan(10)=29:2.Equatingthe!Lpart,(29:2)+1=!2C=(60:3)+1=!C;C=1=2(377rad/s)[(60:3)+(29:2)]=1:48105F:Finally,(60:3)+1=(377rad/s)(1:48105F)L==0:315H:(377rad/s)P37-5Allthreewireshaveemfswhichvarysinusoidallyintime;ifwechooseanytwowiresthephasedierencewillhaveanabsolutevalueof120www.khdaw.com.Wecanthenchooseanytwowiresandexpect(bysymmetry)togetthesameresult.Wechoose1and2.ThepotentialdierenceisthenVV=V(sin!tsin(!t120)):12mWeneedtoaddthesetwosinefunctionstogetjustone.Weuse11sinsin=2sin()cos(+):22Then课后答案网11V1V2=2Vmsin(120)cos(2!t120);22p3=2Vm()cos(!t60);2p=3Vsin(!t+30):mP37-6(a)cos=cos(42)=0:74.(b)Thecurrentleads.(c)Thecircuitiscapacitive.(d)No.Resonancecircuitshaveapowerfactorofone.(e)Theremustbeatleastacapacitorandaresistor.(f)P=(75V)(1:2A)(0:74)=2=33W.158khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comppP37-7(a)!=1=LC=1=(0:988H)(19:3106F)=229rad/s.(b)im=(31:3V)=(5:12)=6:11A.(c)Thecurrentamplitudewillbehalvedwhentheimpedanceisdoubled,orwhenZ=2R.Thisoccurswhen3R2=(!L1=!C)2,or3R2!2=!4L22!2L=C+1=C2:Thesolutiontothisquadraticisp2L+3CR29C2R4+12CR2L!2=;2L2Cso!1=224:6rad/sand!2=233:5rad/s.(d)
!=!=(8:9rad/s)=(229rad/s)=0:039.P37-8(a)Thecurrentamplitudewillbehalvedwhentheimpedanceisdoubled,orwhenZ=2R.Thisoccurswhen3khdaw.comR2=(!L1=!C)2,or3R2!2=!4L22!2L=C+1=C2:Thesolutiontothisquadraticisp2L+3CR29C2R4+12CR2L!2=;2L2CNotethat
!=!+!;withaweebitofalgebra,
!(!+!)=!2!2:+www.khdaw.com+Also,!++!2!.Hence,p9C2R4+12CR2L!
!;2L2Cp!2R9C2R2+12LC!
!;2Lp!R9!2C2R2+12!
!;2Lp
!R9CR2=L+12;课后答案网!p2L!3R;!LassumingthatCR24L=3.P37-9P37-10UseEq.37-46.P37-11(a)Theresistanceofthisbulbis(
V)2(120V)2R===14:4:P(1000W)khdaw.com159若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comThepowerisdirectlyrelatedtothebrightness;ifthebulbistobevariedinbrightnessbyafactorof5thenitwouldhaveaminimumpowerof200W.Thermscurrentthroughthebulbatthispowerwouldbeppirms=P=R=(200W)=(14:4)=3:73A:TheimpedanceofthecircuitmusthavebeenErms(120V)Z===32:2:irms(3:73A)TheinductivereactancewouldthenbeppXL=Z2R2=(32:2)2(14:4)2=28:8:Finally,theinductancewouldbeL=X=!=(28:8)=(2(60:0s1))=7:64H:khdaw.comL(b)Onecoulduseavariableresistor,andsinceitwouldbeinserieswiththelampavalueof32:214:4=17:8wouldwork.Buttheresistorwouldgethot,whileonaveragethereisnopowerradiatedfromapureinductor.www.khdaw.com课后答案网160khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE38-1Themaximumvalueoccurswherer=R;thereB=1RdE=dt.ForrRBishalfofBmaxwhenr=2R.Thenthetwovaluesofrare2:5cmand10:0cm.E38-2ForaparallelplatecapacitorE==0andthe uxisthenE=A=0=q=0.ThendEdqddVid=0==CV=C:dtdtdtdtE38-3UsetheresultsofExercise2,andchangethepotentialdierenceacrosstheplatesofthecapacitorataratedVid(1:0mA)===1:0kV/s:dtC(1:0F)ProvideaconstantcurrenttothecapacitordQddVi==CV=C=id:khdaw.comdtdtdtE38-4SinceEisuniformbetweentheplatesE=EA,regardlessofthesizeoftheregionofinterest.Sincejd=id=A,id1dEdEjd==0=0:AAdtdtE38-5(a)Inthiscaseid=i=1:84A.(b)SinceE=q=0A,dE=dt=i=0A,ordE=dt=(1:84A)=(8:851012Fwww.khdaw.com=m)(1:22m)2=1:401011V=m:(c)id=0dE=dt=0adE=dt.aherereferstotheareaofthesmallersquare.Combinethiswiththeresultsofpart(b)andi=ia=A=(1:84A)(0:61m=1:22m)2=0:46A:dH(d)B~
d~s=i=(4107H=m)(0:46A)=5:78107T
m:0dE38-6SubstituteEq.38-8intotheresultsobtainedinSampleProblem38-1.Outsidethecapacitor=R2E,soER2dE000B==id:课后答案网2rdt2rInsidethecapacitortheenclosed uxis=r2E;butwewantinsteadtodeneiintermsofEdthetotalEinsidethecapacitoraswasdoneabove.Consequently,insidetheconductorrR2dEr000B==id:2R2dt2R2E38-7SincetheelectriceldisuniformintheareaandperpendiculartothesurfaceareawehaveZZZE=E~
dA~=EdA=EdA=EA:ThedisplacementcurrentisthendE122dEid=0A=(8:8510F=m)(1:9m):dtkhdaw.comdt161若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com(a)Intherstregiontheelectricelddecreasesby0.2MV/min4s,so(0:2106V=m)i=(8:851012F=m)(1:9m2)=0:84A:d(4106s)(b)Theelectriceldisconstantsothereisnochangeintheelectric ux,andhencethereisnodisplacementcurrent.(c)Inthelastregiontheelectricelddecreasesby0.4MV/min5s,so(0:4106V=m)i=(8:851012F=m)(1:9m2)=1:3A:d(5106s)HE38-8(a)BecauseofthecircularsymmetryB~
d~s=2rB,whereristhedistancefromthecenterofthecircularplates.Notonlythat,buti=jA=r2j.Equatethesetwoexpressions,dddandB=rj=2=(4107H=m)(0:053m)(1:87101A=m)=2=6:23107T:0dkhdaw.com(b)dE=dt=i=A=j==(1:87101A=m)=(8:851012F=m)=2:111012V=m:d0d0E38-9ThemagnitudeofEisgivenby(162V)E=sin2(60=s)t;(4:8103m)UsingtheresultsfromSampleProblem38-1,00RdEBm=;2dtt=0www.khdaw.com(4107H=m)(8:851012F=m)(0:0321m)(162V)=2(60=s);2(4:8103m)=2:271012T:E38-10(a)Eq.33-13frompage764andEq.33-34frompage762.(b)Eq.27-11frompage618andtheequationfromEx.27-25onpage630.(c)TheequationsfromEx.38-6onpage876.(d)Eqs.34-16and34-17frompage785.E38-11(a)Considerthepath课后答案网abefa.Theclosedlineintegralconsistsoftwoparts:b!eande!f!a!b.ThenIdE~
d~s=dtcanbewrittenasZZdE~
d~s+E~
d~s=abef:b!ee!f!a!bdtNowconsiderthepathbcdeb.Theclosedlineintegralconsistsoftwoparts:b!c!d!eande!b.ThenIdE~
d~s=dtcanbewrittenasZZdE~
d~s+E~
d~s=bcde:b!c!d!ee!bdtkhdaw.com162若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comThesetwoexpressionscanbeaddedtogether,andsinceZZE~
d~s=E~
d~se!bb!ewegetZZdE~
d~s+E~
d~s=(abef+bcde):e!f!a!bb!c!d!edtThelefthandsideofthisisjustthelineintegralovertheclosedpathefadcde;therighthandsideisthenetchangein uxthroughthetwosurfaces.ThenwecansimplifythisexpressionasIdE~
d~s=:dt(b)Doeverythingaboveagain,exceptsubstituteBforE.(c)IftheequationswerenotselfconsistentwewouldarriveatdierentvaluesofEandBdependingonhowwedenedoursurfaces.Thismulti-valuedresultwouldbequiteunphysical.khdaw.comE38-12(a)Considerthepartontheleft.Ithasasharedsurfaces,andtheothersurfacesl.ApplyingEq.I,IZZql=0=E~
dA~=E~
dA~+E~
dA~:slNotethatdA~isdirectedtotherightonthesharedsurface.Considerthepartontheright.Ithasasharedsurfaces,andtheothersurfacesr.ApplyingEq.I,IZZqr=0=E~
dA~=E~
dA~+E~
dA~:srNotethatdA~isdirectedtotheleftonthesharedsurface.www.khdaw.comAddingthesetwoexpressionswillresultinacancelingoutofthepartZE~
dA~ssinceoneisorientedoppositetheother.WeareleftwithZZIqr+ql=E~
dA~+E~
dA~=E~
dA~:0rlE38-13课后答案网E38-14(a)Electricdipoleisbecausethechargesareseparatinglikeanelectricdipole.Magneticdipolebecausethecurrentloopactslikeamagneticdipole.E38-15AseriesLCcircuitwilloscillatenaturallyatafrequency!1f==p22LCWewillneedtocombinethiswithv=f,wherev=cisthespeedofEMwaves.WewanttoknowtheinductancerequiredtoproduceanEMwaveofwavelength=550109m,so2(550109m)2L===5:011021H:42c2C42(3:00108m=s)2(171012F)Thisisasmallinductance!khdaw.com163若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE38-16(a)B=E=c,andBmustbepointinginthenegativeydirectioninorderthatthewavebepropagatinginthepositivexdirection.ThenBx=Bz=0,andB=E=c=(2:34104V=m)=(3:00108m=s)=(7:801013T)sink(xct):yz(b)=2=k=2=(9:72106=m)=6:46107m:E38-17TheelectricandmagneticeldofanelectromagneticwavearerelatedbyEqs.38-15and38-16,E(321V=m)B===1:07pT:c(3:00108m=s)E38-18TakethepartialofEq.38-14withrespecttox,@@E@@B=;@x@x@x@tkhdaw.com@2E@2B=:@x2@x@tTakethepartialofEq.38-17withrespecttot,@@B@@E=00;@t@x@t@t@2B@2E=00:@t@x@t2Equate,andlet=1=c2,then00@2Ewww.khdaw.com1@2E=:@x2c2@t2Repeat,exceptnowtakethepartialofEq.38-14withrespecttot,andthentakethepartialofEq.38-17withrespecttox.E38-19(a)Sincesin(kx!t)isoftheformf(kx!t),thenweonlyneeddopart(b).(b)TheconstantEmdropsoutofthewaveequation,soweneedonlyconcernourselveswithf(kx!t).Lettingg=kx!t,@2f@2f=c2;@t2@x2课后答案网2222@f@g2@f@g=c;@g2@t@g2@x@g@g=c;@t@x!=ck:E38-20Usetherighthandrule.E38-21U=Pt=(1001012W)(1:0109s)=1:0105J:E38-22E=Bc=(28109T)(3:0108m=s)=8:4V=m:Itisinthepositivexdirection.164khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE38-23IntensityisgivenbyEq.38-28,whichissimplyanexpressionofpowerdividedbysurfacearea.TondtheintensityoftheTVsignalat-Centauriweneedtondthedistanceinmeters;r=(4:30light-years)(3:00108m=s)(3:15107s/year)=4:061016m:TheintensityofthesignalwhenithasarrivedatoutnearestneighboristhenP(960kW)292I===4:6310W/m4r24(4:061016m)2E38-24(a)FromEq.38-22,S=cB2=.B=Bsin!t.Thetimeaverageisdenedas0mZT2ZT21cBm2cBmSdt=cos!tdt=:T00T020(b)S=(3:0108m=s)(1:0104T)2=2(4107H=m)=1:2106W=m2:khdaw.comavE38-25I=P=4r2,soppr=P=4I=(1:0103W)=4(130W=m2)=0:78m:E38-26u=E2=2=(cB)2=2=B2=2=u:E000BE38-27(a)IntensityisrelatedtodistancebyEq.38-28.Ifr1istheoriginaldistancefromthestreetlampandI1theintensityatthatdistance,thenPI1=www.khdaw.com2:4r1Thereisasimilarexpressionforthecloserdistancer2=r1162mandtheintensityatthatdistanceI2=1:50I1.Wecancombinethetwoexpressionforintensity,I2=1:50I1;PP=1:50;4r24r221r2=1:50r2;12pr1=1:50(r1162m):课后答案网Thelastlineiseasyenoughtosolveandwendr1=883m.(b)No,wecan"tndthepoweroutputfromthelamp,becausewewereneverprovidedwithanabsoluteintensityreference.pE38-28(a)Em=20cI,sopE=2(4107H=m)(3:00108m=s)(1:38103W=m2)=1:02103V=m:m(b)B=E=c=(1:02103V=m)=(3:00108m=s)=3:40106T:mmE38-29(a)B=E=c=(1:96V=m)=(3:00108m=s)=6:53109T.mm(b)I=E2=2c=(1:96V)2=2(4107H=m)(3:00108m=s)=5:10103W=m2:m0(c)P=4r2I=4(11:2m)2(5:10103W=m2)=8:04W:165khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE38-30(a)TheintensityisP(11012W)I===1:961027W=m2:A4(6:37106m)2ThepowerreceivedbytheAreciboantennaisP=IA=(1:961027W=m2)(305m)2=4=1:41022W:(b)ThepowerofthetransmitteratthecenterofthegalaxywouldbeP=IA=(1:961027W)(2:3104ly)2(9:461015m/ly)2=2:91014W:E38-31(a)TheelectriceldamplitudeisrelatedtotheintensitybyEq.38-26,E2mI=;khdaw.com20corppE=2cI=2(4107H=m)(3:00108m=s)(7:83W=m2)=7:68102V=m:m0(b)ThemagneticeldamplitudeisgivenbyE(7:68102V=m)B=m==2:561010Tmc(3:00108m=s)(c)ThepowerradiatedbythetransmittercanbefoundfromEq.38-28,P=4r2I=4(11:3km)www.khdaw.com2(7:83W=m2)=12:6kW:E38-32(a)Thepowerincidenton(andthenre ectedby)thetargetcraftisP=IA=PA=2r2.110Theintensityofthere ectedbeamisI=P=2r2=PA=42r4.Then210I=(183103W)(0:222m2)=42(88:2103m)4=1:701017W=m2:2(b)UseEq.38-26:ppE=2cI=2(4107H=m)(3:00108m=s)(1:701017W=m2)=1:13107V=m:m0pp(c)B=E=2c=(1:13107V=m)=2(3:00108m=s)=2:661016T:rmsm课后答案网E38-33RadiationpressureforabsorptionisgivenbyEq.38-34,butweneedtondtheenergyabsorbedbeforewecanapplythat.Wearegivenanintensity,asurfacearea,andatime,so
U=(1:1103W=m2)(1:3m2)(9:0103s)=1:3107J:Themomentumdeliveredisp=(
U)=c=(1:3107J)=(3:00108m=s)=4:3102kg
m=s:E38-34(a)F=A=I=c=(1:38103W=m2)=(3:00108m=s)=4:60106Pa:(b)(4:60106Pa)=(101105Pa)=4:551011:E38-35F=A=2P=Ac=2(1:5109W)=(1:3106m2)(3:0108mkhdaw.com=s)=7:7106Pa:166若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE38-36F=A=P=4r2c,soF=A=(500W)=4(1:50m)2(3:00108m=s)=5:89108Pa:E38-37(a)F=IA=c,so(1:38103W=m2)(6:37106m)2F==5:86108N:(3:00108m=s)E38-38(a)AssumingMKSA,theunitsaremFVNmCVsNNs==:smmAmsVmmCmm2s(b)AssumingMKSA,theunitsare22AVNAJN1JJ===:khdaw.comNmAmNCmAmsmmm2sE38-39Wecantreattheobjectashavingtwosurfaces,onecompletelyre ectingandtheothercompletelyabsorbing.IftheentiresurfacehasanareaAthentheabsorbingparthasanareafAwhilethere ectingparthasarea(1f)A.Theaverageforceisthenthesumoftheforceoneachpart,I2IFav=fA+(1f)A;ccwhichcanbewrittenintermsofpressureasFavIwww.khdaw.com=(2f):AcE38-40Wecantreattheobjectashavingtwosurfaces,onecompletelyre ectingandtheothercompletelyabsorbing.IftheentiresurfacehasanareaAthentheabsorbingparthasanareafAwhilethere ectingparthasarea(1f)A.Theaverageforceisthenthesumoftheforceoneachpart,I2IFav=fA+(1f)A;ccwhichcanbewrittenintermsofpressureas课后答案网FavI=(2f):AcTheintensityIisthatoftheincidentbeam;there ectedbeamwillhaveanintensity(1f)I.Eachbeamwillcontributetotheenergydensity|I=cand(1f)I=c,respectively.Addthesetwoenergydensitiestogetthenetenergydensityoutsidethesurface.Theresultis(2f)I=c,whichisthelefthandsideofthepressurerelationabove.E38-41Thebulletdensityis=Nm=V.LetV=Ah;thekineticenergydensityisK=V=1Nmv2=Ah:h=v,however,isthetimetakenforNballstostrikethesurface,sothat2FNmvNmv22KP====:AAtAhV167khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE38-42F=IA=c;P=IA;a=F=m;andv=at.Combine:v=Pt=mc=(10103W)(86400s)=(1500kg)(3108m=s)=1:9103m=s:E38-43TheforceofradiationonthebottomofthecylinderisF=2IA=c.TheforceofgravityonthecylinderisW=mg=HAg:Equating,2I=c=Hg.TheintensityofthebeamisgivenbyI=4P=d2:SolvingforH,8P8(4:6W)7H===4:910m:cgd2(3:0108m=s)(1200kg=m3)(9:8m=s2)(2:6103m)2E38-44F=2IA=c.ThevalueforIisinEx.38-37,amongotherplaces.ThenF=(1:38103W=m2)(3:1106m2)=(3:00108m=s)=29N:P38-1khdaw.comForthetwooutercirclesuseEq.33-13.FortheinnercircleuseE=V=d,Q=CV,C=0A=d,andi=dQ=dt.ThendQ0AdVdEi===0A:dtddtdtThechangein uxisdE=dt=AdE=dt.ThenIB~
d~l=dE=i;000dtsoB=0i=2r.P38-2(a)i=i.Assuming
V=(174103V)sin!t,thenq=C
Vandi=dq=dt=Cd(
V)=dt.dwww.khdaw.comCombine,anduse!=2(50:0=s),i=(1001012F)(174103V)2(50:0=s)=5:47103A:dP38-3(a)i=id=7:63A.(b)d=dt=i==(7:63A)=(8:851012F=m)=8:62105V=m:Ed0(c)i=dq=dt=Cd(
V)=dt;C=0A=d;[d(
V)=dt]m=Em!.Combine,andAAE!(8:851012F=m)(0:182m)2(225V)(128rad/s)d=0=0m==3:48103m:Ci(7:63A)R课后答案网RP38-4(a)q=idt=tdt=t2=2.(b)E===q=A=t2=2R2.000(d)2rB=r2dE=dt,so00B=r(dE=dt)=2=rt=2R2:00(e)CheckExercise38-10!P38-5(a)E~=E^jandB~=Bk^.Then~S=E~B~=0,or~S=EB=mu^i:0Energyonlypassesthroughtheyzfaces;itgoesinonefaceandouttheother.TherateisP=SA=EBa2=mu.0(b)Thenetchangeiszero.khdaw.com168若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP38-6(a)ForasinusoidaltimedependencejdE=dtjm=!Em=2fEm.ThenjdE=dtj=2(2:4109=s)(13103V=m)=1:961014V=m
s:m(b)Usingtheresultofpart(b)ofSampleProblem38-1,171221145B=(410H=m)(8:910F=m)(2:410m)(1:9610V=m
s)=1:310T:22P38-7LookbacktoChapter14foradiscussionontheellipticorbit.Onpage312itispointedoutthattheclosestdistancetothesunisRp=a(1e)whilethefarthestdistanceisRa=a(1+e),whereaisthesemi-majoraxisandetheeccentricity.Thefractionalvariationinintensityis
IIpIa;IIaIpkhdaw.com=1;IaR2a=1;R2p(1+e)2=1:(1e)2Weneedtoexpandthisexpressionforsmalleusing(1+e)21+2e;and(1e)21+2e;andnally(1+2e)21+4e:Combining,
I2(1+2www.khdaw.come)14e:IP38-8Thebeamradiusgrowsasr=(0:440rad)R,whereRisthedistancefromtheorigin.ThebeamintensityisP(3850W)2I===4:310W:r2(0:440rad)2(3:82108m)2P38-9Eq.38-14requires@E@B课后答案网=;@x@tEmkcoskxsin!t=Bm!coskxsin!t;Emk=Bm!:Eq.38-17requires@E@B00=;@t@x00Em!sinkxcos!t=Bmksinkxcos!t;00Em!=Bmk:Dividingoneexpressionbytheother,k2=!2;00khdaw.com169若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comor!1=c=pk00.Notonlythat,butEm=cBm.You"veseenanexpressionsimilartothisbefore,andyou"llseeexpressionssimilartoitagain.(b)We"llassumethatEq.38-21isapplicablehere.Then1EmBmS==sinkxsin!tcoskxcos!t;00E2m=sin2kxsin2!t40cisthemagnitudeoftheinstantaneousPoyntingvector.(c)Thetimeaveragedpower owacrossanysurfaceisthevalueofZTZ1khdaw.com~S
dA~dt;T0whereTistheperiodoftheoscillation.We"lljustglossoveranyconcernsaboutdirection,andassumethatthe~Swillbeconstantindirectionsothatwewill,atmost,needtoconcernourselvesaboutaconstantfactorcos.Wecanthendealwithascalar,insteadofvector,integral,andwecanintegrateitinanyorderwewant.Wewanttodothetintegrationrst,becauseanintegraloversin!tforaperiodT=2=!iszero.Thenwearedone!(d)Thereisnoenergy ow;theenergyremainsinsidethecontainer.P38-10(a)Theelectriceldisparalleltothewireandgivenbywww.khdaw.com2E=V=d=iR=d=(25:0A)(1:00=300m)=8:3310V=m(b)Themagneticeldisinringsaroundthewire.UsingEq.33-13,i(4107H=m)(25A)B=0==4:03103T:2r2(1:24103m)(c)S=EB=0,soS=(8:33102V=m)(4:03103T)=(4107H=m)=267W=m2:课后答案网P38-11(a)We"vealreadycalculatedBpreviously.Itis0iEB=wherei=:2rRTheelectriceldofalongstraightwirehastheformE=k=r,wherekissomeconstant.ButZZb
V=E~
d~s=Edr=kln(b=a):aInthisproblemtheinnerconductorisatthehigherpotential,so
VEk==;ln(b=a)ln(b=a)170khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comandthentheelectriceldisEE=:rln(b=a)Thisisalsoavectoreld,andifEispositivetheelectriceldpointsradiallyoutfromthecentralconductor.(b)ThePoyntingvectoris1~S=E~B~;0E~isradialwhileB~iscircular,sotheyareperpendicular.AssumingthatEispositivethedirectionof~Sisawayfromthebattery.SwitchingthesignofE(connectingthebatteryinreverse)will ipthedirectionofbothE~andB~,so~Swillpickuptwonegativesignsandthereforestillpointawayfromthebattery.ThemagnitudeisEBE2S==02Rln(b=a)r2khdaw.com(c)WewanttoevaluateasurfaceintegralinpolarcoordinatesandsodA=(dr)(rd).Wehavealreadyestablishedthat~Sispointingawayfromthebatteryparalleltothecentralaxis.ThenwecanintegrateZZP=~S
dA~=SdA;ZbZ22E=drdr;a02Rln(b=a)r2Zb2E=dr;aRln(b=a)rE2www.khdaw.com=:R(d)Readpart(b)above.P38-12(a)B~isorientedasringsaroundthecylinder.Ifthethumbisinthedirectionofcurrentthenthengersoftherighthandgripionthedirectionofthemagneticeldlines.E~isdirectedparalleltothewireinthedirectionofthecurrent.~Sisfoundfromthecrossproductofthesetwo,andmustbepointingradiallyinward.(b)ThemagneticeldonthesurfaceisgivenbyEq.33-13:课后答案网B=0i=2a:TheelectriceldonthesurfaceisgivenbyE=V=l=iR=lThenShasmagnitudeiiRi2RS=EB=0==:2al2alR~S
dA~isonlyevaluatedonthesurfaceofthecylinder,nottheendcaps.~SiseverywhereparalleltodA~,sothedotproductreducestoSdA;Sisuniform,soitcanbebroughtoutoftheintegral;RdA=2alonthesurface.Hence,Z~S
dA~=i2R;asitshould.khdaw.com171若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP38-13(a)f=vlambda=(3:00108m=s)=(3:18m)=9:43107Hz.(b)B~mustbedirectedalongthezaxis.ThemagnitudeisB=E=c=(288V=m)=(3:00108m=s)=9:6107T:(c)k=2==2=(3:18m)=1:98=mwhile!=2f,so!=2(9:43107Hz)=5:93108rad/s:(d)I=EmBm=20,so(288V)(9:6107T)I==110W:2(4107H=m)(e)P=I=c=(110W)=(3:00108m=s)=3:67107Pa.P38-14(a)B~isorientedasringsaroundthecylinder.Ifthethumbisinthedirectionofcurrentthenthengersoftherighthandgripionthedirectionofthemagneticeldlines.khdaw.comE~isdirectedparalleltothewireinthedirectionofthecurrent.~Sisfoundfromthecrossproductofthesetwo,andmustbepointingradiallyinward.(b)ThemagnitudeoftheelectriceldisVQQitE====:dCd0A0AThemagnitudeofthemagneticeldontheoutsideoftheplatesisgivenbySampleProblem38-1,00RdE00iR00RB===E:2dtwww.khdaw.com20A2t~ShasmagnitudeEB0R2S==E:02tIntegrating,ZRE2~S
dA~=0E22Rd=Ad0:2ttButEislinearint,sod(E2)=dt=2E2=t;andthenZ~S
dA~=Add1E2:0课后答案网dt2P38-15(a)I=P=A=(5:00103W)=(1:05)2(633109m)2=3:6109W=m2.(b)p=I=c=(3:6109W=m2)=(3:00108m=s)=12Pa(c)F=pA=P=c=(5:00103W)=(3:00108m=s)=1:671011N.(d)a=F=m=F=V,so(1:671011N)a==2:9103m=s2:4(4880kg=m3)(1:05)3(633109)3=3172khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP38-16TheforcefromthesunisF=GMm=r2.Theforcefromradiationpressureis2IA2PAF==:c4r2cEquating,4GMmA=;2P=cso4(6:671011N
m2=kg2)(1:991030kg)(1650kg)A==1:06106m2:2(3:91026W)=(3:0108m=s)That"saboutonesquarekilometer.khdaw.comwww.khdaw.com课后答案网173khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE39-1Bothscalesarelogarithmic;chooseanydatapointfromtherighthandsidesuchasc=f(1Hz)(3108m)=3108m=s;andanotherfromthelefthandsidesuchasc=f(11021Hz)(31013m)=3108m=s:E39-2(a)f=v==(3:0108m=s)=(1:0104)(6:37106m)=4:7103Hz:Ifweassumethatthisisthedatatransmissionrateinbitspersecond(agenerousassumption),thenitwouldtake140daystodownloadaweb-pagewhichwouldtakeonly1secondona56Kmodem!(b)T=1=f=212s=3:5min.E39-3(a)Applyv=f.Thenkhdaw.comf=(3:0108m=s)=(0:0671015m)=4:51024Hz:(b)=(3:0108m=s)=(30Hz)=1:0107m:E39-4Don"tsimplytakereciprocaloflinewidth!f=c=,sof=(c=2).Ignorethenegative,andf=(3:00108m=s)(0:010109m)=(632:8109m)2=7:5109Hz:E39-5(a)WerefertoFig.39-6toanswerthisquestion.Thelimitsareapproximately520nmand620nm.(b)Thewavelengthforwhichtheeyeismostsensitiveis550nm.Thiscorrespondstotoafrequencyofwww.khdaw.comf=c==(3:00108m=s)=(550109m)=5:451014Hz:ThisfrequencycorrespondstoaperiodofT=1=f=1:831015s.E39-6f=c=.Thenumberofcompletepulsesisft,orft=ct==(3:00108m=s)(4301012s)=(520109m)=2:48105:E39-7(a)2(4:34y)=8:68y.(b)2(2:2106y)=4:4106y.课后答案网E39-8(a)t=(150103m)=(3108m=s)=5104s.(b)Thedistancetraveledbythelightis(1:51011m)+2(3:8108m),sot=(1:511011m)=(3108m=s)=503s:(c)t=2(1:31012m)=(3108m=s)=8670s.(d)105465005400BC.E39-9Thisisaquestionofhowmuchtimeittakeslighttotravel4cm,becausethelighttraveledfromtheEarthtothemoon,bouncedoofthere ector,andthentraveledback.Thetimetotravel4cmis
t=(0:04m)=(3108m=s)=0:13ns.NotethatIinterpretedthequestiondierentlythantheanswerinthebackofthebook.174khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE39-10Consideranyincomingray.Thepathoftheraycanbeprojectedontothexyplane,thexzplane,ortheyzplane.Iftheprojectedraysisexactlyre ectedinallthreecasesthenthethreedimensionalincomingraywillbere ectedexactlyreversed.Buttheproblemissymmetric,soitissucienttoshowthatanyplaneworks.NowtheproblemhasbeenreducedtoSampleProblem39-2,sowearedone.E39-11Wewillchoosethemirrortolieinthexyplaneatz=0.Thereisnolossofgeneralityindoingso;wehadtodeneourcoordinatesystemsomehow.Thechoiceisconvenientinthatanynormalisthenparalleltothezaxis.Furthermore,wecanarbitrarilydenetheincidentraytooriginateat(0;0;z1).Lastly,wecanrotatethecoordinatesystemaboutthezaxissothatthere ectedraypassesthroughthepoint(0;y3;z3).Thepointofre ectionforthisrayissomewhereonthesurfaceofthemirror,say(x2;y2;0).Thisdistancetraveledfromthepoint1tothere ectionpoint2ispqkhdaw.comd12=(0x2)2+(0y2)2+(z10)2=x22+y22+z12andthedistancetraveledfromthere ectionpoint2tothenalpoint3ispqd23=(x20)2+(y2y3)2+(0z3)2=x22+(y2y3)2+z32:Theonlypointwhichisfreetomoveisthere ectionpoint,(x2;y2;0),andthatpointcanonlymoveinthexyplane.Fermat"sprinciplestatesthatthere ectionpointwillbesuchtominimizethetotaldistance,qqd12+d23=x22+y22+z12+x22+(y2y3)2+z32:Wedothisminimizationbytakingthepartialderivativewithrespecttobothx2andy2.Butwecandopartbyinspectionalone.Anynon-zerovalueofwww.khdaw.comx2canonlyaddtothetotaldistance,regardlessofthevalueofanyoftheotherquantities.Consequently,x2=0isoneoftheconditionsforminimization.Wearedone!Althoughyouareinvitedtonishtheminimizationprocess,onceweknowthatx2=0wehavethatpoint1,point2,andpoint3alllieintheyzplane.Thenormalisparalleltothezaxis,soitalsoliesintheyzplane.Everythingisthenintheyzplane.E39-12RefertoPage442ofVolume1.E39-13(a)=38.1(b)(1:58)sin(38)=(1:22)sin.Then=arcsin(0:797)=52:9.课后答案网22E39-14n=nsin=sin=(1:00)sin(32:5)=sin(21:0)=1:50.gv12E39-15n=c=v=(3:00108m=s)=(1:92108m=s)=1:56.E39-16v=c=n=(3:00108m=s)=(1:46)=2:05108m=s.E39-17Thespeedoflightinasubstancewithindexofrefractionnisgivenbyv=c=n.AnelectronwillthenemitCerenkovradiationinthisparticularliquidifthespeedexceedsv=c=n=(3:00108m=s)=(1:54)=1:95108m=s:175khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE39-18Sincet=d=v=nd=c,
t=
nd=c.Then
t=(1:000291:00000)(1:61103m)=(3:00108m=s)=1:56109s:E39-19Theangleoftherefractedrayis=90,theangleoftheincidentraycanbefoundby2trigonometry,(1:14m)tan1==1:34;(0:85m)or=53:3.1Wecanusethesetwoangles,alongwiththeindexofrefractionofair,tondthattheindexofrefractionoftheliquidfromEq.39-4,sin(sin90)2n1=n2=(1:00)=1:25:sin1(sin53:3)Therearenounitsattachedtothisquantity.E39-20khdaw.comForanequilateralprism=60.Thensin[+]=2sin[(37)+(60)]=2n===1:5:sin[=2]sin[(60)=2]E39-21p2E39-22t=d=v;butL=d=cos2=1sin2andv=c=n.Combining,nLn2L(1:63)2(0:547m)t=p=p=q=3:07109s:c1sin2cn2sin2www.khdaw.com82221(310m=s)(1:63)sin(24)E39-23Therayoflightfromthetopofthesmokestacktotheliferingis1,wheretan1=L=hwithhtheheightandLthedistanceofthesmokestack.Snell"slawgivesn1sin1=n2sin2,so=arcsin[(1:33)sin(27)=(1:00)]=37:1:1ThenL=htan=(98m)tan(37:1)=74m:1E39-24Thelengthoftheshadowonthesurfaceofthewateris课后答案网x=(0:64m)=tan(55)=0:448m:1Therayoflightwhichformstheend"oftheshadowhasanangleofincidenceof35,sotheraytravelsintothewateratanangleof(1:00)2=arcsinsin(35)=25:5:(1:33)Theraytravelsanadditionaldistancex=(2:00m0:64m)=tan(9025:5)=0:649m2Thetotallengthoftheshadowis(0:448m)+(0:649m)=1:10m:khdaw.com176若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE39-25We"llrelyheavilyonthegureforourarguments.Letxbethedistancebetweenthepointsonthesurfacewheretheverticalraycrossesandthebentraycrosses.q2xdappq1dkhdaw.comInthisexercisewewilltakeadvantageofthefactthat,forsmallangles,sintanInthisapproximationSnell"slawtakesontheparticularlysimpleformn11=n22Thetwoangleshereareconvenientlyfoundfromthegure,x1tan1=;dandx2tan2=:dappInsertingthesetwoanglesintothesimpliedSnell"slaw,aswellassubstitutingwww.khdaw.comn1=nandn2=1:0,n11=n22;xxn=;ddappddapp=:nE39-26(a)Youneedtoaddresstheissueoftotalinternalre ectiontoanswerthisquestion.(b)Rearrangesin[+]=2课后答案网n=sin[=2]=and=(+)=2toget=arcsin(nsin[=2])=arcsin((1:60)sin[(60)=2])=53:1:E39-27UsetheresultsofEx.39-35.Theapparentthicknessofthecarbontetrachloridelayer,asviewedbyanobserverinthewater,isdc;w=nwdc=nc=(1:33)(41mm)=(1:46)=37:5mm:Thetotalthickness"fromthewaterperspectiveisthen(37:5mm)+(20mm)=57:5mm.Theapparentthicknessoftheentiresystemasviewfromtheairisthendapp=(57:5mm)=(1:33)=43:2mmkhdaw.com:177若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE39-28(a)UsetheresultsofEx.39-35.dapp=(2:16m)=(1:33)=1:62m.(b)Needadiagramhere!E39-29(a)n==n=(612nm)=(1:51)=405nm:(b)L=nLn=(1:51)(1:57pm)=2:37pm.Thereisactuallyatypo:thep"inpm"wassupposedtobea.Thismakesahugedierenceforpart(c)!E39-30(a)f=c==(3:00108m=s)=(589nm)=5:091014Hz:(b)n==n=(589nm)=(1:53)=385nm:(c)v=f=(5:091014Hz)(385nm)=1:96108m=s:E39-31(a)ThesecondderivativeofppL=a2+x2+b2+(dx)2iskhdaw.coma2(b2+(d2)2)3=2+b2(a2+x2)3=2:(b2+(d2)2)3=2(a2+x2)3=2Thisisalwaysapositivenumber,sodL=dx=0isaminimum.(a)ThesecondderivativeofppL=n1a2+x2+n2b2+(dx)2isna2(b2+(d2)2)3=2+nb2(a2+x2)3=212:(b2+(d2)2)3=2(a2+x2)3=2www.khdaw.comThisisalwaysapositivenumber,sodL=dx=0isaminimum.E39-32(a)Theangleofincidenceonthefaceacwillbe90.Totalinternalre ectionoccurswhensin(90)>1=n,or<90arcsin[1=(1:52)]=48:9:(b)Totalinternalre ectionoccurswhensin(90)>n=n,orw<90arcsin[(1:33)=(1:52)]=29:0:课后答案网E39-33(a)ThecriticalangleisgivenbyEq.39-17,1n21(1:586)c=sin=sin=72:07:n1(1:667)(b)Criticalanglesonlyexistwhenattempting"totravelfromamediumofhigherindexofrefractiontoamediumoflowerindexofrefraction;inthiscasefromAtoB.E39-34Ifthereisatthewater"sedgethenthelighttravelsalongthesurface,enteringthewaterneartheshwithanangleofincidenceofeectively90.Thentheangleofrefractioninthewaterisnumericallyequivalenttoacriticalangle,sotheshneedstolookupatanangleof=arcsin(1=1:33)=49withthevertical.That"sthesameas41withthehorizontal.178khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE39-35Lightcanonlyemergefromthewaterifithasanangleofincidencelessthanthecriticalangle,or<=arcsin1=n=arcsin1=(1:33)=48:8:cTheradiusofthecircleoflightisgivenbyr=d=tanc,wheredisthedepth.Thediameteristwicethisradius,or2(0:82m)tan(48:8)=1:87m:E39-36Therefractedangleisgivenbynsin=sin(39).Thisraystrikestheleftsurfacewith1anangleofincidenceof90.Totalinternalre ectionoccurswhen1sin(90)=1=n;1butsin(90)=cos,sowecancombineandgettan=sin(39)withsolution=32:2:The111indexofrefractionoftheglassisthenkhdaw.comn=sin(39)=sin(32:2)=1:18:E39-37Thelightstrikesthequartz-airinterfacefromtheinside;itisoriginallywhite",soifthere ectedrayistoappearbluish"(reddish)thentherefractedrayshouldhavebeenreddish"(bluish).Sincepartofthelightundergoestotalinternalre ectionwhiletheotherpartdoesnot,thentheangleofincidencemustbeapproximatelyequaltothecriticalangle.(a)LookatFig.39-11,theindexofrefractionoffusedquartzisgivenasafunctionofthewavelength.Asthewavelengthincreasestheindexofrefractiondecreases.Thecriticalangleisafunctionoftheindexofrefraction;forasubstanceinairthecriticalangleisgivenbysinc=1=n:Asndecreases1=nincreasessocincreases.Forfusedquartz,then,aswavelengthincreasescalsoincreases.Inshort,redlighthasalargercriticalanglethanbluelight.Iftheangleofincidenceismidwaybetweenthecriticalangleofredandthecriticalangleofblue,thenthebluecomponentofthelightwww.khdaw.comwillexperiencetotalinternalre ectionwhiletheredcomponentwillpassthroughasarefractedray.Soyes,thelightcanbemadetoappearbluish.(b)No,thelightcan"tbemadetoappearreddish.Seeabove.(c)Chooseanangleofincidencebetweenthetwocriticalanglesasdescribedinpart(a).Usingavalueofn=1:46fromFig.39-11,=sin1(1=1:46)=43:2:cGettingtheeecttoworkwillrequireconsiderablesensitivity.E39-38(a)Thereneedstobeanopaquespotinthecenterofeachfacesothatnorefractedray课后答案网emerges.Theradiusofthespotwillbelargeenoughtocoverrayswhichmeetthesurfaceatlessthanthecriticalangle.Thismeanstanc=r=d,wheredisthedistancefromthesurfacetothespot,or6.3mm.Since=arcsin1=(1:52)=41:1;cthenr=(6:3mm)tan(41:1)=5:50mm:(b)Thecircleshaveanareaofa=(5:50mm)2=95:0mm2.Eachsidehasanareaof(12:6mm)2;thefractioncoveredisthen(95:0mm2)=(12:6mm)2=0:598:E39-39FoructherelativisticDopplershiftsimpliesto
f=f0u=c=u=0;sou=0
f=(0:211m)
f:khdaw.com179若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE39-40c=f,so0=f
+
f.Then
==
f=f.Furthermore,f0f,fromEq.39-21,isf0u=cforsmallenoughu.Then
ff0u==:f0cE39-41TheDopplertheoryforlightgives1u=c1(0:2)f=f0p=f0p=0:82f0:1u2=c21(0:2)2Thefrequencyisshifteddowntoabout80%,whichmeansthewavelengthisshiftedupbyanadditional25%.Bluelight(480nm)wouldappearyellow/orange(585nm).E39-42UseEq.39-20:1u=c1(0:892)f=f0p=(100Mhz)p=23:9MHz:khdaw.com1u2=c21(0:892)2E39-43(a)Ifthewavelengthisthreetimeslongerthenthefrequencyisone-third,sofortheclassicalDopplershiftf0=3=f0(1u=c);oru=2c.(b)Fortherelativisticshift,1u=cf0=3=f0p;1u2=c2p1u2=c2=www.khdaw.com3(1u=c);c2u2=9(cu)2;0=10u218uc+8c2:Thesolutionisu=4c=5.E39-44(a)f0=f==0.Thisshiftissmall,soweapplytheapproximation:08(462nm)7u=c1=(310m=s)1=1:910m=s:(434nm)(b)Aredshiftcorrespondstoobjectsmovingawayfromus.课后答案网E39-45Thesunrotatesonceevery26daysattheequator,whiletheradiusis7:0108m.Thespeedofapointontheequatoristhen2R2(7:0108m)v===2:0103m=s:T(2:2106s)Thiscorrespondstoavelocityparameterof=u=c=(2:0103m=s)=(3:0108m=s)=6:7106:Thisisacaseofsmallnumbers,sowe"llusetheformulathatyouderivedinExercise39-40:
==(553nm)(6:7106)=3:7103nm:180khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE39-46UseEq.39-23writtenas(1u=c)2=2(1+u=c);0whichcanberearrangedas22(540nm)2(620nm)20u=c===0:137:2+20(540nm)2+(620nm)2Thenegativesignmeansthatyoushouldbegoingtowardtheredlight.E39-47(a)f1=cf=(c+v)andf2=cf=(cv).
f=(f2f)(ff1)=f2+f12f;so
fcc=+2;khdaw.comfc+vcv2v2=;c2v22(8:65105m=s)2=;(3:00108m=s)2(8:65105m=s)2=1:66105:p(b)f=f(cu)=sqrtc2u2andf=f(c+u)=c2u2.12
f=(f2f)(ff1)=f2+f12f;sowww.khdaw.com
f2c=p2;fc2u22(3:00108m=s)=p2;(3:00108m=s)2(8:65105m=s)2=8:3106:E39-48(a)Norelativemotion,soevery6minutes.(b)TheDopplereectatthisspeedis课后答案网1u=c1(0:6)p=p=0:5;1u2=c21(0:6)2thismeansthefrequencyisonehalf,sotheperiodisdoubledto12minutes.(c)IfCsendthesignalattheinstantthesignalfromApasses,thenthetwosignalstraveltogethertoC,soCwouldgetB"ssignalsatthesameratethatitgetsA"ssignals:everysixminutes.E39-49pE39-50ThetransverseDopplereectis=0=1u2=c2.Thenp=(589:00nm)=1(0:122)2=593:43nm:Theshiftis(593:43nm)(589:00nm)=4:43nm.khdaw.com181若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE39-51Thefrequencyobservedbythedetectorfromtherstsourceis(Eq.39-31)pf=f11(0:717)2=0:697f1:Thefrequencyobservedbythedetectorfromthesecondsourceis(Eq.39-30)p1(0:717)20:697f2f=f2=:1+(0:717)cos1+(0:717)cosWeneedtoequatetheseandsolvefor.Then0:697f20:697f1=;1+0:717cos1+0:717cos=f2=f1;cos=(f2=f11)=0:717;=101:1:Subtractfrom180khdaw.comtondtheanglewiththelineofsight.E39-52P39-1ConsiderthetriangleinFig.39-45.Thetruepositioncorrespondstothespeedoflight,theoppositesidecorrespondstothevelocityofearthintheorbit.Then=arctan(29:8103m=s)=(3:00108m=s)=20:500:P39-2ThedistancetoJupiterfrompointxisdx=rjre.ThedistancetoJupiterfrompointyisqd2=re2+r2:www.khdaw.comjThedierenceindistanceisrelatedtothetimeaccordingto(d2d1)=t=c;sop(778109m)2+(150109m)2(778109m)+(150109m)=2:7108m=s:(600s)P39-3sin(30)=(4:0m=s)=sin=(3:0m=s).Then=22:Waterwavestravelmoreslowlyinshallowerwater,whichmeanstheyalwaysbendtowardthenormalastheyapproachland.P39-4(a)Iftherayisnormaltothewater"ssurfacethenitpassesintothewaterunde ected.课后答案网OnceinthewatertheproblemisidenticaltoSampleProblem39-2.There ectedrayinthewaterisparalleltotheincidentrayinthewater,soitalsostrikesthewaternormal,andistransmittednormal.(b)Assumetheraystrikesthewateratanangle1.Itthenpassesintothewateratanangle2,wherenwsin2=nasin1:OncetherayisinthewaterthentheproblemisidenticaltoSampleProblem39-2.There ectedrayinthewaterisparalleltotheincidentrayinthewater,soitalsostrikesthewateratanangle2.Whentheraytravelsbackintotheairittravelswithanangle3,wherenwsin2=nasin3:Comparingthetwoequationsyields1=3,sotheoutgoingrayintheairisparalleltotheincomingray.khdaw.com182若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP39-5(a)AswasdoneinEx.39-25aboveweusethesmallangleapproximationofsintanTheincidentangleis;ifthelightweretogoinastraightlinewewouldexpectittostrikeadistancey1beneaththenormalontherighthandside.Thevariousdistancesarerelatedtotheanglebytany1=t:Thelight,however,doesnotgoinastraightline,itisrefractedaccordingto(thesmallangleapproximationto)Snell"slaw,n11=n22;whichwewillsimplifyfurtherbyletting1=,n2=n,andn1=1,=n2:Thepointwheretherefractedraydoesstrikeisrelatedtotheangleby2tan2=y2=t:Combiningthethreeexpressions,y1=ny2:Thedierence,y1y2istheverticaldistancebetweenthedisplacedrayandtheoriginalrayasmeasuredontheplateglass.Alittlealgebrayieldskhdaw.comy1y2=y1y1=n;=y1(11=n);n1=t:nTheperpendiculardistancexisrelatedtothisdierencebycos=x=(y1y2):Inthesmallangleapproximationcos12=2.Ifwww.khdaw.comissucientlysmallwecanignorethesquareterm,andxy2y1.(b)Remembertouseradiansandnotdegreeswheneverthesmallangleapproximationisapplied.Then(1:52)1x=(1:0cm)(0:175rad)=0:060cm:(1:52)P39-6(a)Atthetoplayer,n1sin1=sin;atthenextlayer,课后答案网n2sin2=n1sin1;atthenextlayer,n3sin3=n2sin2:Combiningallthreeexpressions,n3sin3=sin:(b)=arcsin[sin(50)=(1:00029)]=49:98.Thenshiftis(50)(49:98)=0:02.3P39-7Thebigidea"ofProblem6isthatwhenlighttravelsthroughlayerstheanglethatitmakesinanylayerdependsonlyontheincidentangle,theindexofrefractionwherethatincidentangleoccurs,andtheindexofrefractionatthecurrentpoint.Thatmeansthatlightwhichleavesthesurfaceoftherunwayat90tothenormalwillmakeananglensin90=n(1+ay)sin00khdaw.com183若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comatsomeheightyabovetherunway.Itismildlyentertainingtonotethatthevalueofn0isunim-portant,onlythevalueofa!Theexpression1sin=1ay1+aycanbeusedtondtheanglemadebythecurvedpathagainstthenormalasafunctionofy.Theslopeofthecurveatanypointisgivenbydycos=tan(90)=cot=:dxsinNowweneedtoknowcos.Itispp2cos=1sin2ay:Combiningpdy2ay;khdaw.comdx1ayandnowweintegrate.Wewillignoretheayterminthedenominatorbecauseitwillalwaysbesmallcomparedto1.ThenZdZhdydx=p;002ayrs2h2(1:7m)d===1500m:a(1:5106m1)P39-8TheenergyofaparticleisgivenbyE2=www.khdaw.comp2c2+m2c4.ThisenergyisrelatedtothemasspbyE= mc2.isrelatedtothespeedby=1=1u2=c2.Rearranging,rsu1m2c2=1=1;c2p2+m2c2sp2=:p2+m2c2Sincen=c=uwecanwritethisas课后答案网ssm2c2mc22n=1+=1+:p2pcForthepion,s2(135MeV)n=1+=1:37:(145MeV)Forthemuon,s2(106MeV)n=1+=1:24:(145MeV)184khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP39-9(a)Beforeaddingthedropofliquidprojectthelightrayalongtheanglesothat=0.Increaseslowlyuntiltotalinternalre ectionoccursatangle1.Thenngsin1=1istheequationwhichcanbesolvedtondng.Nowputtheliquidontheglassandrepeattheaboveprocessuntiltotalinternalre ectionoccursatangle2.Thenngsin2=nl:Notethatng0.(d)Rearrangetheformulatosolveforn2,then111n21i=n1+:rroSubstitutingthenumbers,1111n2=(1:0)+;(20)(20)(20)(20)whichhasanysolutionforn2!Sincei<0theimageisvirtual.(e)(1:5)(1:0)+=0:016667;(10)(6)sor=(1:01:5)=(0:016667)=30,andtheimageisvirtual.khdaw.com(f)(1:0)(1:5)(1:0)=0:15;(30)(7:5)soo=(1:5)=(0:15)=10.Theimagewasvirtualsincei<0.(g)(1:0)(1:5)(1:5)2=3:8110;(30)(70)soi=(1:0)=(3:81102)=26,andtheimageisvirtual.(h)SolvingEq.40-10forn2yields1www.khdaw.com=o+1=rn2=n1;1=r1=iso1=(100)+1=(30)n2=(1:5)=1:01=(30)1=(600)andtheimageisreal.E40-19(b)IfthebeamissmallwecanuseEq.40-10.Parallelincomingrayscorrespondtoanobjectatinnity.Solvingforn2yields课后答案网1=o+1=rn2=n1;1=r1=isoifo!1andi=2r,then1=1+1=rn2=(1:0)=2:01=r1=2r(c)Thereisnosolutionifi=r!E40-20Theimagewillbelocatedatapointgivenby111111===:ifo(10cm)(6cm)(15cm)192khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE40-21TheimagelocationcanbefoundfromEq.40-15,111111===;ifo(30cm)(20cm)12cmsotheimageislocated12cmfromthethinlens,onthesamesideastheobject.E40-22Foradoubleconvexlensr1>0andr2<0(seeFig.40-21andtheaccompanyingtext).Thentheproblemstatesthatr2=r1=2.Thelensmaker"sequationcanbeappliedtoget1113(n1)=(n1)=;fr1r2r1sor1=3(n1)f=3(1:51)(60mm)=90mm,andr2=45mm:E40-23Theobjectdistanceisessentiallyo=1,so1=f=1=o+1=iimpliesf=i,andtheimageformsatthefocalpoint.Inreality,however,theobjectdistanceisnotinnite,sothemagnicationisgivenbykhdaw.comm=i=of=o,whereoistheEarth/Sundistance.Thesizeoftheimageisthenh=hf=o=2(6:96108m)(0:27m)=(1:501011m)=2:5mm:ioThefactoroftwoisbecausethesun"sradiusisgiven,andweneedthediameter!E40-24(a)The atsidehasr2=1,so1=f=(n1)=r,whereristhecurvedside.Thenf=(0:20m)=(1:51)=0:40m.(b)1=i=1=f1=o=1=(0:40m)1=(0:40m)=0.Theniis1.E40-25(a)1=f=(1:51)[1=(0:4m)1=(0:4m)]=1=(0:40m).(b)1=f=(1:51)[1=(1)1=(0:4m)]=1=www.khdaw.com(0:80m).(c)1=f=(1:51)[1=(0:4m)1=(0:6m)]=1=(2:40m).(d)1=f=(1:51)[1=(0:4m)1=(0:4m)]=1=(0:40m).(e)1=f=(1:51)[1=(1)1=(0:8m)]=1=(0:80m).(f)1=f=(1:51)[1=(0:6m)1=(0:4m)]=1=(2:40m).E40-26(a)1=f=(n1)[1=(r)1=r],so1=f=2(1n)=r.1=i=1=f1=osoifo=r,then1=i=2(1n)=r1=r=(12n)=r;soi=r=(12n).Forn>0:5theimageisvirtual.(b)Forn>0:5theimageisvirtual;themagnicationis课后答案网m=i=o=r=(12n)=r=1=(2n1):E40-27Accordingtothedenitions,o=f+xandi=f+x0.StartingwithEq.40-15,111+=;oifi+o1=;oif2f+x+x01=;(f+x)(f+x0)f2f2+fx+fx0=f2+fx+fx0+xx0;f2=xx0:193khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE40-28(a)Youcan"tdeterminer1,r2,orn.iisfoundfrom1111==;i+10+20+20theimageisrealandinverted.m=(20)=(20)=1:(b)Youcan"tdeterminer1,r2,orn.Thelensisconvergingsincefispositive.iisfoundfrom1111==;i+10+510theimageisvirtualandupright.m=(10)=(+5)=2:(c)Youcan"tdeterminer1,r2,orn.Sincemispositiveandgreaterthanonethelensisconverging.Thenfispositive.iisfoundfrom1111==;i+10+510theimageisvirtualandupright.khdaw.comm=(10)=(+5)=2:(d)Youcan"tdeterminer1,r2,orn.Sincemispositiveandlessthanonethelensisdiverging.Thenfisnegative.iisfoundfrom1111==;i10+53:3theimageisvirtualandupright.m=(3:3)=(+5)=0:66:(e)fisfoundfrom1111=(1:51)=:f+3030+30Thelensisconverging.iisfoundfromwww.khdaw.com1111==;i+30+1015theimageisvirtualandupright.m=(15)=(+10)=1:5:(f)fisfoundfrom1111=(1:51)=:f30+3030Thelensisdiverging.iisfoundfrom1111课后答案网==;i30+107:5theimageisvirtualandupright.m=(7:5)=(+10)=0:75:(g)fisfoundfrom1111=(1:51)=:f3060120Thelensisdiverging.iisfoundfrom1111==;i120+109:2theimageisvirtualandupright.m=(9:2)=(+10)=0:92:(h)Youcan"tdeterminer1,r2,orn.Uprightimageshavepositivemagnication.iisfoundfromi=(0:5)(10)=5;khdaw.com194若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comfisfoundfrom1111=+=;f+10510sothelensisdiverging.(h)Youcan"tdeterminer1,r2,orn.Realimageshavenegativemagnication.iisfoundfromi=(0:5)(10)=5;fisfoundfrom1111=+=;f+105+3:33sothelensisconverging.E40-29o+i=0:44m=L,so11111L=+=+=;khdaw.comfoioLoo(Lo)whichcanalsobewrittenaso2oL+fL=0.ThishassolutionppLL24fL(0:44m)(0:44m)4(0:11m)(0:44m)o===0:22m:22Thereisonlyonesolutiontothisproblem,butsometimestherearetwo,andothertimestherearenone!E40-30(a)Realimages(fromrealobjects)areonlyproducedbyconverginglenses.(b)Sincehi=h0=2,theni=o=2.Butd=i+www.khdaw.como=o+o=2=3o=2,soo=2(0:40m)=3=0:267m,andi=0:133m.(c)1=f=1=o+1=i=1=(0:267m)+1=(0:133m)=1=(0:0889m).E40-31Stepthroughtheexerciseonelensatatime.Theobjectis40cmtotheleftofaconverginglenswithafocallengthof+20cm.Theimagefromthisrstlenswillbelocatedbysolving111111===;ifo(20cm)(40cm)40cmsoi=40cm.Sinceiispositiveitisarealimage,anditislocatedtotherightoftheconverginglens.Thisimagebecomestheobjectforthediverginglens.课后答案网Theimagefromtheconverginglensislocated40cm-10cmfromthediverginglens,butitislocatedonthewrongside:thediverginglensisintheway"sotherayswhichwouldformtheimagehitthediverginglensbeforetheyhaveachancetoformtheimage.Thatmeansthattherealimagefromtheconverginglensisavirtualobjectinthediverginglens,sothattheobjectdistanceforthediverginglensiso=30cm.Theimageformedbythediverginglensislocatedbysolving111111===;ifo(15cm)(30cm)30cmori=30cm.Thiswouldmeantheimageformedbythediverginglenswouldbeavirtualimage,andwouldbelocatedtotheleftofthediverginglens.Theimageisvirtual,soitisupright.Themagnicationfromtherstlensism1=i=o=(40cm)=(40cm))=khdaw.com1;195若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comthemagnicationfromthesecondlensism2=i=o=(30cm)=(30cm))=1;whichimpliesanoverallmagnicationofm1m2=1.E40-32(a)Theparallelraysoflightwhichstrikethelensoffocallengthfwillconvergeonthefocalpoint.Thispointwillactlikeanobjectforthesecondlens.IfthesecondlensislocatedadistanceLfromtherstthentheobjectdistanceforthesecondlenswillbeLf.NotethatthiswillbeanegativevalueforLy==(2103m)=(1:32104rad)=15m:RE42-22y=D=1:22=a;ory=1:22(500109m)(354103m)=(9:14m=2)=4:73102m:课后答案网E42-23(a)=v=f.NowuseEq.42-11:(1450m=s)=arcsin1:22=6:77:(25103Hz)(0:60m)(b)Followingthesameapproach,(1450m=s)=arcsin1:22(1103Hz)(0:60m)hasnorealsolution,sothereisnominimum.216khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE42-24(a)=v=f.NowuseEq.42-11:(3108m=s)=arcsin1:22=0:173:(220109Hz)(0:55m)Thisistheanglefromthecentralmaximum;theangularwidthistwicethis,or0:35.(b)useEq.42-11:(0:0157m)=arcsin1:22=0:471:(2:33m)Thisistheanglefromthecentralmaximum;theangularwidthistwicethis,or0:94.E42-25Thelinearseparationofthefringesisgivenby
yD=
=or
y=Dddforsucientlysmallkhdaw.comdcomparedto.E42-26(a)dsin=4givesthelocationofthefourthinterferencemaximum,whileasin=givesthelocationoftherstdiractionminimum.Hence,ifd=4atherewillbenofourthinterferencemaximum!(b)Sincedsinmi=migivestheinterferencemaximaandasinmd=mdgivesthediractionminima,andd=4a,thenwhenevermi=4mdtherewillbeamissingmaximum.E42-27(a)Thecentraldiractionenvelopeiscontainedintherange=arcsinwww.khdaw.comaThisanglecorrespondstothemthmaximaoftheinterferencepattern,wheresin=m=d=m=2a:Equating,m=2,sotherearethreeinterferencebands,sincethem=2bandiswashedout"bythediractionminimum.(b)Ifd=athen=andtheexpressionreducesto22sinI=Imcos;课后答案网22sin(2)=Im;2202sin=2Im;0where=20,whichisthesameasreplacingaby2a.E42-28Rememberthatthecentralpeakhasanenvelopewidthtwicethatofanyotherpeak.Ignoringthecentralmaximumthereare(111)=2=5fringesinanyotherpeakenvelope.217khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE42-29(a)Therstdiractionminimumisgivenatananglesuchthatasin=;theorderoftheinterferencemaximumatthatpointisgivenbydsin=m:Dividingoneexpressionbytheotherwegetd=a=m;withsolutionm=(0:150)=(0:030)=5.Thefactthattheanswerisexactly5impliesthatthefthinterferencemaximumissquelchedbythediractionminimum.Thenthereareonlyfourcompletefringesoneithersideofthecentralmaximum.Addthistothecentralmaximumandwegetnineastheanswer.(b)Forthethirdfringem=3,sodsin=3.ThenisEq.42-14is3,whileinEq.42-16isa3a==3;ddsotherelativeintensityofthethirdfringeis,fromEq.42-17,22sin(3a=d)(cos3)=0:255:(3a=d)P42-1khdaw.comy=mD=a.Theny=(10)(632:8109m)(2:65m)=(1:37103m)=1:224102m:Theseparationistwicethis,or2.45cm.P42-2Ifathenthediractionpatternisextremelytight,andthereiseectivelynolightatP.IntheeventthateithershapeproducesaninterferencepatternatPthentheothershapemustproduceanequalbutoppositeelectriceldvectoratthatpointsothatwhenbothpatternsfrombothshapesaresuperimposedtheeldcancel.Buttheintensityistheeldvectorsquared;hencethetwopatternslookidentical.www.khdaw.comP42-3(a)WewanttotakethederivativeofEq.42-8withrespectto,so2dIdsin=Im;ddsincossin=Im2;2sin=Im2(cossin):3Thisequalszerowheneversin课后答案网=0orcos=sin;theformeristhecaseforaminimawhilethelatteristhecaseforthemaxima.Themaximacasecanalsobewrittenastan=:(b)Notethatastheorderofthemaximaincreasesthesolutionsgetcloserandclosertooddintegerstimes=2.Thesolutionsare=0;1:43;2:46;etc.(c)Themvaluesarem==1=2,andcorrespondtom=0:5;0:93;1:96;etc.Thesevalueswillgetcloserandclosertointegersasthevaluesareincreased.218khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP42-4Theoutgoingbeamstrikesthemoonwithacircularspotofradiusr=1:22D=a=1:22(0:69106m)(3:82108m)=(21:3m)=123m:Thelightisnotevenlydistributedoverthiscircle.IfP0isthepowerinthelight,thenZZRP0=Irdrd=2Irdr;0whereRistheradiusofthecentralpeakandIistheangularintensity.Forawecanwritear=D,then2Z=222DsinDP0=2Imd2Im(0:82):a0aThentheintensityatthecenterfallsowithdistancekhdaw.comDas2Im=1:9(a=D)P0Thefractionoflightcollectedbythemirroronthemoonisthen2(21:3m)26P1=P0=1:9(0:10m)=5:610:(0:69106m)(3:82108m)ThefractionoflightcollectedbythemirrorontheEarthisthen2(20:10m)26P2=P1=1:9(1:3m)=5:610:(0:69106m)(3:82www.khdaw.com108m)Finally,P=P=31011.20P42-5(a)Theringisreddishbecauseitoccursattheblueminimum.(b)ApplyEq.42-11forbluelight:d=1:22=sin=1:22(400nm)=sin(0:375)=70m:(c)ApplyEq.42-11forredlight:课后答案网=arcsin(1:22(700nm)=(70m))0:7;whichoccurs3lunarradiifromthemoon.P42-6Thediractionpatternisapropertyofthespeaker,nottheinterferencebetweenthespeak-ers.Thediractionpatternshouldbeunaectedbythephaseshift.Theinterferencepattern,however,shouldshiftupordownasthephaseofthesecondspeakerisvaried.P42-7(a)Themissingfringeat=5isagoodhintastowhatisgoingon.Thereshouldbesomesortofinterferencefringe,unlessthediractionpatternhasaminimumatthatpoint.Thiswouldbetherstminimum,soasin(5)=(440109m)wouldbeagoodmeasureofthewidthofeachslit.Thena=5:05106m.219khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com(b)Ifthediractionpatternenvelopewerenotpresentwecouldexpectthatthefourthinterfer-encemaximabeyondthecentralmaximumwouldoccuratthispoint,andthendsin(5)=4(440109m)yieldingd=2:02105m:(c)ApplyEq.42-17,where=mandaama=sin==m=m=4:ddThenform=1wehave2sin(=4)I1=(7)=5:7;(=4)whileform=2wehave2khdaw.comsin(2=4)I2=(7)=2:8:(2=4)Theseareingoodagreementwiththegure.www.khdaw.com课后答案网220khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE43-1(a)d=(21:5103m)=(6140)=3:50106m.(b)Thereareanumberofanglesallowed:=arcsin[(1)(589109m)=(3:50106m)]=9:7;=arcsin[(2)(589109m)=(3:50106m)]=19:5;=arcsin[(3)(589109m)=(3:50106m)]=30:3;=arcsin[(4)(589109m)=(3:50106m)]=42:3;=arcsin[(5)(589109m)=(3:50106m)]=57:3:E43-2Thedistancebetweenadjacentrulingsis(2)(612109m)d==2:235106m:sin(33:2)Thenumberoflinesisthenkhdaw.comN=D=d=(2:86102m)=(2:235106m)=12;800:E43-3Wewanttondarelationshipbetweentheangleandtheordernumberwhichislinear.We"llplotthedatainthisrepresentation,andthenusealeastsquaresttondthewavelength.Thedatatobeplottedismsinmsin117.60.302-1-17.6-0.302237.30.606-2-37.1-0.603365.20.908-3-65.0-0.906OnmycalculatorIgetthebeststraightlinetaswww.khdaw.com0:302m+8:33104=sin;mwhichmeansthat=(0:302)(1:73m)=522nm:E43-4AlthoughanapproachlikethesolutiontoExercise3shouldbeused,we"llassumethateachmeasurementisperfectanderrorfree.Thenrandomlychoosingthethirdmaximum,dsin(5040109m)sin(20:33)===586109m:课后答案网m(3)E43-5(a)TheprinciplemaximaoccuratpointsgivenbyEq.43-1,sinm=m:dThedierenceofthesineoftheanglebetweenanytwoadjacentordersissinm+1sinm=(m+1)m=:dddUsingtheinformationprovidedwecannddfrom(600109)d===6m:sinm+1sinm(0:30)(0:20)khdaw.com221若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comItdoesn"ttakemuchimaginationtorecognizethatthesecondandthirdordermaximaweregiven.(b)Ifthefourthordermaximaismissingitmustbebecausethediractionpatternenvelopehasaminimumatthatpoint.Anyfourthordermaximashouldhaveoccurredatsin4=0:4.Ifitisadiractionminimathenasinm=mwheresinm=0:4Wecansolvethisexpressionandnd(600109m)a=m=m=m1:5m:sinm(0:4)Theminimumwidthiswhenm=1,ora=1:5m.(c)Thevisibleorderswouldbeintegervaluesofmexceptforwhenmisamultipleoffour.E43-6(a)Findthemaximumintegervalueofm=d==(930nm)=(615nm)=1:5,hencem=1;0;+1;therearethreediractionmaxima.khdaw.com(b)Therstordermaximumoccursat=arcsin(615nm)=(930nm)=41:4:Thewidthofthemaximumis(615nm)4==7:8710rad;(1120)(930nm)cos(41:4)or0:0451.E43-7Thefthordermaximawillbevisibleifd=5;thismeanswww.khdaw.comd(1103m)==635109m:5(315rulings)(5)E43-8(a)Themaximumcouldbetherst,andthendsin(1103m)sin(28)===2367109m:m(200)(1)That"snotvisible.Therstvisiblewavelengthisatm=4,thendsin(1103m)sin(28)课后答案网===589109m:m(200)(4)Thenextisatm=5,thendsin(1103m)sin(28)===469109m:m(200)(5)Tryingm=6resultsinanultravioletwavelength.(b)Yellow-orangeandblue.222khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE43-9Agratingwith400rulings/mmhasaslitseparationof13d==2:510mm:400mm1Tondthenumberofordersoftheentirevisiblespectrumthatwillbepresentweneedonlyconsiderthewavelengthwhichwillbeontheoutsideofthemaxima.Thatwillbethelongerwavelengths,soweonlyneedtolookatthe700nmbehavior.UsingEq.43-1,dsin=m;andusingthemaximumangle90,wendd(2:5106m)m<==3:57;(700109m)sotherecanbeatmostthreeordersoftheentirespectrum.E43-10khdaw.comInthiscased=2a.Sinceinterferencemaximaaregivenbysin=m=dwhilediractionminimaaregivenatsin=m0=a=2m0=dthendiractionminimaoverlapwithinterferencemaximawheneverm=2m0.Consequently,allevenmareatdiractionminimaandthereforevanish.E43-11Ifthesecond-orderspectraoverlapsthethird-order,itisbecausethe700nmsecond-orderlineisatalargeranglethanthe400nmthird-orderline.Startwiththewavelengthsmultipliedbytheappropriateorderparameter,thendividebothsidebyd,andnallyapplyEq.43-1.2(700nm)>www.khdaw.com3(400nm);2(700nm)3(400nm)>;ddsin2;=700>sin3;=400;regardlessofthevalueofd.E43-12Fig.32-2showsthepathlengthdierencefortherighthandsideofthegratingasdsin.Ifthebeamstrikesthegratingatanganglethentherewillbeanadditionalpathlengthdierenceofdsinontherighthandsideofthegure.Thediractionpatternthenhastwocontributionstothepathlengthdierence,theseaddtogive课后答案网d(sin+sinpsi)=m:E43-13E43-14Letdsin=and+20=.Thenii12sin=sincos(20)+cossin(20):211Rearranging,qsin=sincos(20)+1sin2sin(20):211Substitutingtheequationstogetheryieldsarathernastyexpression,21p=cos(20)+1(1=d)2sin(20):ddkhdaw.com223若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comRearranging,
(cos(20))2=d22sin2(20):211Use1=430nmand2=680nm,thensolvefordtondd=914nm.Thiscorrespondsto1090rulings/mm.E43-15Theshortestwavelengthpassesthroughatanangleof=arctan(50mm)=(300mm)=9:46:1Thiscorrespondstoawavelengthof(1103m)sin(9:46)==470109m:1(350)Thelongestwavelengthpassesthroughatanangleof=arctan(60mm)=(300mm)=11:3:khdaw.com2Thiscorrespondstoawavelengthof(1103m)sin(11:3)==560109m:2(350)E43-16(a)
==R==Nm;so
=(481nm)=(620rulings/mm)(5:05mm)(3)=0:0512nm:(b)mmisthelargestintegersmallerthand=www.khdaw.com,orm1=(481109m)(620rulings/mm)=3:35;msom=3ishighestorderseen.E43-17TherequiredresolvingpowerofthegratingisgivenbyEq.43-10(589:0nm)R===982:
(589:6nm)(589:0nm)Ourresolvingpoweristhen课后答案网R=1000.UsingEq.43-11wecanndthenumberofgratinglinesrequired.Wearelookingatthesecond-ordermaxima,soR(1000)N===500:m(2)E43-18(a)N=R=m==m
,so(415:5nm)N==23100:(2)(415:496nm415:487nm)(b)d=w=N,wherewisthewidthofthegrating.Thenm(23100)(2)(415:5109m)=arcsin=arcsin=27:6:d(4:15102m)khdaw.com224若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE43-19N=R=m==m
,so(656:3nm)N==3650(1)(0:180nm)E43-20StartwithEq.43-9:mdsin=tanD===:dcosdcosE43-21(a)WendtherulingspacingbyEq.43-1,m(3)(589nm)d===9:98m:sinmsin(10:2)(b)TheresolvingpowerofthegratingneedstobeatleastR=1000forthethird-orderline;seetheworkforEx.43-17above.ThenumberoflinesrequiredisgivenbyEq.43-11,khdaw.comR(1000)N===333;m(3)sothewidthofthegrating(oratleastthepartthatisbeingused)is333(9:98m)=3:3mm.E43-22(a)Condition(1)issatisedifd2(600nm)=sin(30)=2400nm:Thedispersionismaximalforthesmallestd,sod=2400nm.(b)Toremovethethirdorderrequiresd=3awww.khdaw.com,ora=800nm.E43-23(a)Theanglesoftherstthreeordersare(1)(589109m)(40000)=arcsin=18:1;1(76103m)(2)(589109m)(40000)=arcsin=38:3;2(76103m)(3)(589109m)(40000)=arcsin=68:4:3(76103m)Thedispersionforeachorderis课后答案网(1)(40000)360D==3:2102=nm;1(76106nm)cos(18:1)2(2)(40000)360D==7:7102=nm;2(76106nm)cos(38:3)2(3)(40000)360D==2:5101=nm:3(76106nm)cos(68:4)2(b)R=Nm,soR1=(40000)(1)=40000;R2=(40000)(2)=80000;R3=(40000)(3)=120000:khdaw.com225若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE43-24d=m=2sin,so(2)(0:122nm)d==0:259nm:2sin(28:1)E43-25Braggre ectionisgivenbyEq.43-122dsin=m;wheretheanglesaremeasurednotagainstthenormal,butagainsttheplane.Thevalueofddependsonthefamilyofplanesunderconsideration,butitisatneverlargerthana0,theunitcelldimension.Wearelookingforthesmallestangle;thiswillcorrespondtothelargestdandthesmallestm.Thatmeansm=1andd=0:313nm.Thentheminimumangleis(1)(29:31012m)=sin1=2:68:2(0:313109m)E43-26khdaw.com2d==sin1and2d=2=sin2.Then=arcsin[2sin(3:40)]=6:81:2E43-27WeapplyEq.43-12toeachofthepeaksandndtheproductm=2dsin:Thefourvaluesare26pm,39pm,52pm,and78pm.Thelasttwovaluesaretwicethersttwo,sothewavelengthsare26pmand39pm.www.khdaw.comE43-28(a)2dsin=m,so(3)(96:7pm)d==171pm:2sin(58:0)(b)=2(171pm)sin(23:2)=(1)=135pm:E43-29Theangleagainstthefaceofthecrystalis9051:3=38:7.Thewavelengthis=2(39:8pm)sin(38:7)=(1)=49:8pm:E43-30If>2d课后答案网then=2d>1.But=2d=sin=m:Thismeansthatsin>m,butthesinefunctioncanneverbegreaterthanone.E43-31Therearetoomanyunknowns.Itisonlypossibletodeterminetheratiod=.E43-32Awavelengthwillbediractedifm=2dsin.Thepossiblesolutionsare3=2(275pm)sin(47:8)=(3)=136pm;4=2(275pm)sin(47:8)=(4)=102pm:226khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE43-33WeuseEq.43-12torstndd;m(1)(0:261109m)d===1:451010m:2sin2sin(63:8)disthespacingbetweentheplanesinFig.43-28;itcorrespondtohalfofthediagonaldistancebetweentwocellcenters.Then(2d)2=a2+a2;00orppa=2d=2(1:451010m)=0:205nm:0E43-34Diractionoccurswhen2dsin=m.Theanglesinthiscasearethengivenby(0:125109m)sin=m=(0:248)m:2(0:252109m)Therearefoursolutionstothisequation.Theyare14khdaw.com:4,29:7,48:1,and82:7.Theyinvolverotatingthecrystalfromtheoriginalorientation(9042:4=47:6)byamounts47:614:4=33:2;47:629:7=17:9;47:648:1=0:5;47:682:7=35:1:P43-1Sincetheslitsaresonarrowweonlyneedtoconsiderinterferenceeects,notdiractioneects.Therearethreewaveswhichcontributeatanypoint.Thephaseanglebetweenadjacentwww.khdaw.comwavesis=2dsin=:Wecanaddtheelectriceldvectorsaswasdoneinthepreviouschapters,orwecandoitinadierentorderasisshowninthegurebelow.课后答案网ThenthevectorssumtoE(1+2cos):Weneedtosquarethisquantity,andthennormalizeitsothatthecentralmaximumisthemaximum.Then(1+4cos+4cos2)I=Im:9khdaw.com227若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP43-2(a)SolveforI=Im=2,thisoccurswhen3p=1+2cos;2or=0:976rad.Thecorrespondinganglexis(0:976)x==:2d2d6:44dBut
=2x,so
:3:2d(b)Forthetwoslitpatternthehalfwidthwasfoundtobe
==2d:Thehalfwidthinthethreeslitcaseissmaller.P43-3(a)and(b)Aplotoftheintensityquicklyrevealsthatthereisanalternationoflargemaximum,thenasmallermaximum,etc.Thelargemaximaareatkhdaw.com=2n,thesmallermaximaarehalfwaybetweenthosevalues.(c)Theintensityatthesesecondarymaximaisthen(1+4cos+4cos2)ImI=Im=:99Notethattheminimaarenotlocatedhalf-waybetweenthemaxima!P43-4Coveringupthemiddleslitwillresultinatwoslitapparatuswithaslitseparationof2d.Thehalfwidth,asfoundinProblem41-5,isthenwww.khdaw.com
==2(2d);==4d;whichisnarrowerthanbeforecoveringupthemiddleslitbyafactorof3:2=4=0:8.P43-5(a)IfNislargewecantreatthephasorsassummingtoforma exibleline"oflengthNE.Wethenassume(incorrectly)thatthesecondarymaximaoccurwhentheloopwrapsaroundonitselfasshownintheguresbelow.Notethattheresultantphasoralwayspointsstraightup.Thisisn"tright,butitisclosetoreality.课后答案网228khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comThelengthoftheresultantdependsonhowmanyloopsthereare.Fork=0therearenone.Fork=1thereareoneandahalfloops.Thecircumferenceoftheresultingcircleis2NE=3,thediameterisNE=3.Fork=2therearetwoandahalfloops.Thecircumferenceoftheresultingcircleis2NE=5,thediameterisNE=5.Thepatternforhigherkissimilar:thecircumferenceis2NE=(2k+1),thediameterisNE=(k+1=2):Theintensityatthisapproximate"maximaisproportionaltotheresultantsquared,or(NE)2Ik/:(k+1=2)22butIisproportionalto(NE)2,som1Ik=Im:(k+1=2)22(b)Nearthemiddlethevectorssimplyfoldbackononeanother,leavingaresultantofE.Then(NE)2khdaw.comI/(E)2=;kN2soImIk=;N2(c)Lethavethevalueswhichresultinsin=1,andthenthetwoexpressionsareidentical!P43-6(a)v=f,sov=f+f.Assumingv=0,wehavef=f==.Ignorethenegativesign(wedon"tneedithere).Thenwww.khdaw.comfcR===;

f
fandthencc
f==:RNm(b)Therayonthetopgetsthererst,therayonthebottommusttravelanadditionaldistanceofNdsin.Ittakesatime
t=Ndsin=ctodothis.(c)Sincem=d课后答案网sin,thetworesultingexpressioncanbemultipliedtogethertoyieldcNdsin(
f)(
t)==1:NmcThisisalmost,butnotquite,oneofHeisenberg"suncertaintyrelations!P43-7(b)WesketchparallellineswhichconnectcenterstoformalmostanyrighttrianglesimilartotheoneshownintheFig.43-18.Thetrianglewillhavetwosideswhichhaveintegermultipleplengthsofthelatticespacinga0.Thehypotenuseofthetrianglewillthenhavelengthh2+k2a0,wherehandkaretheintegers.InFig.43-18h=2whilek=1.Thenumberofplaneswhichcutthediagonalisequaltoh2+k2if,andonlyif,handkarerelativelyprime.Theinter-planarspacingisthenph2+k2a0a0d==p:h2+k2h2+k2229khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com(a)Thenextvespacingsarethenph=1;k=1;d=a0=2;ph=1;k=2;d=a0=5;ph=1;k=3;d=a0=10;ph=2;k=3;d=a0=13;ph=1;k=4;d=a0=17:P43-8Themiddlelayercellswillalsodiractabeam,butthisbeamwillbeexactlyoutofphasewiththetoplayer.Thetwobeamswillthencanceloutexactlybecauseofdestructiveinterference.khdaw.comwww.khdaw.com课后答案网230khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE44-1(a)Thedirectionofpropagationisdeterminedbyconsideringtheargumentofthesinefunction.Astincreasesymustdecreasetokeepthesinefunctionlooking"thesame,sothewaveispropagatinginthenegativeydirection.(b)Theelectriceldisorthogonal(perpendicular)tothemagneticeld(soEx=0)andthedirectionofmotion(soEy=0);Consequently,theonlynon-zerotermisEz.ThemagnitudeofEwillbeequaltothemagnitudeofBtimesc.Since~S=E~B~=0,whenB~pointsinthepositivexdirectionthenE~mustpointinthenegativezdirectioninorderthat~Spointinthenegativeydirection.ThenEz=cBsin(ky+!t):(c)Thepolarizationisgivenbythedirectionoftheelectriceld,sothewaveislinearlypolarizedinthezdirection.E44-2Letonewavebepolarizedinthexdirectionandtheotherintheydirection.ThenthenetelectriceldisgivenbyE2=E2+E2,orxy
E2=E2sin2(kz!t)+sin2(kz!t+);khdaw.com0whereisthephasedierence.Wecanconsideranypointinspace,includingz=0,andthenaveragetheresultoverafullcycle.Sincemerelyshiftstheintegrationlimits,thentheresultisindependentof.Consequently,therearenointerferenceeects.E44-3(a)ThetransmittedintensityisI=2=6:1103W=m2.Themaximumvalueoftheelectric0eldisppEm=20cI=2(1:26106H=m)(3:00108m=s)(6:1103W=m2)=2:15V=m:(b)Theradiationpressureiscausedbytheabsorbedhalfoftheincidentlight,sowww.khdaw.comp=I=c=(6:1103W=m2)=(3:00108m=s)=2:031011Pa:E44-4Therstsheettransmitshalftheoriginalintensity,thesecondtransmitsanamountpro-portionaltocos2.ThenI=(I=2)cos2,or0pp=arccos2I=I=arccos2(I=3)=I35:3:000E44-5Therstsheetpolarizestheun-polarizedlight,halfoftheintensityistransmitted,soI=1I.120课后答案网ThesecondsheettransmitsaccordingtoEq.44-1,2121I2=I1cos=I0cos(45)=I0;24andthetransmittedlightispolarizedinthedirectionofthesecondsheet.Thethirdsheetis45tothesecondsheet,sotheintensityofthelightwhichistransmittedthroughthethirdsheetis2121I3=I2cos=I0cos(45)=I0:48231khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE44-6Thetransmittedintensitythroughtherstsheetisproportionaltocos2,thetransmittedintensitythroughthesecondsheetisproportionaltocos2(90)=sin2.ThenI=Icos2sin2=(I=4)sin22;00or1p1p=arcsin4I=I=arcsin4(0:100I)=I=19:6:00022Notethat70:4isalsoavalidsolution!E44-7Therstsheettransmitshalfoftheoriginalintensity;eachoftheremainingsheetstransmitsanamountproportionaltocos2,where=30.ThenI12
316=cos=(cos(30))=0:211I022E44-8Therstsheettransmitsanamountproportionaltocos2,where=58:8.Thesecondsheettransmitsanamountproportionaltocoskhdaw.com2(90)=sin2.ThenI=Icos2sin2=(43:3W=m2)cos2(58:8)sin2(58:8)=8:50W=m2:0E44-9Sincetheincidentbeamisunpolarizedtherstsheettransmits1/2oftheoriginalintensity.Thetransmittedbeamthenhasapolarizationsetbytherstsheet:58:8tothevertical.Thesecondsheetishorizontal,whichputsit31:2totherstsheet.Thenthesecondsheettransmitscos2(31:2)oftheintensityincidentonthesecondsheet.Thenalintensitytransmittedbythesecondsheetcanbefoundfromtheproductoftheseterms,1
I=(43:3W=m2)coswww.khdaw.com2(31:2)=15:8W=m2:2E44-10=arctan(1:53=1:33)=49:0.pE44-11(a)Theangleforcompletepolarizationofthere ectedrayisBrewster"sangle,andisgivenbyEq.44-3(sincetherstmediumisair)=tan1n=tan1(1:33)=53:1:p(b)Sincetheindexofrefractiondepends(slightly)onfrequency,thensodoesBrewster"sangle.课后答案网E44-12(b)Sincer+p=90,p=90(31:8)=58:2.(a)n=tan=tan(58:2)=1:61.pE44-13Theanglesarebetween=tan1n=tan1(1:472)=55:81:pand=tan1n=tan1(1:456)=55:52:pE44-14Thesmallestpossiblethicknesstwillallowforonehalfawavelengthphasedierencefortheoandewaves.Then
nt==2,ort=(525109m)=2(0:022)=1:2105m:232khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE44-15(a)Theincidentwaveisat45totheopticalaxis.Thismeansthattherearetwocomponents;assumetheyoriginallypointinthe+yand+zdirection.Whentheytravelthroughthehalfwaveplatetheyarenowoutofphaseby180;thismeansthatwhenonecomponentisinthe+ydirectiontheotherisinthezdirection.Ineectthepolarizationhasbeenrotatedby90.(b)Sincethehalfwaveplatewilldelayonecomponentsothatitemerges180later"thanitshould,itwillineectreversethehandednessofthecircularpolarization.(c)Pretendthatanunpolarizedbeamcanbebrokenintotwoorthogonallinearlypolarizedcomponents.Botharethenrotatedthrough90;butwhenrecombineditlooksliketheoriginalbeam.Assuch,thereisnoapparentchange.E44-16Thequarterwaveplatehasathicknessofx==4
n,sothenumberofplatesthatcanbecutisgivenbyN=(0:250103m)4(0:181)=(488109m)=371:P44-1Intensityisproportionaltotheelectriceldsquared,sotheoriginalintensityreachingtheeyeisI,withcomponentsI=(2:3)2I,andthenkhdaw.com0hvI0=Ih+Iv=6:3IvorIv=0:16I0:Similarly,I=(2:3)2I=0:84I.hv0(a)Whenthesun-batherisstandingonlytheverticalcomponentpasses,while(b)whenthesun-batherislyingdownonlythehorizontalcomponentpasses.P44-2TheintensityofthetransmittedlightwhichwasoriginallyunpolarizedisreducedtoIu=2,regardlessoftheorientationofthepolarizingsheet.Theintensityofthetransmittedlightwhichwasoriginallypolarizedisbetween0andIp,dependingontheorientationofthepolarizingsheet.ThenthemaximumtransmittedintensityisIu=2+www.khdaw.comIp,whiletheminimumtransmittedintensityisIu=2.Theratiois5,soIu=2+IpIp5==1+2;Iu=2IuorIp=Iu=2.Thenthebeamis1/3unpolarizedand2/3polarized.P44-3Eachsheettransmitsafraction22cos=cos:N课后答案网ThereareNsheets,sothefractiontransmittedthroughthestackisN2cos:NWewanttoevaluatethisinthelimitasN!1.AsNgetslargerwecanuseasmallangleapproximationtothecosinefunction,12cosx1xforx12Thethetransmittedintensityis22N11:2N2khdaw.com233若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comThisexpressioncanalsobeexpandedinabinomialexpansiontoget1212N;2N2whichinthelimitasN!1approaches1.Thestackthentransmitsallofthelightwhichmakesitpasttherstlter.Assumingthelightisoriginallyunpolarized,thenthestacktransmitshalftheoriginalintensity.P44-4(a)Stackseveralpolarizingsheetssothattheanglebetweenanytwosheetsissucientlysmall,butthetotalangleis90.(b)Thetransmittedintensityfractionneedstobe0:95.Eachsheetwilltransmitafractioncos2,where=90=N,withNthenumberofsheets.Thenwewanttosolve
N0:95=cos2(90=N)forN.ForlargeenoughN,willbesmall,sowecanexpandthecosinefunctionaskhdaw.comcos2=1sin212;so
N0:951(=2N)21N(=2N)2;whichhassolutionN=2=4(0:05)=49:P44-5Sincepassingthroughaquarterwaveplatetwicecanrotatethepolarizationofalinearlypolarizedwaveby90,thenifthelightpassesthroughapolarizer,throughtheplate,re ectsothecoin,thenthroughtheplate,andthroughthepolarizer,itwouldbepossiblethatwhenitpassesthroughthepolarizerthesecondtimeitis90tothepolarizerandnolightwillpass.Youwon"tseethecoin.www.khdaw.comOntheotherhandifthelightpassesrstthroughtheplate,thenthroughthepolarizer,thenisre ected,thepassesagainthroughthepolarizer,allthere ectedlightwillpassthroughhepolarizerandeventuallyworkitswayoutthroughtheplate.Sothecoinwillbevisible.Hence,sideAmustbethepolarizingsheet,andthatsheetmustbeat45totheopticalaxis.P44-6(a)Thedisplacementofarayisgivenbytank=yk=t;sotheshiftis课后答案网
y=t(tanetano):Solvingforeachangle,1e=arcsinsin(38:8)=24:94;(1:486)1o=arcsinsin(38:8)=22:21:(1:658)Theshiftisthen
y=(1:12102m)(tan(24:94)tan(22:21))=6:35104m:(b)Thee-raybendslessthantheo-ray.(c)Therayshavepolarizationswhichareperpendiculartoeachother;theo-wavebeingpolarizedalongthedirectionoftheopticaxis.(d)Oneray,thentheother,woulddisappear.khdaw.com234若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP44-7ThemethodisoutlineinSampleProblem44-24;useapolarizingsheettopickouttheo-rayorthee-ray.khdaw.comwww.khdaw.com课后答案网235khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE45-1(a)TheenergyofaphotonisgivenbyEq.45-1,E=hf,sohcE=hf=:Puttinginbest"numbers(6:626068761034J
s)hc=(2:99792458108m=s)=1:23984106eV
m:(1:6021764621019C)Thismeansthathc=1240eV
nmisaccuratetoalmostonepartin8000!(b)E=(1240eV
nm)=(589nm)=2:11eV.E45-2UsingtheresultsofExercise45-1,(1240eV
nm)==2100nm;(0:60eV)whichisintheinfrared.khdaw.comE45-3UsingtheresultsofExercise45-1,(1240eV
nm)E1==3:31eV;(375nm)and(1240eV
nm)E2==2:14eV;(580nm)Thedierenceis
E=(3:307eV)(2:138eV)=1www.khdaw.com:17eV:E45-4P=E=t,so,usingtheresultofExercise45-1,(1240eV
nm)P=(100=s)=230eV/s:(540nm)That"sasmall3:681017W.E45-5Whentalkingabouttheregionsinthesun"sspectrumitismorecommontorefertowavelengthsthanfrequencies.SowewillusetheresultsofExercise45-1(a),andsolve课后答案网=hc=E=(1240eV
nm)=E:TheenergiesarebetweenE=(1:01018J)=(1:61019C)=6:25eVandE=(1:01016J)=(1:61019C)=625eV.Theseenergiescorrespondtowavelengthsbetween198nmand1.98nm;thisistheultravioletrange.E45-6TheenergyperphotonisE=hf=hc=.Theintensityispowerperarea,whichisenergypertimeperarea,soPEnhchcnI====:AAtAtAtButR=n=tistherateofphotonsperunittime.SincehandcareconstantsandIandAareequalforthetwobeams,wehaveR1=1=R2=2,orR1=R2=1=2:khdaw.com236若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE45-7(a)Sincethepoweristhesame,thebulbwiththelargerenergyperphotonwillemitsfewerphotonspersecond.Sincelongerwavelengthshavelowerenergies,thebulbemitting700nmmustbegivingomorephotonspersecond.(b)Howmanymorephotonspersecond?IfE1istheenergyperphotonforoneofthebulbs,thenN1=P=E1isthenumberofphotonspersecondemitted.ThedierenceisthenPPPN1N2==(12);E1E2hcor(130W)9920N1N2=((70010m)(40010m))=1:9610:(6:631034J
s)(3:00108m=s)E45-8UsingtheresultsofExercise45-1,theenergyofonephotonis(1240eV
nm)E==1:968eV;khdaw.com(630nm)ThetotallightenergygivenobythebulbisE=Pt=(0:932)(70W)(730hr)(3600s/hr)=1:71108J:tThenumberofphotonsisE(1:71108J)t26n===5:4310:E0(1:968eV)(1:61019J=eV)E45-9ApplyWien"slaw,Eq.45-4,maxT=2898www.khdaw.comm
K;so(2898m
K)6T==9110K:(321012m)Actually,thewavelengthwassupposedtobe32m.Thenthetemperaturewouldbe91K.E45-10ApplyWien"slaw,Eq.45-4,maxT=2898m
K;so(2898m
K)==1:45m:(0:0020K)Thisisintheradioregion,near207ontheFMdial.课后答案网E45-11ThewavelengthofthemaximumspectralradiancyisgivenbyWien"slaw,Eq.45-4,maxT=2898m
K:Applyingtoeachtemperatureinturn,(a)=1:06103m,whichisinthemicrowave;(b)=9:4106m,whichisintheinfrared;(c)=1:6106m,whichisintheinfrared;(d)=5:0107m,whichisinthevisible;(e)=2:91010m,whichisinthex-ray;(f)=2:91041m,whichisinahardgammaray.237khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE45-12(a)ApplyWien"slaw,Eq.45-4,maxT=2898m
K:;so(2898m
K)7==5:0010m:(5800K)That"sblue-green.(b)ApplyWien"slaw,Eq.45-4,maxT=2898m
K:;so(2898m
K)T==5270K:(550109m)E45-13I=T4andP=IA.ThenP=(5:67108W=m2
K4)(1900K)4(0:5103m)2=0:58W:E45-14SinceI/T4,doublingTresultsina24=16timesincreaseinI.Thenthenewpowerleveliskhdaw.com(16)(12:0mW)=192mW:E45-15(a)WewanttoapplyEq.45-6,2c2h1R(;T)=:5ehc=kT1Weknowtheratioofthespectralradianciesattwodierentwavelengths.Dividingtheaboveequationattherstwavelengthbythesameequationatthesecondwavelength,
5ehc=1kT113:5=www.khdaw.com
;5ehc=2kT12where1=200nmand2=400nm.Wecanconsiderablysimplifythisexpressionifweletx=ehc=2kT;becausesince2=21wewouldhaveehc=1kT=e2hc=2kT=x2:Thenweget521x11课后答案网3:5==(x+1):2x132WewillusetheresultsofExercise45-1fortheexponentsandthenrearrangetogethc(3:10eV)T===7640K:1kln(111)(8:62105eV=K)ln(111)(b)Themethodisthesame,exceptthatinsteadof3.5wehave1/3.5;thismeanstheequationforxis11=(x+1);3:532withsolutionx=8:14,sothenhc(3:10eV)T===17200K:1kln(8:14)(8:62105eV=K)ln(8:14)khdaw.com238若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE45-16hf=,so(5:32eV)15f===1:2810Hz:h(4:141015eV
s)E45-17We"llusetheresultsofExercise45-1.Visibleredlighthasanenergyof(1240eV
nm)E==1:9eV:(650nm)Thesubstancemusthaveaworkfunctionlessthanthistoworkwithredlight.Thismeansthatonlycesiumwillworkwithredlight.Visiblebluelighthasanenergyof(1240eV
nm)E==2:75eV:(450nm)Thismeansthatbarium,lithium,andcesiumwillworkwithbluelight.E45-18khdaw.comSinceKm=hf,K=(4:141015eV
s)(3:191015Hz)(2:33eV)=10:9eV:mE45-19(a)UsetheresultsofExercise45-1tondtheenergyofthecorrespondingphoton,hc(1240eV
nm)E===1:83eV:(678nm)Sincethisenergyislessthanthantheminimumenergyrequiredtoremoveanelectronthenthephoto-electriceectwillnotoccur.www.khdaw.com(b)Thecut-owavelengthisthelongestpossiblewavelengthofaphotonthatwillstillresultinthephoto-electriceectoccurring.Thatwavelengthis(1240eV
nm)(1240eV
nm)===544nm:E(2:28eV)Thiswouldbevisibleasgreen.E45-20(a)SinceKm=hc=,(1240eV
nm)课后答案网Km=(4:2eV)=2:0eV:(200nm)(b)Theminimumkineticenergyiszero;theelectronjustbarelymakesitothesurface.(c)Vs=Km=q=2:0V:(d)Thecut-owavelengthisthelongestpossiblewavelengthofaphotonthatwillstillresultinthephoto-electriceectoccurring.Thatwavelengthis(1240eV
nm)(1240eV
nm)===295nm:E(4:2eV)E45-21Km=qVs=4:92eV.ButKm=hc=,so(1240eV
nm)==172nm:(4:92eV+2:28eV)khdaw.com239若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE45-22(a)Km=qVsandKm=hc=.WehavetwodierentvaluesforqVsand,sosubtractingthisequationfromitselfyieldsq(Vs;1Vs;2)=hc=1hc=2:Solvingfor2,hc2=;hc=1q(Vs;1Vs;2)(1240eV
nm)=;(1240eV
nm)=(491nm)(0:710eV)+(1:43eV)=382nm:(b)Km=qVsandKm=hc=,so=(1240eV
nm)=(491nm)(0:710eV)=1:82eV:E45-23khdaw.com(a)ThestoppingpotentialisgivenbyEq.45-11,hV0=f;eeso(1240eV
nm)(1:85eVV0==1:17V:e(410nme(b)Thesearenotrelativisticelectrons,sopppv=2K=m=c2K=mc2=c2(1:17eV)=(0:511106eV)=2:14103c;orv=64200m=s.www.khdaw.comE45-24Itwillhavebecomethestoppingpotential,orhV0=f;eeso(4:141015eV
m)(2:49eV)V=(6:331014=s)=0:131V:0(1:0e)(1:0e)E45-25课后答案网E45-26(a)UsingtheresultsofExercise45-1,(1240eV
nm)==62pm:(20103eV)(b)Thisisinthex-rayregion.E45-27(a)UsingtheresultsofExercise45-1,(1240eV
nm)E==29;800eV:(41:6103nm)(b)f=c==(3108m=s)=(41:6pm)=7:211018=s:(c)p=E=c=29;800eV=c=2:98104eV=c:khdaw.com240若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE45-28(a)E=hf,so(0:511106eV)f==1:231020=s:(4:141015eV
s)(b)=c=f=(3108m=s)=(1:231020=s)=2:43pm:(c)p=E=c=(0:511106eV)=c.E45-29Theinitialmomentumofthesystemisthemomentumofthephoton,p=h=.Thismomentumisimpartedtothesodiumatom,sothenalspeedofthesodiumisv=p=m,wheremisthemassofthesodium.Thenh(6:631034J
s)v===2:9cm/s:m(589109m)(23)(1:71027kg)E45-30(a)=h=mc=hc=mc2,soC(1240eV
nm)C==2:43pm:khdaw.com(0:511106eV)(c)SinceE=hf=hc=,and=h=mc=hc=mc2,thenE=hc==mc2:(b)Seepart(c).E45-31ThechangeinthewavelengthofaphotonduringComptonscatteringisgivenbyEq.45-17,0h=+(1cos):mcwww.khdaw.comWe"llusetheresultsofExercise45-30tosavesometime,andleth=mc=C,whichis2.43pm.(a)For=35,0=(2:17pm)+(2:43pm)(1cos35)=2:61pm:(b)For=115,0=(2:17pm)+(2:43pm)(1cos115)=5:63pm:E45-32(a)We"llusetheresultsofExercise45-1:课后答案网(1240eV
nm)==2:43pm:(0:511106eV)(b)ThechangeinthewavelengthofaphotonduringComptonscatteringisgivenbyEq.45-17,0h=+(1cos):mcWe"llusetheresultsofExercise45-30tosavesometime,andleth=mc=C,whichis2.43pm.0=(2:43pm)+(2:43pm)(1cos72)=4:11pm:(c)We"llusetheresultsofExercise45-1:(1240eV
nm)E==302keV:(4:11pm)khdaw.com241若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE45-33ThechangeinthewavelengthofaphotonduringComptonscatteringisgivenbyEq.45-17,0h=(1cos):mcWearenotusingtheexpressionwiththeform
because
and
Earenotsimplyrelated.Thewavelengthisrelatedtofrequencybyc=f,whilethefrequencyisrelatedtotheenergybyEq.45-1,E=hf.Then
E=EE0=hfhf0;11=hc;00=hc:0IntothislastexpressionwesubstitutetheComptonformula.Thenh2(1cos)
E=:khdaw.comm0NowE=hf=hc=,andwecandividethisonbothsidesoftheaboveequation.Also,0=c=f0,andwecansubstitutethisintotherighthandsideoftheaboveequation.Bothofthesestepsresultin
Ehf0=(1cos):Emc2Notethatmc2istherestenergyofthescatteringparticle(usuallyanelectron),whilehf0istheenergyofthescatteredphoton.E45-34Thewavelengthisrelatedtofrequencybyc=f,whilethefrequencyisrelatedtotheenergybyEq.45-1,E=hf.Thenwww.khdaw.com
E=EE0=hfhf0;11=hc;00=hc;0
E
=;E+
But
E=E=3=4,so课后答案网3+3
=4
;or
=3.E45-35Themaximumshiftoccurswhen=180,soh(1240eV
nm)15
m=2=2=2:6410m:mc938MeV)E45-36SinceE=hffrequencyshiftsareidenticaltoenergyshifts.ThenwecanusetheresultsofExercise45-33toget(0:9999)(6:2keV)(0:0001)=(1cos);(511keV)whichhassolution=7:4.(b)(0:0001)(6:2keV)=0:62eV:khdaw.com242若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE45-37(a)ThechangeinwavelengthisindependentofthewavelengthandisgivenbyEq.45-17,hc(1240eV
nm)3
=(1cos)=2=4:8510nm:mc2(0:511106eV)(b)Thechangeinenergyisgivenbyhchc
E=;fi11=hc;i+
i11=(1240eV
nm)=42:1keV(9:77pm)+(4:85pm)(9:77pm)(c)Thisenergywenttotheelectron,sothenalkineticenergyoftheelectronis42:1keV.E45-38khdaw.comFor=90
=h=mc.Then
Ehf0=1=1;Ehf+
h=mc=:+h=mcButh=mc=2:43pmfortheelectron(seeExercise45-30).(a)
E=E=(2:43pm)=(3:00cm+2:43pm)=8:11011:(b)
E=E=(2:43pm)=(500nm+2:43pm)=4:86106:(c)
E=E=(2:43pm)=(0:100nm+2:43pm)=0www.khdaw.com:0237:(d)
E=E=(2:43pm)=(1:30pm+2:43pm)=0:651:E45-39WecanusetheresultsofExercise45-33toget(0:90)(215keV)(0:10)=(1cos);(511keV)whichhassolution=42=6.E45-40(a)Acrudeestimateisthatthephotonscan"tarrivemorefrequentlythanonceevery108s.Thatwouldprovideanemissionrateof10课后答案网8=s.(b)Thepoweroutputwouldbe8(1240eV
nm)8P=(10)=2:310eV/s;(550nm)whichis3:61011W!E45-41WecanfollowtheexampleofSampleProblem45-6,andapply=0(1v=c):(a)Solvingfor0,(588:995nm)0==588:9944nm:(1(300m=s)(3108m=s)khdaw.com243若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com(b)ApplyingEq.45-18,h(6:61034J
s)
v===3102m=s:m(22)(1:71027kg)(590109m)(c)Emittinganotherphotonwillslowthesodiumbyaboutthesameamount.E45-42(a)(430m=s)=(0:15m=s)2900interactions.(b)Iftheargonaveragesaspeedof220m=s,thenitrequiresinteractionsattherateof(2900)(220m=s)=(1:0m)=6:4105=sifitisgoingtoslowdownintime.P45-1TheradiantintensityisgivenbyEq.45-3,I=T4.ThepowerthatisradiatedthroughtheopeningisP=IA,whereAistheareaoftheopening.Butenergygoesbothwaysthroughtheopening;itisthekhdaw.comdierencethatwillgivethenetpowertransfer.Then
P=(II)A=AT4T4:net0101Putinthenumbers,and
P=(5:67108W=m2
K4)(5:20104m2)(488K)4(299K)4=1:44W:netP45-2(a)I=T4andP=IA.ThenT4=P=A,ors(100W)T=4=3248K:(5:67108W=m2
K4)(0:28www.khdaw.com103m)(1:8102m)That"s2980C.(b)TheratethatenergyisradiatedoisgivenbydQ=dt=mCdT=dt.Themassisfoundfromm=V,whereVisthevolume.ThiscanbecombinedwiththepowerexpressiontoyieldAT4=VCdT=dt;whichcanbeintegratedtoyieldVC33
t=(1=T21=T1):3APuttinginnumbers,课后答案网(19300kg=m3)(0:28103m)(132J=kgC)
t=[1=(2748K)31=(3248K)3];3(5:67108W=m2
K4)(4)=20ms:P45-3Lightfromthesunwillheat-up"thethinblackscreen.Asthetemperatureofthescreenincreasesitwillbegintoradiateenergy.Whentherateofenergyradiationfromthescreenisequaltotherateatwhichtheenergyfromthesunstrikesthescreenwewillhaveequilibrium.Weneedrsttondanexpressionfortherateatwhichenergyfromthesunstrikesthescreen.ThetemperatureofthesunisT.TheradiantintensityisgivenbyEq.45-3,I=T4.TheSSStotalpowerradiatedbythesunistheproductofthisradiantintensityandthesurfaceareaofthesun,soP=4r2T4;SSSkhdaw.com244若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comwhererSistheradiusofthesun.AssumingthatthelensisonthesurfaceoftheEarth(areasonableassumption),thenwecanndthepowerincidentonthelensifweknowtheintensityofsunlightatthedistanceoftheEarthfromthesun.ThatintensityisPSPSIE==;A4RE2whereREistheradiusoftheEarth"sorbit.Combining,24rSIE=TSREThetotalpowerincidentonthelensisthen24rS2Plens=IEAlens=TSrl;REwherekhdaw.comrlistheradiusofthelens.Alloftheenergythatstrikesthelensisfocusedontheimage,sothepowerincidentonthelensisalsoincidentontheimage.Thescreenradiatesasthetemperatureincreases.TheradiantintensityisI=T4,whereTisthetemperatureofthescreen.Thepowerradiatedisthisintensitytimesthesurfacearea,soP=IA=2r2T4:iThefactorof2"isbecausethescreenhastwosides,whileriistheradiusoftheimage.SetthisequaltoPlens,2244rS22riT=TSrl;www.khdaw.comREor2414rSrlT=TS:2REriTheradiusoftheimageofthesundividedbytheradiusofthesunisthemagnicationofthelens.Butmagnicationisalsorelatedtoimagedistancedividedbyobjectdistance,sorii=jmj=;rSoDistancesshouldbemeasuredfromthelens,butsincethesunissofarfromtheEarth,wewon"tbefaroinstatingo课后答案网RE.Sincetheobjectissofarfromthelens,theimagewillbevery,veryclosetothefocalpoint,sowecanalsostateif.Thenrif=;rSREsotheexpressionforthetemperatureofthethinblackscreenisconsiderablysimpliedto2414rlT=TS:2fNowwecanputinsomeofthenumbers.s1(1:9cm)T=(5800K)=1300K:21=4(26cm)245khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP45-4ThederivativeofRwithrespecttois232(hc)ch2chekT10+:(hc)(hc)6(ekT1)7(ekT1)2kTOhh,that"sugly.Settingitequaltozeroallowsconsiderablesimplication,andweareleftwith(5x)ex=5;wherex=hc=kT.Thesolutionisfoundnumericallytobex=4:965114232.Then(1240eV
nm)2:898103m
K==:(4:965)(8:62105eV/K)TTP45-5(a)IftheplanethasatemperatureT,thentheradiantintensityoftheplanetwillbeIT4,andtherateofenergyradiationfromtheplanetwillbekhdaw.comP=4R2T4;whereRistheradiusoftheplanet.Asteadystateplanettemperaturerequiresthattheenergyfromthesunarriveatthesamerateastheenergyisradiatedfromtheplanet.TheintensityoftheenergyfromthesunadistancerfromthesunisP=4r2;sunandthetotalpowerincidentontheplanetisthen2PsunP=R:www.khdaw.com4r2Equating,242Psun4RT=R;4r24PsunT=:16r2(b)UsingthelastequationandthenumbersfromProblem3,s1(6:96108m)课后答案网T=p(5800K)=279K:2(1:51011m)That"sabout43F.P45-6(a)Changevariablesassuggested,then=hc=xkTandd=(hc=x2kT)dx.Integrate(notetheswappingofthevariablesofintegrationpicksupaminussign):Z2c2h(hc=x2kT)dxI=;(hc=xkT)5ex1Z2k4T4x3dx=;h3c2ex125k4=T4:15h3c2khdaw.com246若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP45-7(a)P=E=t=nhf=t=(hc=)(n=t),wheren=tistherateofphotonemission.Then(100W)(589109m)n=t==2:961020=s:(6:631034J
s)(3:00108m=s)(b)The uxatadistanceristheratedividedbytheareaofthesphereofradiusr,sos(2:961020=s)7r==4:810m:4(1104=m2
s)(c)Thephotondensityisthe uxdividedbythespeedoflight;thedistanceisthens(2:961020=s)r==280m:4(1106=m3)(3108m=s)(d)The uxisgivenby(2:961020=s)khdaw.com=5:891018=m2
s:4(2:0m)2Thephotondensityis(5:891018m2
s)=(3:00108m=s)=1:961010=m3:P45-8Momentumconservationrequiresp=pe;whileenergyconservationrequireswww.khdaw.comE+mc2=E:eSquarebothsidesoftheenergyexpressionandE2+2Emc2+m2c4=E2=p2c2+m2c4;eeE2+2Emc2=p2c2;ep2c2+2Emc2=p2c2:eButthemomentumexpressioncanbeusedhere,andtheresultis2Emc2=0:课后答案网Notlikely.P45-9(a)SinceqvB=mv2=r,v=(q=m)rBThekineticenergyof(non-relativistic)electronswillbe11q2(rB)2K=mv2=;22mor1(1:61019C)K=(188106T
m)2=3:1103eV:2(9:11031kg)(b)UsetheresultsofExercise45-1,(1240eV
nm)34=3:110eV=1:4410eV:(71103nm)khdaw.com247若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP45-10P45-11(a)Themaximumvalueof
is2h=mc.Themaximumenergylostbythephotonisthenhchc
E=;fi11=hc;i+
i2h=mc=hc;(+2h=mc)whereinthelastlinewewrotefori.Theenergygiventotheelectronisthenegativeofthis,so2h2Kmax=:khdaw.comm(+2h=mc)Multiplyingthroughby12=(E=hc)2weget2E2Kmax=:mc2(1+2hc=mc2)orE2Kmax=:mc2=2+E(b)Theansweris(17:5keV)2Kmax=www.khdaw.com=1:12keV:(511eV)=2+(17:5keV)课后答案网248khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE46-1(a)ApplyEq.46-1,=h=p.Themomentumofthebulletisp=mv=(0:041kg)(960m=s)=39kg
m=s;sothecorrespondingwavelengthis=h=p=(6:631034J
s)=(39kg
m=s)=1:71035m:(b)Thislengthismuchtoosmalltobesignicant.Howmuchtoosmall?Iftheradiusofthegalaxywereonemeter,thisdistancewouldcorrespondtothediameterofaproton.E46-2(a)=h=pandp2=2m=K,thenhc(1240eV
nm)1:226nm=p=pp=p:2mc2K2(511keV)KK(b)K=eV,sokhdaw.comr1:226nm1:5V=p=nm:eVVpE46-3Fornon-relativisticparticles=h=pandp2=2m=K,so=hc=2mc2K.(a)Fortheelectron,(1240eV
nm)=p=0:0388nm:2(511keV)(1:0keV)(c)Fortheneutron,(1240MeV
fm)=pwww.khdaw.com=904fm:2(940MeV)(0:001MeV)(b)Forultra-relativisticparticlesKEpc,sohc(1240eV
nm)===1:24nm:E(1000eV)E46-4Fornon-relativisticparticlesp=h=andp2=2m=K,soK=(hc)2=2mc22.Then(1240eV
nm)2K==4:34106eV:2(5:11106eV)(589nm)2课后答案网E46-5(a)ApplyEq.46-1,p=h=.Theprotonspeedwouldthenbehhc(1240MeV
fm)v==c=c=0:0117c:mmc2(938MeV)(113fm)Thisisgood,becauseitmeanswewerejustiedinusingthenon-relativisticequations.Thenv=3:51106m=s.(b)Thekineticenergyofthiselectronwouldbe1212K=mv=(938MeV)(0:0117)=64:2keV:22Thepotentialthroughwhichitwouldneedtobeacceleratedis64.2kV.249khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.compE46-6(a)K=qVandp=2mK.Thenpp=2(22)(932MeV=c2)(325eV)=3:65106eV=c:(b)=h=p,sohc(1240eV
nm)4===3:3910nm:pc(3:65106eV=c)cpE46-7(a)Fornon-relativisticparticles=h=pandp2=2m=K,so=hc=2mc2K.Forthealphaparticle,(1240MeV
fm)=p=5:2fm:2(4)(932MeV)(7:5MeV)(b)Sincethewavelengthofthealphaisconsiderablysmallerthanthedistancetothenucleuswecanignorethewavenatureofthealphaparticle.E46-8khdaw.com(a)Fornon-relativisticparticlesp=h=andp2=2m=K,soK=(hc)2=2mc22.Then(1240keV
pm)2K==15keV:2(511keV)(10pm)2(b)Forultra-relativisticparticlesKEpc,sohc(1240keV
pm)E===124keV:(10pm)E46-9Therelativisticrelationshipbetweenenergyandmomentumiswww.khdaw.comE2=p2c2+m2c4;andiftheenergyisverylarge(comparedtomc2),thenthecontributionofthemasstotheaboveexpressionissmall,andE2p2c2:ThenfromEq.46-1,hhchc(1240MeV
fm)2=====2:510fm:ppcE(50103MeV)课后答案网pE46-10(a)K=3kT=2,p=2mK,and=h=p,sohhc=p=p;3mkT3mc2kT(1240MeV
fm)=p=74pm:3(4)(932MeV)(8:621011MeV/K)(291K)(b)pV=NkT;assumingthateachparticleoccupiesacubeofvolumed3=Vthentheinter-0particlespacingisd,sosp(1:381023J=K)(291K)33d=V=N==3:4nm:(1:01105Pa)250khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE46-11p=mvandp=h=,som=h=v.Takingtheratio,mev44==(1:81310)(3)=5:43910:meveThemassoftheunknownparticleisthen(0:511MeV=c2)m==939:5MeV:(5:439104)Thatwouldmakeitaneutron.pE46-12(a)Fornon-relativisticparticles=h=pandp2=2m=K,so=hc=2mc2K.Fortheelectron,(1240eV
nm)=p=1:0nm:2(5:11105eV)(1:5eV)khdaw.comForultra-relativisticparticlesKEpc,soforthephotonhc(1240eV
nm)===830nm:E(1:5eV)(b)Electronswithenergiesthathighareultra-relativistic.Boththephotonandtheelectronwillthenhavethesamewavelength;hc(1240MeV
fm)===0:83fm:E(1:5GeV)E46-13(a)Theclassicalexpressionforkineticenergyiswww.khdaw.compp=2mK;sohhc(1240keV
pm)==p=p=7:76pm:p2mc2K2(511keV)(25:0keV)(a)Therelativisticexpressionformomentumispppc=sqrtE2m2c4=(mc2+K)2m2c4=K2+2mc2K:Then课后答案网hc(1240keV
pm)==p=7:66pm:pc(25:0keV)2+2(511keV)(25:0keV)E46-14Wewanttomatchthewavelengthofthegammatothatoftheelectron.Forthegamma,=hc=E.Fortheelectron,K=p2=2m=h2=2m2:Combining,h2E22K=2mh2c2E=2mc2:Withnumbers,(136keV)2K==18:1keV:2(511keV)Thatwouldrequireanacceleratingpotentialof18:1kV.khdaw.com251若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE46-15Firstndthewavelengthoftheneutrons.Fornon-relativisticparticles=h=pandpp2=2m=K,so=hc=2mc2K.Then(1240keV
pm)=p=14pm:2(940103keV)(4:2103keV)Braggre ectionoccurswhen2dsin=,so=arcsin(14pm)=2(73:2pm)=5:5:E46-16ThisismerelyaBraggre ectionproblem.Then2dsin=m,or=arcsin(1)(11pm)=2(54:64pm)=5:78;=arcsin(2)(11pm)=2(54:64pm)=11:6;=arcsin(3)(11pm)=2(54:64pm)=17:6:E46-17khdaw.com(a)Sincesin52=0:78,then2(=d)=1:57>1,sothereisnodiractionorderotherthantherst.(b)Foranacceleratingpotentialof54voltswehave=d=0:78.Increasingthepotentialwillincreasethekineticenergy,increasethemomentum,anddecreasethewavelength.dwon"tchange.Thekineticenergyisincreasedbyafactorof60=54=1:11,themomentumincreasesbyafactorofp1:11=1:05,sothewavelengthchangesbyafactorof1=1:05=0:952.Thenewangleisthen=arcsin(0:9520:78)=48:E46-18Firstndthewavelengthoftheelectrons.Fornon-relativisticparticles=h=pandpp2=2m=K,so=hc=2mc2K.Thenwww.khdaw.com(1240keV
pm)=p=62:9pm:2(511keV)(0:380keV)ThisisnowaBraggre ectionproblem.Then2dsin=m,or=arcsin(1)(62:9pm)=2(314pm)=5:74;=arcsin(2)(62:9pm)=2(314pm)=11:6;=arcsin(3)(62:9pm)=2(314pm)=17:5;课后答案网=arcsin(4)(62:9pm)=2(314pm)=23:6;=arcsin(5)(62:9pm)=2(314pm)=30:1;=arcsin(6)(62:9pm)=2(314pm)=36:9;=arcsin(7)(62:9pm)=2(314pm)=44:5;=arcsin(8)(62:9pm)=2(314pm)=53:3;=arcsin(9)(62:9pm)=2(314pm)=64:3:Buttheoddordersvanish(seeChapter43foradiscussiononthis).E46-19Since
f

t1=2,wehave
f=1=2(0:23s)=0:69=s:252khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE46-20Since
f

t1=2,wehave
f=1=2(0:10109s)=1:61010=s:Thebandwidthwouldn"ttinthefrequencyallocation!E46-21ApplyEq.46-9,h4:141015eV
s)
E==7:6105eV:2
t2(8:71012s)Thisismuchsmallerthanthephotonenergy.E46-22ApplyHeisenbergtwice:4:141015eV
s)
E==5:49108eV:khdaw.com12(12109s)and4:141015eV
s)
E==2:86108eV:22(23109s)Thesumis
E=8:4108eV.transitionE46-23ApplyHeisenberg:6:631034J
s)
p==8:81024kg
m=s:2(121012m)www.khdaw.comE46-24
p=(0:5kg)(1:2s)=0:6kg
m=s.Thepositionuncertaintywouldthenbe(0:6J=s)
x==0:16m:2(0:6kg
m=s)E46-25Wewantv
v,whichmeansp
p.ApplyEq.46-8,andhh
x:2
p2pAccordingtoEq.46-1,thedeBrogliewavelengthisrelatedtothemomentumby课后答案网=h=p;so
x:2E46-26(a)Aparticleconnedina(onedimensional)boxofsizeLwillhaveapositionuncertaintyofnomorethan
xL.Themomentumuncertaintywillthenbenolessthanhh
p:2
x2Lso(6:631034J
s)
p=11024kg
m=s:2(1010m)khdaw.com253若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com(b)Assumingthatp
p,wehavehp;2Landthentheelectronwillhavea(minimum)kineticenergyofp2h2E:2m82mL2or(hc)2(1240keV
pm)2E==0:004keV:82mc2L282(511keV)(100pm)2E46-27(a)Aparticleconnedina(onedimensional)boxofsizeLwillhaveapositionuncer-taintyofnomorethan
xL.Themomentumuncertaintywillthenbenolessthanhh
p:khdaw.com2
x2Lso(6:631034J
s)
p=11020kg
m=s:2(1014m)(b)Assumingthatp
p,wehavehp;2Landthentheelectronwillhavea(minimum)kineticenergyofp2h2E:2mwww.khdaw.com82mL2or(hc)2(1240MeV
fm)2E==381MeV:82mc2L282(0:511MeV)(10fm)2Thisissolargecomparedtothemassenergyoftheelectronthatwemustconsiderrelativisticeects.Itwillbeveryrelativistic(3810:5!),sowecanuseE=pcaswasderivedinExercise9.Thenhc(1240MeV
fm)E===19:7MeV:2L2(10fm)Thisisthetotalenergy;sowesubtract0.511MeVtoget课后答案网K=19MeV.E46-28WewanttondLwhenT=0:01.ThismeanssolvingEE2kLT=161e;U0U0(5:0eV)(5:0eV)2k0L(0:01)=161e;(6:0eV)(6:0eV)0=2:22e2kL;ln(4:5103)=2(5:12109=m)L;5:31010m=L:254khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE46-29Thewavenumber,k,isgivenby2pk=2mc2(U0E):hc(a)Fortheprotonmc2=938MeV,so2p1k=2(938MeV)(10MeV3:0MeV)=0:581fm:(1240MeV
fm)Thetransmissioncoecientisthen(3:0MeV)(3:0MeV)2(0:581fm1)(10fm)5T=161e=3:010:(10MeV)(10MeV)(b)Forthedeuteronmc2=2938MeV,so2p1k=2(2)(938MeV)(10MeV3:0MeV)=0:821fm:khdaw.com(1240MeV
fm)Thetransmissioncoecientisthen(3:0MeV)(3:0MeV)2(0:821fm1)(10fm)7T=161e=2:510:(10MeV)(10MeV)E46-30Thewavenumber,k,isgivenby2pk=2mc2(U0E):hcwww.khdaw.com(a)Fortheprotonmc2=938MeV,so2pk=2(938MeV)(6:0eV5:0eV)=0:219pm1:(1240keV
pm)WewanttondT.ThismeanssolvingEE2kLT=161e;U0U0(5:0eV)(5:0eV)2(0:2191012)(0:7109)=161e;课后答案网(6:0eV)(6:0eV)=1:610133:Acurrentof1kAcorrespondstoN=(1103C=s)=(1:61019C)=6:31021=sprotonsperseconds.Thetimerequiredforoneprotontopassisthent=1=(6:31021=s)(1:610133)=9:910110s:That"s10104years!255khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP46-1Wewillinterpretlowenergytomeannon-relativistic.Thenhh==p:p2mnKThediractionpatternisthengivenbypdsin=m=mh=2mnK;wheremisdiractionorderwhilemnistheneutronmass.WewanttoinvestigatethespreadbytakingthederivativeofwithrespecttoK,mhdcosd=pdK:22mnK3Dividethisbytheoriginalequation,andthencosdKd=:khdaw.comsin2KRearrange,changethedierentialtoadierence,andthen
K
=tan:2KWedroppedthenegativesignoutoflaziness;buttheanglesareinradians,soweneedtomultiplyby180=toconverttodegrees.P46-2www.khdaw.comP46-3WewanttosolveEE2kLT=161e;U0U0forE.Unfortunately,Eiscontainedinksince2pk=2mc2(U0E):hcWecandothisbyiteration.ThemaximumpossiblevalueforEE课后答案网1U0U0is1=4;usingthisvaluewecangetanestimatefork:(0:001)=16(0:25)e2kL;ln(2:5104)=2k(0:7nm);5:92=nm=k:NowsolveforE:E=U(hc)2k28mc22;0(1240eV
nm)2(5:92=nm)2=(6:0eV);82(5:11105eV)=4:67eV:khdaw.com256若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comPutthisvalueforEbackintothetransmissionequationtondanewk:EE2kLT=161e;U0U0(4:7eV)(4:7eV)2kL(0:001)=161e;(6:0eV)(6:0eV)ln(3:68104)=2k(0:7nm);5:65=nm=k:NowsolveforEusingthisnew,improved,valuefork:E=U(hc)2k28mc22;0(1240eV
nm)2(5:65=nm)2=(6:0eV);82(5:11105eV)=4:78eV:Keepatit.You"lleventuallystoparoundkhdaw.comE=5:07eV.P46-4(a)AonepercentincreaseinthebarrierheightmeansU0=6:06eV.Fortheelectronmc2=5:11105eV,so2pk=2(5:11105eV)(6:06eV5:0eV)=5:27nm1:(1240eV
nm)WewanttondT.ThismeanssolvingEE2kLT=161e;U0U0www.khdaw.com(5:0eV)(5:0eV)2(5:27)(0:7)=161e;(6:06eV)(6:06eV)=1:44103:That"sa16%decrease.(b)AonepercentincreaseinthebarrierthicknessmeansL=0:707nm.Fortheelectronmc2=5:11105eV,so2pk=2(5:11105eV)(6:0eV5:0eV)=5:12nm1:(1240eV课后答案网
nm)WewanttondT.ThismeanssolvingEE2kLT=161e;U0U0(5:0eV)(5:0eV)2(5:12)(0:707)=161e;(6:0eV)(6:0eV)=1:59103:That"sa8.1%decrease.(c)AonepercentincreaseintheincidentenergymeansE=5:05eV.Fortheelectronmc2=5:11105eV,so2pk=2(5:11105eV)(6:0eV5:05eV)=4:99nm1:(1240eV
nm)khdaw.com257若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comWewanttondT.ThismeanssolvingEE2kLT=161e;U0U0(5:05eV)(5:05eV)2(4:99)(0:7)=161e;(6:0eV)(6:0eV)=1:97103:That"sa14%increase.P46-5First,theruleforexponentsei(a+b)=eiaeib:ThenapplyEq.46-12,ei=cos+isin,khdaw.comcos(a+b)+isin(a+b)=(cosa+isina)(sinb+isinb):Expandtherighthandside,rememberingthati2=1,cos(a+b)+isin(a+b)=cosacosb+icosasinb+isinacosbsinasinb:Sincetherealpartofthelefthandsidemustequaltherealpartoftherightandtheimaginarypartofthelefthandsidemustequaltheimaginarypartoftheright,weactuallyhavetwoequations.Theyarecos(a+b)=cosacosbsinasinbandwww.khdaw.comsin(a+b)=cosasinb+sinacosb:P46-6课后答案网258khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE47-1(a)Thegroundstateenergylevelwillbegivenbyh2(6:631034J
s)2E===3:11010J:18mL28(9:111031kg)(1:41014m)2Theansweriscorrect,buttheunitsmakeitalmostuseless.Wecandividebytheelectronchargetoexpressthisinelectronvolts,andthenE=1900MeV.Notethatthisisanextremelyrelativisticquantity,sotheenergyexpressionlosesvalidity.(b)Wecanrepeatwhatwedidabove,orwecanapplyatrick"thatisoftenusedinsolvingtheseproblems.Multiplyingthetopandthebottomoftheenergyexpressionbyc2weget(hc)2E1=8(mc2)L2Then(1240MeV
fm)2E1==1:0MeV:8(940MeV)(14fm)2(c)Findinganneutroninsidethenucleusseemsreasonable;butndingtheelectronwouldnot.Theenergyofsuchanelectronisconsiderablylargerthanbindingenergiesoftheparticlesinthekhdaw.comnucleus.E47-2Solven2(hc)2En=8(mc2)L2forL,thennhcL=p;8mc2En(3)(1240eVwww.khdaw.com
nm)=p;8(5:11105eV)(4:7eV)=0:85nm:E47-3SolveforE4E1:42(hc)212(hc)2E4E1=;8(mc2)L28(mc2)L2(161)(1240eV
nm)2=;8(5:11105)(0:253nm)2课后答案网=88:1eV:E47-4SinceE/1=L2,doublingthewidthofthewellwilllowerthegroundstateenergyto(1=2)2=1=4,or0.65eV.E47-5(a)SolveforE2E1:22h212h2E2E1=;8mL28mL2(3)(6:631034J
s)2=;8(40)(1:671027kg)(0:2m)2=6:21041J:(b)K=3kT=2=3(1:381023J=K)(300K)=2=6:211021:Theratiois11020.(c)T=2(6:21041J)=3(1:381023J=K)=3:01018K:khdaw.com259若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE47-6(a)Thefractionaldierenceis(En+1En)=En,or
Eh2h2h2n=(n+1)2n2=n2;En8mL28mL28mL2(n+1)2n2=;n22n+1=:n2(b)Asn!1thefractionaldierencegoestozero;thesystembehavesasifitiscontinuous.E47-7(a)Wewilltakeadvantageofthetrick"thatwasdevelopedinpart(b)ofExercise47-1.Then(hc)2(1240eV
nm)2E=n2=(15)2=8:72keV:n8mc2L8(0:511106eV)(0:0985nm)2(b)ThemagnitudeofthemomentumisexactlyknownbecauseE=p2=2m.Thismomentumisgivenbykhdaw.compppc=2mc2E=2(511keV)(8:72keV)=94:4keV:Whatwedon"tknowisinwhichdirectiontheparticleismoving.Itisbouncingbackandforthbetweenthewallsofthebox,sothemomentumcouldbedirectedtowardtherightortowardtheleft.Theuncertaintyinthemomentumisthen
p=pwhichcanbeexpressedintermsoftheboxsizeLbyrpn2h2nh
p=p=2mE==:www.khdaw.com4L22L(c)Theuncertaintyinthepositionis98.5pm;theelectroncouldbeanywhereinsidethewell.E47-8Theprobabilitydistributionfunctionis222xP2=sin:LLWewanttointegrateoverthecentralthird,orZL=6222xP=sindx;课后答案网L=6LLZ=312=sind;=3=0:196:E47-9(a)Maximumprobabilityoccurswhentheargumentofthecosine(sine)functionisk([k+1=2]).Thisoccurswhenx=NL=2nforoddN.(b)Minimumprobabilityoccurswhentheargumentofthecosine(sine)functionis[k+1=2](k).Thisoccurswhenx=NL=2nforevenN.khdaw.com260若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE47-10InExercise47-21weshowthatthehydrogenlevelscanbewrittenasE=(13:6eV)=n2:n(a)TheLymanseriesistheserieswhichendsonE1.TheleastenergeticstatestartsonE2.ThetransitionenergyisEE=(13:6eV)(1=121=22)=10:2eV:21Thewavelengthishc(1240eV
nm)===121:6nm:
E(10:2eV)(b)Theserieslimitis0E=(13:6eV)(1=12)=13:6eV:1Thewavelengthishc(1240eV
nm)===91:2nm:khdaw.com
E(13:6eV)E47-11Thegroundstateofhydrogen,asgivenbyEq.47-21,isme4(9:1091031kg)(1:6021019C)4E===2:1791018J:182h28(8:8541012F=m)2(6:6261034J
s)20IntermsofelectronvoltsthegroundstateenergyisE=(2:1791018J)=(1:6021019C)=13:60eV:1E47-12InExercise47-21weshowthatthehydrogenlevelscanbewrittenaswww.khdaw.comE=(13:6eV)=n2:n(c)Thetransitionenergyis
E=EE=(13:6eV)(1=121=32)=12:1eV:31(a)Thewavelengthishc(1240eV
nm)===102:5nm:课后答案网
E(12:1eV)(b)Themomentumisp=E=c=12:1eV=c:E47-13InExercise47-21weshowthatthehydrogenlevelscanbewrittenasE=(13:6eV)=n2:n(a)TheBalmerseriesistheserieswhichendsonE2.TheleastenergeticstatestartsonE3.ThetransitionenergyisEE=(13:6eV)(1=221=32)=1:89eV:32Thewavelengthishc(1240eV
nm)===656nm:
E(1:89eV)khdaw.com261若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com(b)ThenextenergeticstatestartsonE4.ThetransitionenergyisEE=(13:6eV)(1=221=42)=2:55eV:42Thewavelengthishc(1240eV
nm)===486nm:
E(2:55eV)(c)ThenextenergeticstatestartsonE5.ThetransitionenergyisEE=(13:6eV)(1=221=52)=2:86eV:52Thewavelengthishc(1240eV
nm)===434nm:
E(2:86eV)(d)ThenextenergeticstatestartsonE6.ThetransitionenergyisEE=(13:6eV)(1=221=62)=3:02eV:62Thewavelengthiskhdaw.comhc(1240eV
nm)===411nm:
E(3:02eV)(e)ThenextenergeticstatestartsonE7.ThetransitionenergyisEE=(13:6eV)(1=221=72)=3:12eV:72Thewavelengthishc(1240eV
nm)===397nm:
E(3:12eV)E47-14InExercise47-21weshowthatthehydrogenlevelscanbewrittenaswww.khdaw.comE=(13:6eV)=n2:nThetransitionenergyishc(1240eV
nm)
E===10:20eV:(121:6nm)ThismustbepartoftheLymanseries,sothehigherstatemustbeEn=(10:20eV)(13:6eV)=3:4eV:Thatwouldcorrespondto课后答案网n=2.E47-15Thebindingenergyistheenergyrequiredtoremovetheelectron.Iftheenergyoftheelectronisnegative,thenthatnegativeenergyisameasureoftheenergyrequiredtosettheelectronfree.Therstexcitedstateiswhenn=2inEq.47-21.Itisnotnecessarytore-evaluatetheconstantsinthisequationeverytime,instead,westartfromE1En=whereE1=13:60eV:n2Thentherstexcitedstatehasenergy(13:6eV)E2==3:4eV:(2)2Thebindingenergyisthen3.4eV.khdaw.com262若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE47-16r=an2,son0pn=(847pm)=(52:9pm)=4:E47-17(a)Theenergyofthisphotonishc(1240eV
nm)E===0:96739eV:(1281:8nm)Thenalstateofthehydrogenmusthaveanenergyofnomorethan0:96739,sothelargestpossiblenofthenalstateispn<13:60eV=0:96739eV=3:75;sothenalnis1,2,or3.Theinitialstateisonlyslightlyhigherthanthenalstate.Thejumpfromn=2ton=1istoolarge(seeExercise15),anyotherinitialstatewouldhavealargerenergydierence,son=1isnotthenalstate.khdaw.comSowhatlevelmightbeaboven=2?We"lltrypn=13:6eV=(3:4eV0:97eV)=2:36;whichissofarfrombeinganintegerthatwedon"tneedtolookfarther.Then=3statehasenergy13:6eV=9=1:51eV.Thentheinitialstatecouldbepn=13:6eV=(1:51eV0:97eV)=5:01;whichiscloseenoughto5thatwecanassumethetransitionwasn=5ton=3.(b)ThisbelongstothePaschenseries.www.khdaw.comE47-18InExercise47-21weshowthatthehydrogenlevelscanbewrittenasE=(13:6eV)=n2:n(a)Thetransitionenergyis
E=EE=(13:6eV)(1=121=42)=12:8eV:41(b)Alltransitionsn!mareallowedforn4andm4.Weonlyknowthatkhdaw.comms=1=2.269若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE47-46Thereare2n2statesinashelln,soifn=5thereare50states.E47-47Eachisinthen=1shell,thel=0angularmomentumstate,andtheml=0state.Butoneisinthestatems=+1=2whiletheotherisinthestatems=1=2.E47-48ApplyEq.47-31:(+1=2)=arccosp=54:7(1=2)(1=2+1)and(1=2)=arccosp=125:3:(1=2)(1=2+1)E47-49Allofthestatementsaretrue.E47-50khdaw.comTherearenpossiblevaluesforl(startat0!).Foreachvalueoflthereare2l+1possiblevaluesforml.Ifn=1,thesumis1.Ifn=2,thesumis1+3=4.Ifn=3,thesumis1+3+5=9.Thepatternisclear,thesumisn2.Buttherearetwospinstates,sothenumberofstatesis2n2.P47-1Wecansimplifytheenergyexpressionas
h2E=En2+n2+n2whereE=:0xyz08mL2Tondthelowestenergylevelsweneedtofocusonthevaluesofnx,ny,andnz.Itdoesn"ttakemuchimaginationtorealizethattheset(1www.khdaw.com;1;1)willresultinthesmallestvalueforn2+n2+n2.Thenextchoiceistosetoneofthevaluesequalto2,andtrytheset(2;1;1).xyzThenitstartstogetharder,asthenextlowestmightbeeither(2;2;1)or(3;1;1).Theonlywaytondoutistotry.I"lltabulatetheresultsforyou:nnnn2+n2+n2Mult.nnnn2+n2+n2Mult.xyzxyzxyzxyz111313211462116332217322193411183311113331193222121421216课后答案网Wearenowinapositiontostatethevelowestenergylevels.Thefundamentalquantityis(hc)2(1240eV
nm)2E===6:02106eV:08mc2L28(0:511106eV)(250nm)2Thevelowestlevelsarefoundbymultiplyingthisfundamentalquantitybythenumbersinthetableabove.P47-2(a)Writethestatesbetween0andL.Thenallstates,oddoreven,canbewrittenwithprobabilitydistributionfunction22nxP(x)=sin;LL270khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comwendtheprobabilityofndingtheparticleintheregion0xL=3isZL=322nxP=cosdx;0LL1sin(2n=3)=1:32n=3(b)Ifn=1usetheformulaandP=0:196.(c)Ifn=2usetheformulaandP=0:402.(d)Ifn=3usetheformulaandP=0:333.(e)Classicallytheprobabilitydistributionfunctionisuniform,sothereisa1/3chanceofndingitintheregion0toL/3.P47-3TheregionofinterestissmallcomparedtothevariationinP(x);assuchwecanapproxi-matetheprobabilitywiththeexpressionP=P(x)
x:khdaw.com(b)Evaluating,224xP=sin
x;LL224(L=8)=sin(0:0003L);LL=0:0006:(b)Evaluating,224xP=sin
x;LLwww.khdaw.com224(3L=16)=sin(0:0003L);LL=0:0003:P47-4(a)P=,or2P=A2e2m!x=h:0(b)Integrating,Z121=A2e2m!x=hdx;01课后答案网rZ12hu2=A0edu;2m!1rhp=A2pi;02m!r42m!=A0:h(c)x=0.P47-5Wewillwantanexpressionford20:dx2271khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comDoingthemathonederivativeatatime,d2dd0=0;dx2dxdxdm!x2=h=A0(2m!x=h)e;dx22=A(2m!x=h)2em!x=h+A(2m!=h)em!x=h;00
2=(2m!x=h)2(2m!=h)Aem!x=h;0
=(2m!x=h)2(2m!=h):0Inthelastlinewefactoredout0.Thiswillmakeourliveseasierlateron.NowwewanttogotoSchr•odinger"sequation,andmakesomesubstitutions.h2d282mdx20+U0=E0;h2
(2m!x=h)2(2m!=h)+U=E;khdaw.com82m000h2
(2m!x=h)2(2m!=h)+U=E;82mwhereinthelastlinewedividedthroughby0.Nowforsomealgebra,h2
U=E+(2m!x=h)2(2m!=h);82mm!2x2h!=E+:24ButwearegiventhatE=h!=4,sothissimpliestowww.khdaw.comm!2x2U=2whichlookslikeaharmonicoscillatortypepotential.P47-6Assumetheelectronisoriginallyinthestaten.Theclassicalfrequencyoftheelectronisf0,wheref0=v=2r:Accordingtoelectrostaticsanduniformcircularmotion,课后答案网mv2=r=e2=4r2;0orsse2e4e2v====:40mr420h2n220hnThene21me2me42E1f0=2hn2h2n2=42h3n3=hn3000HereE1=13:6eV.Photonfrequencyisrelatedtoenergyaccordingtof=
Enm=h,where
Enmistheenergyoftransitionfromstatendowntostatem.ThenE111f=;hn2m2khdaw.com272若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comwhereE1=13:6eV.Combiningthefractionsandlettingm=n,whereisaninteger,Em2n12f=;hm2n2E1(nm)(m+n)=;hm2n2E1(2n+)=;h(n+)2n2E1(2n);h(n)2n22E1==f0:hn3P47-7Weneedtousethereducedmassofthemuonsincethemuonandprotonmassesaresoclosetogether.Then(207)(1836)khdaw.comm=me=186me:(207)+(1836)(a)ApplyEq.47-201/2:a=a0=(186)=(52:9pm)=(186)=0:284pm:(b)ApplyEq.47-21:E=E1(186)=(13:6eV)(186)=2:53keV:(c)=(1240keV
pm)=(2:53pm)=490pm:www.khdaw.comP47-8(a)Thereducedmassoftheelectronis(1)(1)m=me=0:5me:(1)+(1)Thespectrumissimilar,exceptforthisadditionalfactorof1/2;hencepos=2H:(b)apos=a0=(186)=(52:9pm)=(1=2)=105:8pm.Thisisthedistancebetweentheparticles,buttheyarebothrevolvingaboutthecenterofmass.Theradiusisthenhalfthisquantity,or52:9pm.课后答案网P47-9Thisproblemisn"treallythatmuchofaproblem.Startwiththemagnitudeofavectorintermsofthecomponents,L2+L2+L2=L2;xyzandthenrearrange,L2+L2=L2L2:xyzAccordingtoEq.47-28L2=l(l+1)h2=42,whileaccordingtoEq.47-30L=mh=2.Substitutezlthatintotheequation,and
h2L2+L2=l(l+1)h2=42m2h2=42=l(l+1)m2:xyll42Takethesquarerootofbothsidesofthisexpression,andwearedone.khdaw.com273若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comThemaximumvalueformlisl,whiletheminimumvalueis0.Consequently,qqpL2x+L2y=l(l+1)m2h=2l(l+1)h=2;landqqpL2x+L2y=l(l+1)m2h=2lh=2:lP47-10Assumethat(0)isareasonableestimatefor(r)everywhereinsidethesmallsphere.Thene012==:a3a300Theprobabilityofndingitinasphereofradius1:11015misZ1:11015m4r2dr4(1:11015m)3==1:21014:khdaw.com0a303(5:291011m)3P47-11Assumethat(0)isareasonableestimatefor(r)everywhereinsidethesmallsphere.Then(2)2e012==:32a38a300Theprobabilityofndingitinasphereofradius1:11015misZ1:11015m4r2dr1(1:11015m)3==1:51015:08a306(5:291011m)3www.khdaw.comP47-12(a)Thewavefunctionsquaredise2r=a02=a30Theprobabilityofndingitinasphereofradiusr=xa0isZxa04r2e2r=a0drP=;a300Zx=4x2e2xdx;课后答案网0=1e2x(1+2x+2x2):(b)Letx=1,thenP=1e2(5)=0:323:P47-13WewanttoevaluatethedierencebetweenthevaluesofPatx=2andx=2.Then

P(2)P(1)=1e4(1+2(2)+2(2)2)1e2(1+2(1)+2(1)2);=5e213e4=0:439:274khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP47-14UsingtheresultsofProblem47-12,0:5=1e2x(1+2x+2x2);ore2x=1+2x+2x2:Theresultisx=1:34,orr=1:34a0.P47-15Theprobabilityofndingitinasphereofradiusr=xa0isZxa0r2(2r=a)2er=a0dr0P=8a300Zx122x=x(2x)edx80khdaw.com=1ex(y4=8+y2=2+y+1):Theminimumoccursatx=2,soP=1e2(2+2+2+1)=0:0527:www.khdaw.com课后答案网275khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE48-1Thehighestenergyx-rayphotonwillhaveanenergyequaltothebombardingelectrons,asisshowninEq.48-1,hcmin=eVInserttheappropriatevaluesintotheaboveexpression,(4:141015eV
s)(3:00108m=s)1240109eV
mmin==:eVeVTheexpressionisthen1240109V
m1240kV
pmmin==:VVSolongaswearecertainthattheV"willbemeasuredinunitsofkilovolts,wecanwritethisasmin=1240pm=V:E48-2khdaw.comf=c==(3:00108m=s)=(31:11012m)=9:6461018=s.Planck"sconstantisthenE(40:0keV)15h===4:1410eV
s:f(9:6461018=s)E48-3ApplyingtheresultsofExercise48-1,(1240kV
pm)
V==9:84kV:(126pm)E48-4(a)ApplyingtheresultsofExercise48-1,www.khdaw.com(1240kV
pm)min==35:4pm:(35:0kV)(b)ApplyingtheresultsofExercise45-1,(1240keV
pm)K==49:6pm:(25:51keV)(0:53keV)(c)ApplyingtheresultsofExercise45-1,(1240keV
pm)课后答案网K==56:5pm:(25:51keV)(3:56keV)E48-5(a)Changingtheacceleratingpotentialofthex-raytubewilldecreasemin.Thenewvaluewillbe(usingtheresultsofExercise48-1)min=1240pm=(50:0)=24:8pm:(b)Kdoesn"tchange.Itisapropertyoftheatom,notapropertyoftheacceleratingpotentialofthex-raytube.TheonlywayinwhichtheacceleratingpotentialmightmakeadierenceisifK<minforwhichcasetherewouldnotbeaKline.(c)Kdoesn"tchange.Seepart(b).276khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE48-6(a)ApplyingtheresultsofExercise45-1,(1240keV
pm)
E==64:2keV:(19:3pm)(b)Thisisthetransitionn=2ton=1,so
E=(13:6eV)(1=121=22)=10:2eV:E48-7ApplyingtheresultsofExercise45-1,(1240keV
pm)
E==19:8keV:(62:5pm)and(1240keV
pm)
E==17:6keV:khdaw.com(70:5pm)Thedierenceis
E=(19:8keV)(17:6keV)=2:2eV:E48-8SinceE=hf=hc=,and=h=mc=hc=mc2,thenE=hc==mc2:or
V=E=e=mc2=e=511kV:E48-9The50.0keVelectronmakesacollisionandloseshalfofitsenergytoaphoton,thenthewww.khdaw.comphotonhasanenergyof25.0keV.Theelectronisnowa25.0keVelectron,andonthenextcollisionagainlosesloseshalfofitsenergytoaphoton,thenthisphotonhasanenergyof12.5keV.Onthethirdcollisiontheelectronlosestheremainingenergy,sothisphotonhasanenergyof12.5keV.Thewavelengthsofthesephotonswillbegivenby(1240keV
pm)=;EwhichisavariationofExercise45-1.E48-10(a)Thex-raywillneedtoknockfreeaKshellelectron,soitmusthaveanenergyofatleast69.5keV.课后答案网(b)ApplyingtheresultsofExercise48-1,(1240kV
pm)min==17:8pm:(69:5kV)(c)ApplyingtheresultsofExercise45-1,(1240keV
pm)K==18:5pm:(69:5keV)(2:3keV)ApplyingtheresultsofExercise45-1,(1240keV
pm)K==21:3pm:(69:5keV)(11:3keV)khdaw.com277若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE48-11(a)ApplyingtheresultsofExercise45-1,(1240keV
pm)EK==19:7keV:(63pm)AgainapplyingtheresultsofExercise45-1,(1240keV
pm)EK==17:5keV:(71pm)(b)ZrorNb;theotherswillnotsignicantlyabsorbeitherline.E48-12ApplyingtheresultsofExercise45-1,(1240keV
pm)K==154:5pm:(8:979keV)(0:951keV)ApplyingtheBraggre ectionrelationship,khdaw.com(154:5pm)d===282pm:2sin2sin(15:9)pE48-13Plotthedata.TheplotshouldlookjustlikeFig48-4.Notethattheverticalaxisisppf,whichisrelatedtothewavelengthaccordingtof=c=:E48-14RememberthattheminEq.48-4referstotheelectron,notthenucleus.ThismeansppthattheconstantCinEq.48-5isthesameforallelements.Sincef=c=,wehave21Zwww.khdaw.com21=:2Z11ForGaandNbthewavelengthratioisthen2Nb(31)1==0:5625:Ga(41)1E48-15(a)Thegroundstatequestionisfairlyeasy.Then=1shelliscompletelyoccupiedbythersttwoelectrons.Sothethirdelectronwillbeinthen=2state.Thelowestenergyangularmomentumstateinanyshellisthessub-shell,correspondingtol=0.Thereisonlyonechoiceformlinthiscase:ml=0.Thereisnowayatthislevelofcoveragetodistinguishbetweentheenergy课后答案网ofeitherthespinuporspindownconguration,sowe"llarbitrarilypickspinup.(b)Determiningthecongurationfortherstexcitedstatewillrequiresomethought.WecouldassumethatoneoftheKshellelectrons(n=1)ispromotedtotheLshell(n=2).OrwecouldassumethattheLshellelectronispromotedtotheMshell.OrwecouldassumethattheLshellelectronremainsintheLshell,butthattheangularmomentumvalueischangedtol=1.Thequestionthatwewouldneedtoansweriswhichofthesepossibilitieshasthelowestenergy.Theansweristhelastchoice:increasingthelvalueresultsinasmallincreaseintheenergyofmulti-electronatoms.E48-16RefertoSampleProblem47-6:a(1)2(5:291011m)r=0==5:751013m:1Z(92)khdaw.com278若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE48-17Wewillassumethattheorderingoftheenergyoftheshellsandsub-shellsisthesame.Thatorderingis1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p<8s:Ifthereisnospinthessub-shellwouldhold1electron,thepsub-shellwouldhold3,thedsub-shell5,andthefsub-shell7.inertgasesoccurwhenapsub-shellhaslled,sotherstthreeinertgaseswouldbeelement1(Hydrogen),element1+1+3=5(Boron),andelement1+1+3+1+3=9(Fluorine).Isthereapattern?Yes.Thenewinertgaseshavehalfoftheatomicnumberoftheoriginalinertgases.Thefactorofone-halfcomesaboutbecausetherearenolongertwospinstatesforeachsetofn;l;mlquantumnumbers.Wecansavetimeandsimplydividetheatomicnumbersoftheremaininginertgasesinhalf:element18(Argon),element27(Cobalt),element43(Technetium),element59(Praseodymium).E48-18khdaw.comThepatternis2+8+8+18+18+32+32+?or2(12+22+22+33+33+42+42+x2)Theunknownisprobablyx=5,thenextnobleelementisprobably118+2
52=168:E48-19(a)ApplyEq.47-23,whichcanbewrittenas(13www.khdaw.com:6eV)Z2En=:n2Forthevalenceelectronofsodiumn=3,s(5:14eV)(3)2Z==1:84;(13:6eV)whileforthevalenceelectronofpotassiumn=4,s(4:34eV)(4)2Z==2:26;课后答案网(13:6eV)(b)TheratioswiththeactualvaluesofZare0.167and0.119,respectively.E48-20(a)Therearethreemlstatesallowed,andtwomsstates.TherstelectroncanbeinanyoneofthesesixcombinationsofM1andm2.Thesecondelectron,givennoexclusionprinciple,couldalsobeinanyoneofthesesixstates.Thetotalis36.Unfortunately,thisiswrong,becausewecan"tdistinguishelectrons.Ofthistotalof36,sixinvolvetheelectronsbeinginthesamestate,while30involvetheelectronbeingindierentstates.Butiftheelectronsareindierentstates,thentheycouldbeswapped,andwewon"tknow,sowemustdividethisnumberbytwo.Thetotalnumberofdistinguishablestatesisthen(30=2)+6=21:(b)Six.Seetheabovediscussion.khdaw.com279若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE48-21(a)TheBohrorbitsarecircularorbitsofradiusr=an2(Eq.47-20).Theelectronisn0orbitingwheretheforceise2Fn=;40r2nandthisforceisequaltothecentripetalforce,somv2e2=:rn40rn2wherevisthevelocityoftheelectron.Rearranging,se2v=:40mrnThetimeittakesfortheelectrontomakeoneorbitcanbeusedtocalculatethecurrent,skhdaw.comqeee2i===:t2rn=v2rn40mrnThemagneticmomentofacurrentloopisthecurrenttimestheareaoftheloop,sosee2=iA=r2;n2rn40mrnwhichcanbesimpliedtosee2=rn:24www.khdaw.com0mrnButr=an2,son0sea0e2=n:240mThismightnotlookright,buta=h2=me2,sotheexpressioncansimplifyto00reh2eh=n=n=nB:242m24m(b)Inrealitythemagneticmomentsdependontheangularmomentumquantumnumber,not课后答案网theprinciplequantumnumber.AlthoughtheBohrtheorycorrectlypredictsthemagnitudes,itdoesnotcorrectlypredictwhenthesevalueswouldoccur.E48-22(a)ApplyEq.48-14:dBz24325Fz=z=(9:2710J=T)(1610T=m)=1:510N:dz(b)a=F=m,
z=at2=2,andt=y=v.ThenyFy2(1:51025N)(0:82m)2
z===3:2105m:2mvy22(1:671027kg)(970m=s)2E48-23a=(9:271024J=T)(1:4103T=m)=(1:71025kg)=7khdaw.com:6104m=s2:280若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE48-24(a)
U=2B,or
U=2(5:79105eV=T)(0:520T)=6:02105eV:(b)f=E=h=(6:02105eV)(4:141015eV
s)=1:451010Hz:(c)=c=f=(3108m=s)=(1:451010Hz)=2:07102m:E48-25TheenergychangecanbederivedfromEq.48-13;wemultiplybyafactorof2becausethespiniscompletely ipped.Then
E=2B=2(9:271024J=T)(0:190T)=3:521024J:zzThecorrespondingwavelengthishc(6:631034J
s)(3:00108m=s)===5:65102m:E(3:521024J)Thisissomewherenearthemicrowaverange.khdaw.comE48-26ThephotonhasanenergyE=hc=.ThisenergyisrelatedtothemagneticeldinthevicinityoftheelectronaccordingtoE=2B;sohc(1240eV
nm)B===0:051T:22(5:79105J=T)(21107nm)E48-27ApplyingtheresultsofExercise45-1,www.khdaw.com(1240eV
nm)E==1:55eV:(800nm)Theproductionrateisthen(5:0103W)R==2:01016=s:(1:55eV)(1:61019J=eV)E48-28(a)x=(3108m=s)(121012s)=3:6103m:(b)ApplyingtheresultsofExercise45-1,课后答案网(1240eV
nm)E==1:786eV:(694:4nm)ThenumberofphotonsinthepulseisthenN=(0:150J)=(1:786eV)(1:61019J=eV)=5:251017:E48-29Weneedtondouthowmany10MHzwidesignalscantbetweenthetwowavelengths.Thelowerfrequencyisc(3:00108m=s)f===4:291014Hz:1700109m)1281khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comThehigherfrequencyisc(3:00108m=s)f===7:501014Hz:1400109m)1Thenumberofsignalsthatcanbesentinthisrangeisff(7:501014Hz)(4:291014Hz)217==3:2110:(10MHz)(10106Hz)That"squiteanumberoftelevisionchannels.E48-30ApplyingtheresultsofExercise45-1,(1240eV
nm)E==1:960eV:(632:8nm)Thenumberofphotonsemittedinoneminuteisthenkhdaw.com(2:3103W)(60s)N==4:41017:(1:960eV)(1:61019J=eV)E48-31ApplyEq.48-19.E13E11=2(1:2eV):.Theratioisthenn13(2:4eV)=(8:62105eV/K)(2000K)7=e=910:n11E48-32(a)Populationinversionmeansthatthehigherenergystateismorepopulated;thiscanwww.khdaw.comonlyhappeniftheratioinEq.48-19isgreaterthanone,whichcanonlyhappeniftheargumentoftheexponentispositive.Thatwouldrequireanegativetemperature.(b)Ifn2=1:1n1thentheratiois1:1,so(2:26eV5T==2:7510K:(8:62105eV/K)ln(1:1)E48-33(a)AtthermalequilibriumthepopulationratioisgivenbyNeE2=kT2==e
E=kT:课后答案网N1eE1=kTBut
Ecanbewrittenintermsofthetransitionphotonwavelength,sothisexpressionbecomesN=Nehc=kT:21Puttinginthenumbers,5N=(4:01020)e(1240eV
nm)=(582nm)(8:6210eV=K)(300K))=6:621016:2That"seectivelynone.(b)Ifthepopulationoftheupperstatewere7:01020,theninasinglelaserpulsehc(6:631034J
s)(3:00108m=s)E=N=(7:01020)=240J:(582109m)khdaw.com282若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE48-34Theallowedwavelengthinastandingwavechamberaren=2L=n.Forlargenwecanwrite2L2L2Ln+1=:n+1nn2Thewavelengthdierenceisthen2L2n
==;n22Lwhichinthiscaseis(533109m)2
==1:71012m:2(8:3102m)E48-35(a)Thecentraldiskwillhaveanangleasmeasuredfromthecentergivenbydsin=(1:22);andsincetheparallelraysofthelaserarefocusedonthescreeninadistancef,wealsohaveR=f=sin.Combining,andrearranging,khdaw.com1:22fR=:d(b)R=1:22(3:5cm)(515nm)=(3mm)=7:2106m.(c)I=P=A=(5:21W)=(1:5mm)2=7:37105W=m2:(d)I=P=A=(0:84)(5:21W)=(7:2m)2=2:71010W=m2:E48-36P48-1Let1bethewavelengthoftherstphoton.Then2=1+130pm.Thetotalenergytransferedtothetwophotonsisthenhcwww.khdaw.comhcE1+E2=+=20:0keV:12Wecansolvethisfor1,20:0keV11=+;hc11+130pm21+130pm=;1(1+130pm)whichcanalsobewrittenas1(1+130pm)=(62pm)(21+130pm);课后答案网2+(6pm)(8060pm2)=0:11Thisequationhassolutions1=86:8pmand92:8pm:Onlythepositiveanswerhasphysicalmeaning.Theenergyofthisrstphotonisthen(1240keV
pm)E1==14:3keV:(86:8pm)(a)Afterthisrstphotonisemittedtheelectronstillhasakineticenergyof20:0keV14:3keV=5:7keV:(b)Wefoundtheenergyandwavelengthoftherstphotonabove.Theenergyofthesecondphotonmustbe5.7keV,withwavelength2=(86:8pm)+130pm=217pmkhdaw.com:283若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP48-2Originally,1=p=2:412:1(2:73108m=s)2=(3108m=s)2TheenergyoftheelectronisE= mc2=(2:412)(511keV)=1232keV:0UponemittingthephotonthenewenergyisE=(1232keV)(43:8keV)=1189keV;sothenewgammafactoris=(1189keV)=(511keV)=2:326;andthenewspeedispv=c11=(2:326)2=(0:903)c:P48-3khdaw.comSwitchtoareferenceframewheretheelectronisoriginallyatrest.Momentumconservationrequires0=p+pe=0;whileenergyconservationrequiresmc2=E+E:eRearrangetoE=mc2E:eSquarebothsidesofthisenergyexpressionandE22Emc2+m2c4=E2=p2c2+m2c4;www.khdaw.comeeE22Emc2=p2c2;ep2c22Emc2=p2c2:eButthemomentumexpressioncanbeusedhere,andtheresultis2Emc2=0:Notlikely.P48-4(a)IntheBohrtheorywecanassumethattheKshellelectronssee"anucleuswithchargeZ.TheLshellelectrons,however,areshieldedbytheoneelectroninthe课后答案网Kshellandsotheysee"anucleuswithchargeZ1.Finally,theMshellelectronsareshieldedbytheoneelectronintheKshellandtheeightelectronsintheKshell,sotheysee"anucleuswithchargeZ9.Thetransitionwavelengthsarethen1
EE(Z1)2110==;hchc2212E(Z1)230=:hc4and1
EE011==;hchc3212E(Z9)280=:hc9khdaw.com284若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comTheratioofthesetwowavelengthsis27(Z1)2=:32(Z9)2Notethattheformulainthetexthasthesquareinthewrongplace!P48-5(a)E=hc=;theenergydierenceisthen11
E=hc;1221=hc:21hc=
:21Sinceandaresoclosetogetherwecantreattheproductasbeingeither2or2.Thenkhdaw.com121212(1240eV
nm)3
E=(0:597nm)=2:110eV:(589nm)2(b)Thesameenergydierenceexistsinthe4s!3pdoublet,so(1139nm)2
=(2:1103eV)=2:2nm:(1240eV
nm)P48-6(a)WecanassumethattheKshellelectronsees"anucleusofchargeZ1,sincetheotherelectronintheshellscreensit.Then,accordingtothederivationleadingtoEq.47-22,www.khdaw.comrK=a0=(Z1):(b)Theoutermostelectronsees"anucleusscreenedbyalloftheotherelectrons;assuchZ=1,andtheradiusisr=a0P48-7Weassumeinthiscrudemodelthatoneelectronmovesinacircularorbitattractedtotheheliumnucleusbutrepelledfromtheotherelectron.LookbacktoSampleProblem47-6;weneedtousesomeoftheresultsfromthatSampleProblemtosolvethisproblem.Thefactorofe2inEq.47-20(theexpressionfortheBohrradius)andthefactorof(e2)2inEq.47-21(theexpressionfortheBohrenergylevels)wasfromtheCoulombforcebetweenthesingle课后答案网electronandthesingleprotoninthenucleus.Thisforceise2F=:40r2Inourapproximationtheforceofattractionbetweentheoneelectronandtheheliumnucleusis2e2F1=:40r2Thefactoroftwoisbecausetherearetwoprotonsintheheliumnucleus.Thereisalsoarepulsiveforcebetweentheoneelectronandtheotherelectron,e2F2=;40(2r)2khdaw.com285若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comwherethefactorof2risbecausethetwoelectronsareonoppositesideofthenucleus.Thenetforceontherstelectroninourapproximationisthen2e2e2F1F2=;40r240(2r)2whichcanberearrangedtoyielde21e27Fnet=2=:40r2440r24Itisapparentthatweneedtosubstitute7e2=4foreveryoccurrenceofe2.(a)ThegroundstateradiusoftheheliumatomwillthenbegivenbyEq.47-20withtheappropriatesubstitution,h240r==a0:m(7e2=4)7khdaw.com(b)TheenergyofoneelectroninthisgroundstateisgivenbyEq.47-21withthesubstitutionof7e2=4foreveryoccurrenceofe2,thenm(7e2=4)249me4E==:840h216840h2Wealreadyevaluatedalloftheconstantstobe13.6eV.Onelastthing.Therearetwoelectrons,soweneedtodoubletheaboveexpression.Thegroundstateenergyofaheliumatominthisapproximationis49E0=2(13:6eV)=83:3eV:16www.khdaw.com(c)Removingoneelectronwillallowtheremainingelectrontomoveclosertothenucleus.TheenergyoftheremainingelectronisgivenbytheBohrtheoryforHe+,andisEHe+=(4)(13:60eV)=54:4eV;sotheionizationenergyis83.3eV-54.4eV=28.9eV.Thiscompareswellwiththeacceptedvalue.P48-8ApplyingEq.48-19:(3:2eV)4T==1:010K:课后答案网(8:62105eV=K)ln(6:11013=2:51015)P48-9sinr=R,whereristheradiusofthebeamonthemoonandRisthedistancetothemoon.Then1:22(600109m)(3:82108m)r==2360m:(0:118m)Thebeamdiameteristwicethis,or4740m.P48-10(a)N=2L=n,or2(6102m)(1:75)N==3:03105:(694109)286khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com(b)N=2nLf=c,soc(3108m=s)
f===1:43109=s:2nL2(1:75)(6102m)Notethatthetraveltimetoandfrois
t=2nL=c!(c)
f=fisthen
f(694109)===3:3106:f2nL2(1:75)(6102m)khdaw.comwww.khdaw.com课后答案网287khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE49-1(a)Equation49-2ispp82m3=282(mc2)3=2n(E)=E1=2=E1=2:h3(hc)3Wecanevaluatethisbysubstitutinginallknownquantities,p82(0:511106eV)3=2n(E)=E1=2=(6:811027m3
eV3=2)E1=2:(1240109eV
m)3Onceagain,wesimpliedtheexpressionbywritinghcwhereverwecould,andthenusinghc=1240109eV
m.(b)Then,ifE=5:00eV,n(E)=(6:811027m3
eV3=2)(5:00eV)1=2=1:521028m3
eV1:E49-2ApplytheresultsofEx.49-1:khdaw.comn(E)=(6:811027m3
eV3=2)(8:00eV)1=2=1:931028m3
eV1:E49-3Monovalentmeansonlyoneelectronisavailableasaconductingelectron.Henceweneedonlycalculatethedensityofatoms:NN(19:3103kg=m3)(6:021023mol1)=A==5:901028=m3:VAr(0:197kg=mol)E49-4Usetheidealgaslaw:pV=NkT.ThenN283238p=kT=(8:4910m)(1:3810J=
K)(297K)=3:4810Pa:Vwww.khdaw.comE49-5(a)Theapproximatevolumeofasinglesodiumatomis(0:023kg=mol)293V1==3:9310m:(6:021023part/mol)(971kg=m3)Thevolumeofthesodiumionsphereis4123303V2=(9810m)=3:9410m:3ThefractionalvolumeavailableforconductionelectronsisVV(3:931029m3)(3:941030m3)12课后答案网V=(3:931029m3)=90%:1(b)Theapproximatevolumeofasinglecopperatomis(0:0635kg=mol)293V1==1:1810m:(6:021023part/mol)(8960kg=m3)Thevolumeofthecopperionsphereis4123303V2=(9610m)=3:7110m:3ThefractionalvolumeavailableforconductionelectronsisVV(1:181029m3)(3:711030m3)12==69%:V1(1:181029m3)(c)Sodium,sincemoreofthevolumeisavailablefortheconductionelectron.khdaw.com288若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE49-6(a)ApplyEq.49-6:hi5p=1=e(0:0730eV)=(8:6210eV=K)(0K)+1=0:(b)ApplyEq.49-6:hi5p=1=e(0:0730eV)=(8:6210eV=K)(320K)+1=6:62102:E49-7ApplyEq.49-6,rememberingtousetheenergydierence:hi5p=1=e(1:1)eV)=(8:6210eV=K)(273K)+1=1:00;hi5p=1=e(0:1)eV)=(8:6210eV=K)(273K)+1=0:986;hi5p=1=e(0:0)eV)=(8:6210eV=K)(273K)+1=0:5;hikhdaw.com(0:1)eV)=(8:62105eV=K)(273K)p=1=e+1=0:014;hi5p=1=e(1:1)eV)=(8:6210eV=K)(273K)+1=0:0:(b)Invertingtheequation,
ET=;kln(1=p1)so(0:1eV)T==700K(8:62105eV=K)ln(1www.khdaw.com=(0:16)1)E49-8Theenergydierencesareequal,exceptforthesign.Then11+=;e+
E=kt+1e
E=kt+1e
E=2kte+
E=2kt+=;e+
E=2kt+e
E=2kte
E=2kt+e+
E=2kte
E=2kt+e+
E=2kt=1:e
E=2kt+e+
E=2ktE49-9TheFermienergyisgivenbyEq.49-5,课后答案网22=3h3nEF=;8mwherenisthedensityofconductionelectrons.Forgoldwehave(19:3g/cm3)(6:021023part/mol)2233n==5:9010elect./cm=59elect./nm(197g/mol)TheFermienergyisthen!2=3(1240eV
nm)23(59electrons/nm3)EF==5:53eV:8(0:511106eV)289khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE49-10CombinetheresultsofEx.49-1andEq.49-6:pCEno=:e
E=kt+1Thenforeachoftheenergieswehavep(6:811027m3
eV3=2)(4eV)n==1:361028=m3
eV;oe(3:06eV)=(8:62105eV=K)(1000K)+1p(6:811027m3
eV3=2)(6:75eV)n==1:721028=m3
eV;oe(0:31eV)=(8:62105eV=K)(1000K)+1p(6:811027m3
eV3=2)(7eV)n==9:021027=m3
eV;oe(0:06eV)=(8:62105eV=K)(1000K)+1p(6:811027m3
eV3=2)(7:25eV)n==1:821027=m3
eV;oe(0:19eV)=(8:62105eV=K)(1000K)+1p(6:811027m3
eV3=2)(9eV)khdaw.comn==3:431018=m3
eV:oe(1:94eV)=(8:62105eV=K)(1000K)+1E49-11Solven2(hc)2En=8(mc2)L2forn=50,sincetherearetwoelectronsineachlevel.Then(50)2(1240eV
nm)2E==6:53104eV:f8(5:11105eV)(0:12nm)2E49-12WeneedtobemuchhigherthanT=(7www.khdaw.com:06eV)=(8:62105eV=K)=8:2104K:E49-13Equation49-5is22=3h3nEF=;8mandifwecollecttheconstants,22=3h33=23=2EF=n=An;8mwhere,ifwemultiplythetopandbottomby课后答案网c222=3922=3(hc)3(124010eV
m)3192A===3:6510m
eV:8mc28(0:511106eV)E49-14(a)InvertingEq.49-6,
E=kTln(1=p1);so
E=(8:62105eV=K)(1050K)ln(1=(0:91)1)=0:209eV:ThenE=(0:209eV)+(7:06eV)=6:85eV:(b)ApplytheresultsofEx.49-1:n(E)=(6:811027m3
eV3=2)(6:85eV)1=2=1:781028m3
eV1:(c)n=np=(1:781028m3
eV1)(0:910)=1:621028m3
eV1:okhdaw.com290若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE49-15Equation49-5is22=3h3nEF=;8mandifwerearrange,3h3E3=2=pn;F162m3=2Equation49-2isthenp82m3=23n(E)=E1=2=nE3=2E1=2:h32FE49-16ph=1p,so1ph=1;e
E=kT+1e
E=kTkhdaw.com=;e
E=kT+11=:1+e
E=kTE49-17ThestepstosolvethisexerciseareequivalenttothestepsforExercise49-9,exceptnowtheironatomseachcontribute26electronsandwehavetondthedensity.First,thedensityism(1:991030kg)===1:84109kg=m34r3=34(6:3710www.khdaw.com6m)3=3Then(26)(1:84106g/cm3)(6:021023part/mol)293n==5:110elect./cm;(56g/mol)83=5:110elect./nmTheFermienergyisthen!2=3(1240eV
nm)23(5:1108elect./nm3)EF==230keV:课后答案网8(0:511106eV)E49-18First,thedensityism2(1:991030kg)===9:51017kg=m34r3=34(10103m)3=3Thenn=(9:51017kg=m3)=(1:671027kg)=5:691044=m3:TheFermienergyisthen!2=3(1240MeV
fm)23(5:69101/fm3)EF==137MeV:8(940MeV)291khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE49-19E49-20(a)EF=7:06eV,so3(8:62105eV
K)(0K)f==0;2(7:06eV)(b)f=3(8:62105eV
K)(300K)=2(7:06eV)=0:0055:(c)f=3(8:62105eV
K)(1000K)=2(7:06eV)=0:0183:E49-21UsingtheresultsofExercise19,2fEF2(0:0130)(4:71eV)T===474K:3k3(8:62105eV
K)E49-22khdaw.comf=3(8:62105eV
K)(1235K)=2(5:5eV)=0:029:E49-23(a)Monovalentmeansonlyoneelectronisavailableasaconductingelectron.Henceweneedonlycalculatethedensityofatoms:NN(10:5103kg=m3)(6:021023mol1)=A==5:901028=m3:VAr(0:107kg=mol)(b)UsingtheresultsofEx.49-13,E=(3:651019m2
eV)(5:901028=m3)2=3=5:5eV:Fpwww.khdaw.com(c)v=2K=m,orpv=2(5:5eV)(5:11105eV=c2)=1:4108m=s:(d)=h=p,or(6:631034J
s)==5:21012m:(9:111031kg)(1:4108m=s)E49-24(a)Bivalentmeanstwoelectronsareavailableasaconductingelectron.Henceweneedtodoublethecalculationofthedensityofatoms:课后答案网33231NNA2(7:1310kg=m)(6:0210mol)293===1:3210=m:VAr(0:065kg=mol)(b)UsingtheresultsofEx.49-13,E=(3:651019m2
eV)(1:321029=m3)2=3=9:4eV:Fp(c)v=2K=m,orpv=2(9:4eV)(5:11105eV=c2)=1:8108m=s:(d)=h=p,or(6:631034J
s)==4:01012m:(9:111031kg)(1:8108m=s)khdaw.com292若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE49-25(a)RefertoSampleProblem49-5wherewelearnthatthemeanfreepathcanbewrittenintermsofFermispeedvFandmeantimebetweencollisionsas=vF:TheFermispeedisppv=c2E=mc2=c2(5:51eV)=(5:11105eV)=4:64103c:FFThetimebetweencollisionsism(9:111031kg)===3:741014s:ne2(5:861028m3)(1:601019C)2(1:62108
m)WefoundnbylookinguptheanswersfromExercise49-23inthebackofthebook.Themeanfreepathisthen=(4:64103)(3:00108m=s)(3:741014s)=52nm:khdaw.com(b)Thespacingbetweentheioncoresisapproximatedbythecuberootofvolumeperatom.Thisatomicvolumeforsilveris(108g/mol)233V==1:7110cm:(6:021023part/mol)(10:5g/cm3)Thedistancebetweentheionsisthenp3l=V=0:257nm:Theratiois=l=190www.khdaw.com:E49-26(a)ForT=1000Kwecanusetheapproximation,sofordiamond5p=e(5:5eV)=2(8:6210eV=K)(1000K)=1:41014;whileforsilicon,5p=e(1:1eV)=2(8:6210eV=K)(1000K)=1:7103;(b)ForT=4Kwecanusethesameapproximation,butnow
EkTandtheexponentialfunctiongoestozero.课后答案网E49-27(a)EEF0:67eV=2=0:34eV:.Theprobabilitythestateisoccupiedisthenhi5p=1=e(0:34)eV)=(8:6210eV=K)(290K)+1=1:2106:(b)EEF0:67eV=2=0:34eV:.Theprobabilitythestateisunoccupiedisthen1p,orhi5p=11=e(0:34)eV)=(8:6210eV=K)(290K)+1=1:2106:E49-28(a)EEF0:67eV=2=0:34eV:.Theprobabilitythestateisoccupiedisthenhi5p=1=e(0:34)eV)=(8:6210eV=K)(289K)+1=1:2106:293khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE49-29(a)Thenumberofsiliconatomsperunitvolumeis(6:021023part/mol)(2:33g/cm3)223n==4:9910part./cm:(28:1g/mol)Ifoneoutof1:0eex7arereplacedthentherewillbeanadditionalchargecarrierdensityof4:991022part./cm3=1:0107=4:991015part./cm3=4:991021m3:(b)Theratiois(4:991021m3)=(21:51016m3)=1:7105:Theextrafactoroftwoisbecauseallofthechargecarriersinsilicon(holesandelectrons)arechargecarriers.E49-30Sinceoneoutofevery5106siliconatomsneedstobereplaced,thenthemassofphos-phoruswouldbekhdaw.comm=130=2:1107g:510628pE49-31l=31=1022=m3=4:6108m:E49-32TheatomdensityofgermaniumisNN(5:32103kg=m3)(6:021023mol1)=A==1:631028=m3:VAr(0:197kg=mol)Theatomdensityoftheimpurityiswww.khdaw.com(1:631028=m3)=(1:3109)=1:251019:Theaveragespacingispl=31=1:251019=m3=4:3107m:E49-33Therstoneisaninsulatorbecausethelowerbandislledandbandgapissolarge;thereisnoimpurity.Thesecondoneisanextrinsicn-typesemiconductor:itisasemiconductorbecausethelowerbandislledandthebandgapissmall;itisextrinsicbecausethereisanimpurity;sincetheimpuritylevelisclosetothetopofthebandgaptheimpurityisadonor.课后答案网Thethirdsampleisanintrinsicsemiconductor:itisasemiconductorbecausethelowerbandislledandthebandgapissmall.Thefourthsampleisaconductor;althoughthebandgapislarge,thelowerbandisnotcompletelylled.Thefthsampleisaconductor:theFermilevelisabovethebottomoftheupperband.Thesixthoneisanextrinsicp-typesemiconductor:itisasemiconductorbecausethelowerbandislledandthebandgapissmall;itisextrinsicbecausethereisanimpurity;sincetheimpuritylevelisclosetothebottomofthebandgaptheimpurityisanacceptor.E49-346:62105eV=1:1eV=6:0105electron-holepairs.E49-35(a)R=(1V)=(501012A)=21010.(b)R=(0:75V)=(8mA)=90.khdaw.com294若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE49-36(a)Aregionwithsomepotentialdierenceexiststhathasagapbetweenthechargedareas.(b)C=Q=
V.UsingtheresultsinSampleProblem49-9forqand
V,n0eAd=2C==20A=d:n0ed2=40E49-37(a)Applythateversousefulformulahc(1240eV
nm)===225nm:E(5:5eV)Whyisthisamaximum?Becauselongerwavelengthswouldhavelowerenergy,andsonotenoughtocauseanelectrontojumpacrossthebandgap.(b)Ultraviolet.E49-38khdaw.comApplythateversousefulformulahc(1240eV
nm)E===4:20eV:(295nm)E49-39Thephotonenergyishc(1240eV
nm)E===8:86eV:(140nm)whichisenoughtoexcitetheelectronsthroughthebandgap.Assuch,thephotonwillbeabsorbed,whichmeansthecrystalisopaquetothiswavelength.www.khdaw.comE49-40P49-1WecancalculatetheelectrondensityfromEq.49-5,23=28mcEFn=;3(hc)263=28(0:51110eV)(11:66eV)=;课后答案网3(1240eV
nm)23=181electrons/nm:Fromthiswecalculatethenumberofelectronsperparticle,3(181electrons/nm)(27:0g/mol)=3:01;(2:70g/cm3)(6:021023particles/mol)whichwecanreasonablyapproximateas3.295khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP49-2AtabsolutezeroallstatesbelowEFarelled,annoneabove.UsingtheresultsofEx.49-15,ZEF1Eav=En(E)dE;n0ZEF33=23=2=EFEdE;2033=225=2=EFEF;253=EF:5P49-3(a)Thetotalnumberofconductionelectronis(0:0031kg)(6:021023mol1)n==2:941022:khdaw.com(0:0635kg=mol)Thetotalenergyis322234E=(7:06eV)(2:9410)=1:2410eV=210J:5(b)Thiswilllighta100Wbulbfort=(2104J)=(100W)=200s:P49-4(a)Firstdotheeasypart:nc=Ncp(Ec),sowww.khdaw.comNc:e(EcEF)=kT+1ThenusetheresultsofEx.49-16,andwriteNvnv=Nv[1p(Ev)]=:e(EvEF)=kT+1Sinceeachelectronintheconductionbandmusthaveleftaholeinthevalenceband,thenthesetwoexpressionsmustbeequal.(b)Iftheexponentialsdominatethenwecandropthe+1ineachdenominator,and课后答案网NcNv=;e(EcEF)=kTe(EvEF)=kTNc(Ec2EF+Ev)=kT=e;Nv1EF=(Ec+Ev+kTln(Nc=Nv)):2P49-5(a)WewanttouseEq.49-6;althoughwedon"tknowtheFermienergy,wedoknowthedierencesbetweentheenergiesinquestion.Intheun-dopedsiliconEEF=0:55eVforthebottomoftheconductionband.ThequantitykT=(8:62105eV=K)(290K)=0:025eV;whichisagoodnumbertoremember|atroomtemperaturekTis1/40ofanelectron-volt.khdaw.com296若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comThen110p==2:810:e(0:55eV)=(0:025eV)+1InthedopedsiliconEEF=0:084eVforthebottomoftheconductionband.Then12p==3:410:e(0:084eV)=(0:025eV)+1(b)ForthedonorstateEEF=0:066eV,so1p==0:93:e(0:066eV)=(0:025eV)+1P49-6(a)InvertingEq.49-6,EEF=kTln(1=p1);soE=(1:1eV0:11eV)(8:62105eV=K)(290K)ln(1=(4:8105)1)=0:74eVkhdaw.comFabovethevalenceband.(b)EEF=(1:1eV)(0:74eV)=0:36eV,so17p==5:610:e(0:36eV)=(0:025eV)+1P49-7(a)Plotthegraphwithaspreadsheet.ItshouldlooklikeFig.49-12.(b)kT=0:025eVwhenT=290K.Theratioisthenie(0:5eV)=(0:025www.khdaw.comeV)+1f8==4:910:ire(0:5eV)=(0:025eV)+1P49-8课后答案网297khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE50-1WewanttofollowtheexamplesetinSampleProblem50-1.ThedistanceofclosestapproachisgivenbyqQd=;40K(2)(29)(1:601019C)2=;4(8:851012C2=Nm2)(5:30MeV)(1:601013J=MeV)=1:571014m:That"sprettyclose.E50-2(a)Thegoldatomcanbetreatedasapointparticle:q1q2F=;40r2(2)(79)(1:601019C)2=;khdaw.com4(8:851012C2=Nm2)(0:16109m)2=1:4106N:(b)W=Fd,so(5:3106eV)(1:61019J=eV)d==6:06107m:(1:4106N)That"s1900goldatomdiameters.E50-3TakeanapproachsimilartoSampleProblem50-1:qQwww.khdaw.comK=;40d(2)(79)(1:601019C)2=;4(8:851012C2=Nm2)(8:781015m)(1:601019J=eV)=2:6107eV:E50-4Allarestableexcept88Rband239Pb.E50-5WecanmakeanestimateofthemassnumberAfromEq.50-1,课后答案网1=3R=R0A;whereR0=1:2fm.Ifthemeasurementsindicatearadiusof3.6fmwewouldhave33A=(R=R0)=((3:6fm)=(1:2fm))=27:E50-6E50-7ThemassnumberofthesunisA=(1:991030kg)=(1:671027kg)=1:21057:TheradiuswouldbepR=(1:21015m)31:21057=1:3104m:298khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE50-8239Puiscomposedof94protonsand23994=145neutrons.ThecombinedmassofthefreeparticlesisM=Zmp+Nmn=(94)(1:007825u)+(145)(1:008665u)=240:991975u:ThebindingenergyisthedierenceEB=(240:991975u239:052156u)(931:5MeV/u)=1806:9MeV;andthebindingenergypernucleonisthen(1806:9MeV)=(239)=7:56MeV:E50-962Niiscomposedof28protonsand6228=34neutrons.Thecombinedmassofthefreeparticlesiskhdaw.comM=Zmp+Nmn=(28)(1:007825u)+(34)(1:008665u)=62:513710u:ThebindingenergyisthedierenceEB=(62:513710u61:928349u)(931:5MeV/u)=545:3MeV;andthebindingenergypernucleonisthen(545:3MeV)=(62)=8:795MeV:E50-10(a)Multiplyeachby1=1:007825,som1H=1www.khdaw.com:00000;m12C=11:906829;andm238U=236:202500:E50-11(a)Sincethebindingenergypernucleonisfairlyconstant,theenergymustbeproportionaltoA.(b)Coulombrepulsionactsbetweenpairsofprotons;thereareZprotonsthatcanbechosenasrstinthepair,andZ1protonsremainingthatcanmakeupthepartnerinthepair.ThatmakesforZ(Z1)pairs.Theelectrostaticenergymustbeproportionaltothis.课后答案网(c)Z2growsfasterthanA,whichisroughlyproportionaltoZ.E50-12Solve(0:7899)(23:985042)+x(24:985837)+(0:2101x)(25:982593)=24:305forx.Theresultisx=0:1001,andthentheamount26Mgis0:1100.299khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE50-13TheneutronconnedinanucleusofradiusRwillhaveapositionuncertaintyontheorderof
xR.Themomentumuncertaintywillthenbenolessthanhh
p:2
x2RAssumingthatp
p,wehavehp;2Randthentheneutronwillhavea(minimum)kineticenergyofp2h2E:2m82mR2ButR=RA1=3,so0(hc)2E:82mc2R02A2=3Foranatomwithkhdaw.comA=100weget(1240MeV
fm)2E=0:668MeV:82(940MeV)(1:2fm)2(100)2=3Thisisaboutafactorof5or10lessthanthebindingenergypernucleon.E50-14(a)Toremoveaproton,E=[(1:007825)+(3:016049)(4:002603)](931:5MeV)=19:81MeV:Toremoveaneutron,www.khdaw.comE=[(1:008665)+(2:014102)(3:016049)](931:5MeV)=6:258MeV:Toremoveaproton,E=[(1:007825)+(1:008665)(2:014102)](931:5MeV)=2:224MeV:(b)E=(19:81+6:258+2:224)MeV=28:30MeV:(c)(28:30MeV)=4=7:07MeV:E50-15(a)
=[(1:007825)(1)](931:5MeV)=7:289MeV:(b)
=[(1:008665)课后答案网(1)](931:5MeV)=8:071MeV:(c)
=[(119:902197)(120)](931:5MeV)=91:10MeV:E50-16(a)E=(Zm+Nmm)c2.Substitutethedenitionformassexcess,mc2=Ac2+
,BHNandE=Z(c2+
)+N(c2+
)Ac2
;BHN=Z
H+N
N
:(b)For197Au,EB=(79)(7:289MeV)+(19779)(8:071MeV)(31:157MeV)=1559MeV;andthebindingenergypernucleonisthen(1559MeV)=(197)=7:92MeV:khdaw.com300若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE50-17Thebindingenergyof63CuisgivenbyM=Zmp+Nmn=(29)(1:007825u)+(34)(1:008665u)=63:521535u:ThebindingenergyisthedierenceEB=(63:521535u62:929601u)(931:5MeV/u)=551:4MeV:Thenumberofatomsinthesampleis(0:003kg)(6:021023mol1)n==2:871022:(0:0629kg=mol)Thetotalenergyisthen(2:871022)(551:4MeV)(1:61019J=eV)=2:531012J:E50-18khdaw.com(a)Forultra-relativisticparticlesE=pc,so(1240MeV
fm)==2:59fm:(480MeV)(b)Yes,sincethewavelengthissmallerthannuclearradii.E50-19Wewilldothisonetheeasywaybecausewecan.Thismethodwon"tworkexceptwhenthereisanintegernumberofhalf-lives.Theactivityofthesamplewillfalltoone-halfoftheinitialdecayrateafteronehalf-life;itwillfalltoone-halfofone-half(one-fourth)aftertwohalf-lives.Sotwohalf-liveshaveelapsed,foratotalof(2)(140d)=280d.www.khdaw.comE50-20N=N(1=2)t=t1=2,so0N=(481019)(0:5)(26)=(6:5)=3:01019:E50-21(a)t1=2=ln2=(0:0108/h)=64:2h:(b)N=N(1=2)t=t1=2,so0N=N=(0:5)(3)=0:125:0(c)N=N(1=2)t=t1=2,so0课后答案网N=N=(0:5)(240)=(64:2)=0:0749:0E50-22(a)=(dN=dt)=N,or=(12=s)=(2:51018)=4:81018=s:(b)t1=2=ln2=,sot=ln2=(4:81018=s)=1:441017s;1=2whichis4.5billionyears.301khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE50-23(a)Thedecayconstantfor67GacanbederivedfromEq.50-8,ln2ln261===2:46110s:t1=2(2:817105s)TheactivityisgivenbyR=N,sowewanttoknowhowmanyatomsarepresent.Thatcanbefoundfrom1u1atom223:42g=3:07710atoms:1:66051024g66:93uSotheactivityisR=(2:461106=s1)(3:0771022atoms)=7:5721016decays/s:(b)After1:728105stheactivitywouldhavedecreasedto615R=Ret=(7:5721016decays/s)e(2:46110=s)(1:72810s)=4:9491016decays/s:0E50-24N=Net,but=ln2=t,sokhdaw.com01=2t=t11=2N=Neln2t=t1=2=N(2)t=t1=2=N:0002E50-25Theremaining223isN=(4:71021)(0:5)(28)=(11:43)=8:61020:Thenumberofdecays,eachofwhichproducedanalphaparticle,is(4:71021)(8:6www.khdaw.com1020)=3:841021:E50-26Theamountremainingafter14hoursism=(5:50g)(0:5)(14)=(12:7)=2:562g:Theamountremainingafter16hoursism=(5:50g)(0:5)(16)=(12:7)=2:297g:Thedierenceistheamountwhichdecayedduringthetwohourinterval:(2:562g)(2:297g)=0:265g:课后答案网E50-27(a)ApplyEq.50-7,R=Ret:0WerstneedtoknowthedecayconstantfromEq.50-8,ln2ln271===5:61810s:t1=2(1:234106s)Andthethetimeisfoundfrom1Rt=ln;R01(170counts/s)=ln;(5:618107s1)(3050counts/s)=5:139106s59:5days:khdaw.com302若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comNotethatcounts/sisnotthesameasdecays/s.Notalldecayeventswillbepickedupbyadetectorandrecordedasacount;weareassumingthatwhateverscalingfactorwhichconnectstheinitialcountratetotheinitialdecayrateisvalidatlatertimesaswell.Suchanassumptionisareasonableassumption.(b)Thepurposeofsuchanexperimentwouldbetomeasuretheamountofphosphorusthatistakenupinaleaf.Buttheactivityofthetracerdecayswithtime,andsowithoutacorrectionfactorwewouldrecordthewrongamountofphosphorusintheleaf.ThatcorrectionfactorisR0=R;weneedtomultiplythemeasuredcountsbythisfactortocorrectforthedecay.InthiscaseRt(5:618107s1)(3:007105s)=e=e=1:184:R0E50-28Thenumberofparticlesof147Smis(0:001kg)(6:021023mol1)n=(0:15)=6:1431020:(0:147kg=mol)Thedecayconstantiskhdaw.com=(120=s)=(6:1431020)=1:951019=s:Thehalf-lifeist=ln2=(1:951019=s)=3:551018s;1=2or110Gy.E50-29Thenumberofparticlesof239Puis(0:012kg)(6:021023mol1)n==3:0231022:0(0:239kg=mol)www.khdaw.comThenumberwhichdecayishinn=(3:0251022)1(0:5)(20000)=(24100)=1:321022:0Themassofheliumproducedisthen(0:004kg=mol)(1:321022)m==8:78105kg:(6:021023mol1)E50-30LetR33=(R33+R32)=x,wherex0=0:1originally,andwewanttondoutatwhattimex=0:9.Rearranging,课后答案网(R33+R32)=R33=1=x;soR32=R33=1=x1:SinceR=R(0:5)t=t1=2wecanwritearatio011=11(0:5)t=t32t=t33:xx0Putinsomeofthenumbers,and11ln[(1=9)=(9)]=ln[0:5]t;14:325:3whichhassolutiont=209d.khdaw.com303若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE50-31E50-32(a)N=N=(0:5)(4500)=(82)=3:01017.0(b)N=N=(0:5)(4500)=(0:034)=0.0E50-33TheQvaluesareQ3=(235:043923232:0380503:016029)(931:5MeV)=9:46MeV;Q4=(235:043923231:0362974:002603)(931:5MeV)=4:68MeV;Q5=(235:043923230:0331275:012228)(931:5MeV)=1:33MeV:OnlyreactionswithpositiveQvaluesareenergeticallypossible.E50-34khdaw.com(a)Forthe14Cdecay,Q=(223:018497208:98107514:003242)(931:5MeV)=31:84MeV:Forthe4Hedecay,Q=(223:018497219:0094754:002603)(931:5MeV)=5:979MeV:E50-35Q=(136:907084136:905821)(931:5MeV)=1:17MeV:E50-36Q=(1:0086651:007825)(931:5MeV)=0:782MeV:www.khdaw.comE50-37(a)Thekineticenergyofthiselectronissignicantcomparedtotherestmassenergy,sowemustuserelativitytondthemomentum.ThetotalenergyoftheelectronisE=K+mc2,themomentumwillbegivenbypppc=E2m2c4=K2+2Kmc2;p=(1:00MeV)2+2(1:00MeV)(0:511MeV)=1:42MeV:ThedeBrogliewavelengthisthenhc(1240MeV
fm)===873fm:课后答案网pc(1:42MeV)(b)TheradiusoftheemittingnucleusisR=RA1=3=(1:2fm)(150)1=3=6:4fm:0(c)Thelongestwavelengthstandingwaveonastringxedateachendistwicethelengthofthestring.Althoughtherulesforstandingwavesinaboxareslightlymorecomplicated,itisafairassumptionthattheelectroncouldnotexistasastandingawaveinthenucleus.(d)Seepart(c).304khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE50-38Theelectronisrelativistic,soppc=E2m2c4;p=(1:71MeV+0:51MeV)2(0:51MeV)2;=2:16MeV:Thisisalsothemagnitudeofthemomentumoftherecoiling32S.Non-relativisticrelationsareK=p2=2m,so(2:16MeV)2K==78:4eV:2(31:97)(931:5MeV)E50-39N=mN=Mwillgivethenumberofatomsof198Au;R=Nwillgivetheactivity;Ar=ln2=t1=2willgivethedecayconstant.Combining,NMrRt1=2Mrm==:khdaw.comNAln2NAThenforthesampleinquestion(250)(3:71010=s)(2:693)(86400s)(198g/mol)m==1:02103g:ln2(6:021023=mol)E50-40R=(8722=60s)=(3:71010=s)=3:93109Ci:E50-41Theradiationabsorbeddose(rad)isrelatedtotheroentgenequivalentman(rem)bythequalityfactor,soforthechestx-ray(25mrem)www.khdaw.com=29mrad:(0:85)Thisiswellbeneaththeannualexposureaverage.Eachradcorrespondstothedeliveryof105J/g,sotheenergyabsorbedbythepatientis512(0:029)(10J/g)(88kg)=1:2810J:2E50-42(a)(75kg)(102J=kg)(0:024rad)=1:8102J:(b)(0:024rad)(12)=0:29rem:课后答案网E50-43R=R(0:5)t=t1=2,so055R=(3:94Ci)(2)(6:04810s)=(1:8210s)=39:4Ci:0E50-44(a)N=mNA=MR,so(2103g)(6:021023/mol)N==5:081018:(239g/mol)(b)R=N=ln2N=t1=2,soR=ln2(5:081018)=(2:411104y)(3:15107s/y)=4:64106=s:(c)R=(4:64106=s)=(3:71010decays/s
Ci)=1:25104Cikhdaw.com:305若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE50-45Thehospitalusesa6000Cisource,andthatisalltheinformationweneedtondthenumberofdisintegrationspersecond:(6000Ci)(3:71010decays/s
Ci)=2:221014decays/s:Wearetoldthehalflife,buttondthenumberofradioactivenucleipresentwewanttoknowthedecayconstant.Thenln2ln291===4:1710s:t1=2(1:66108s)Thenumberof60ConucleiisthenR(2:221014decays/s)N===5:321022:(4:17109s1)E50-46Theannualequivalentdoesiskhdaw.com(12104rem/h)(20h/week)(52week/y)=1:25rem:E50-47(a)N=mNA=MRandMR=(226)+2(35)=296,so(1101g)(6:021023/mol)N==2:031020:(296g/mol)(b)R=N=ln2N=t1=2,soR=ln2(2:031020)=(1600y)(3:15107s/y)=2:8109Bq:(c)(2:8109)=(3:71010)=76mCi:www.khdaw.comE50-48R=N=ln2N=t1=2,so(4:6106)(3:71010=s)(1:28109y)(3:15107s/y)N==9:91021;ln2N=mNA=MR,so(40g/mol)(9:91021)m==0:658g:(6:021023/mol)课后答案网E50-49WecanapplyEq.50-18tondtheageoftherock,t1=2NFt=ln1+;ln2NI(4:47109y)(2:00103g)=(206g/mol)=ln1+;ln2(4:20103g)=(238g/mol)=2:83109y:306khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE50-50Thenumberofatomsof238Uoriginallypresentis(3:71103g)(6:021023/mol)N==9:381018:(238g/mol)Thenumberremainingafter260millionyearsisN=(9:381018)(0:5)(260My)=(4470My)=9:011018:Thedierencedecaysintolead(eventually),sothemassofleadpresentshouldbe(206g/mol)(0:371018)m==1:27104g:(6:021023/mol)E50-51WecanapplyEq.50-18tondtheageoftherock,t1=2NFt=ln1+;khdaw.comln2NI(4:47109y)(150106g)=(206g/mol)=ln1+;ln2(860106g)=(238g/mol)=1:18109y:InvertingEq.50-18tondthemassof40Koriginallypresent,NFt=t=21=21;NIso(sincetheyhavethesameatomicmass)themassofwww.khdaw.com40Kis(1:6103g)m==1:78103g:2(1:18)=(1:28)1E50-52(a)Thereisanexcessprotonontheleftandanexcessneutron,sotheunknownmustbeadeuteron,ord.(b)We"veaddedtwoprotonsbutonlyone(net)neutron,sotheelementisTiandthemassnumberis43,or43Ti.(c)Themassnumberdoesn"tchangebutweswappedoneprotonforaneutron,so7Li.E50-53Dothemath:课后答案网Q=(58:933200+1:00782558:9343521:008665)(931:5MeV)=1:86MeV:E50-54Thereactionsare201Hg( ;)197Pt;197Au(n;p)197Pt;196Pt(n; )197Pt;198Pt( ;n)197Pt;196Pt(d;p)197Pt;198Pt(p;d)197Pt:307khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE50-55WewillwritethesereactionsinthesamewayasEq.50-20representsthereactionofEq.50-19.Itishelpfultoworkbackwardsbeforeproceedingbyaskingthefollowingquestion:whatnucleiwillwehaveifwesubtractoneoftheallowedprojectiles?Thegoalis60Co,whichhas27protonsand6027=33neutrons.1.Removingaprotonwillleave26protonsand33neutrons,whichis59Fe;butthatnuclideisunstable.2.Removinganeutronwillleave27protonsand32neutrons,whichis59Co;andthatnuclideisstable.3.Removingadeuteronwillleave26protonsand32neutrons,whichis58Fe;andthatnuclideisstable.Itlooksasifonly59Co(n)60Coand58Fe(d)60Coarepossible.If,however,weallowforthepossibilityofotherdaughterparticlesweshouldalsoconsidersomeofthefollowingreactions.1.Swappinganeutronforaproton:60Ni(n,p)60Co.khdaw.com2.Usinganeutrontoknockoutadeuteron:61Ni(n,d)60Co.3.Usinganeutrontoknockoutanalphaparticle:63Cu(n,)60Co.4.Usingadeuterontoknockoutanalphaparticle:62Ni(d,)60Co.E50-56(a)Thepossibleresultsare64Zn,66Zn,64Cu,66Cu,61Ni,63Ni,65Zn,and67Zn.(b)Thestableresultsare64Zn,66Zn,61Ni,and67Zn.E50-57E50-58Theresultingreactionsare194Pt(d,)192www.khdaw.comIr,196Pt(d,)194Ir,and198Pt(d,)196Ir.E50-59E50-60Shellsoccuratnumbers2,8,20,28,50,82.Theshellsoccurseparatelyforprotonsandneutrons.ToanswerthequestionyouneedtoknowbothZandN=AZoftheisotope.(a)Filledshellsare18O,60Ni,92Mo,144Sm,and207Pb.(b)Onenucleonoutsideashellare40K,91Zr,121Sb,and143Nd.(c)Onevacancyinashellare13C,40K,49Ti,205Tl,and207Pb.E50-61(a)Thebindingenergyofthisneutroncanbefoundbyconsideringthe课后答案网Qvalueofthereaction90Zr(n)91Zrwhichis(89:904704+1:00866590:905645)(931:5MeV)=7:19MeV:(b)ThebindingenergyofthisneutroncanbefoundbyconsideringtheQvalueofthereaction89Zr(n)90Zrwhichis(88:908889+1:00866589:904704)(931:5MeV)=12:0MeV:Thisneutronisboundmoretightlythattheoneinpart(a).(c)Thebindingenergypernucleonisfoundbydividingthebindingenergybythenumberofnucleons:(401:007825+511:00866590:905645)(931:5MeV)=8:69MeV:91Theneutronintheoutsideshellof91Zrislesstightlyboundthantheaveragenucleoninkhdaw.com91Zr.308若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP50-1Beforedoinganythingweneedtoknowwhetherornotthemotionisrelativistic.Therestmassenergyofanparticleismc2=(4:00)(931:5MeV)=3:73GeV;andsincethisismuchgreaterthanthekineticenergywecanassumethemotionisnon-relativistic,andwecanapplynon-relativisticmomentumandenergyconservationprinciples.Theinitialvelocityoftheparticleisthenpppv=2K=m=c2K=mc2=c2(5:00MeV)=(3:73GeV)=5:18102c:Foranelasticcollisionwherethesecondparticleisatoriginallyatrestwehavethenalvelocityoftherstparticleasm2m12(4:00u)(197u)2v1;f=v1;i=(5:1810c)=4:9710c;m2+m1(4:00u)+(197u)whilethenalvelocityofthesecondparticleiskhdaw.com2m122(4:00u)3v2;f=v1;i=(5:1810c)=2:0610c:m2+m1(4:00u)+(197u)(a)Thekineticenergyoftherecoilingnucleusis1213262K=mv=m(2:0610c)=(2:1210)mc22=(2:12106)(197)(931:5MeV)=0:389MeV:(b)Energyconservationisthefastestwaytoanswerthisquestion,sinceitisanelasticcollision.www.khdaw.comThen(5:00MeV)(0:389MeV)=4:61MeV:P50-2Thegammaraycarriesawayamassequivalentenergyofm=(2:2233MeV)=(931:5MeV/u)=0:002387u:TheneutronmasswouldthenbemN=(2:0141021:007825+0:002387)u=1:008664u:P50-3(a)Therearefoursubstates:课后答案网mjcanbe+3/2,+1/2,-1/2,and-3/2.(b)
E=(2=3)(3:26)(3:15108eV=T)(2:16T)=1:48107eV:(c)=(1240eV
nm)=(1:48107eV)=8:38m:(d)Thisisintheradioregion.P50-4(a)Thechargedensityis=3Q=4R3.Thechargeontheshellofradiusrisdq=4r2dr.Thepotentialatthesurfaceofasolidsphereofradiusrisqr2V==:40r30Theenergyrequiredtoaddalayerofchargedqis42r4dU=Vdq=dr;30khdaw.com309若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comwhichcanbeintegratedtoyield42R53Q2U==:30200R(b)For239Pu,3(94)2(1:61019C)U==1024106eV:20(8:851012F=m)(7:451015m)(c)Theelectrostaticenergyis10:9MeVperproton.P50-5ThedecayrateisgivenbyR=N,whereNisthenumberofradioactivenucleipresent.IfRexceedsPthennucleiwilldecayfasterthantheyareproduced;butthiswillcauseNtodecrease,whichmeansRwilldecreaseuntilitisequaltoP.IfRislessthanPthennucleiwillbeproducedfasterthantheyaredecaying;butthiswillcauseNtoincrease,whichmeansRwillincreaseuntilitisequaltoP.IneithercaseequilibriumoccurswhenR=P,anditisastableequilibriumbecauseitisapproachednomatterwhichsideislarger.Thenkhdaw.comP=R=Natequilibrium,soN=P=.P50-6(a)A=N;atequilibriumA=P,soP=8:881010=s.(b)(8:881010=s)(1e0:269t),wheretisinhours.Thefactor0:269comesfromln(2)=(2:58)=.(c)N=P==(8:881010=s)(3600s/h)=(0:269/h)=1:191015:(d)m=NMr=NA,or(1:191015)(55:94g/mol)m=www.khdaw.com=1:10107g:(6:021023/mol)P50-7(a)A=N,soln2mNln2(1103g)(6:021023/mol)A7A===3:6610=s:t1=2Mr(1600)(3:15107s)(226g/mol)(b)Theratemustbethesameifthesystemisinsecularequilibrium.(c)N=P==t1=2P=ln2,so(3:82)(86400s)(3:66107=s)(222g/mol)m课后答案网==6:43109g:(6:021023/mol)ln2P50-8ThenumberofwatermoleculesinthebodyisN=(6:021023/mol)(70103g)=(18g/mol)=2:341027:Therearetenprotonsineachwatermolecule.TheactivityisthenA=(2:341027)ln2=(11032y)=1:62105/y:Thetimebetweendecaysisthen1=A=6200y:310khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP50-9Assumingthe238Unucleusisoriginallyatrestthetotalinitialmomentumiszero,whichmeansthemagnitudesofthenalmomentaoftheparticleandthe234Thnucleusareequal.Theparticlehasanalvelocityofpppv=2K=m=c2K=mc2=c2(4:196MeV)=(4:0026931:5MeV)=4:744102c:Sincethemagnitudesofthenalmomentaarethesame,the234Thnucleushasanalvelocityof2(4:0026u)4(4:74410c)=8:11310c:(234:04u)Thekineticenergyofthe234Thnucleusis1214272K=mv=m(8:11310c)=(3:29110)mc22=(3:291107)(234:04)(931:5MeV)=71:75keV:khdaw.comTheQvalueforthereactionisthen(4:196MeV)+(71:75keV)=4:268MeV;whichagreeswellwiththeSampleProblem.P50-10(a)TheQvalueisQ=(238:0507834:002603234:043596)(931:5MeV)=4:27MeV:(b)TheQvaluesforeachsteparewww.khdaw.comQ=(238:050783237:0487241:008665)(931:5MeV)=6:153MeV;Q=(237:048724236:0486741:007825)(931:5MeV)=7:242MeV;Q=(236:048674235:0454321:008665)(931:5MeV)=5:052MeV;Q=(235:045432234:0435961:007825)(931:5MeV)=5:579MeV:(c)ThetotalQforpart(b)is24:026MeV.Thedierencebetween(a)and(b)is28:296MeV.ThebindingenergyforthealphaparticleisE=[2(1:007825)+2(1:008665)4:002603](931:5MeV)=28:296MeV:课后答案网P50-11(a)Theemittedpositronleavestheatom,sothemassmustbesubtracted.Butthedaughterparticlenowhasanextraelectron,sothatmustalsobesubtracted.Hencethefactor2me.(b)TheQvalueisQ=[11:01143411:0093052(0:0005486)](931:5MeV)=0:961MeV:P50-12(a)Capturinganelectronisequivalenttonegativebetadecayinthatthetotalnumberofelectronsisaccountedforonboththeleftandrightsidesoftheequation.ThelossoftheKshellelectron,however,mustbetakenintoaccountasthisenergymaybesignicant.(b)TheQvalueisQ=(48:94851748:947871)(931:5MeV)(0:00547MeV)=0khdaw.com:596MeV:311若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP50-13Thedecayconstantfor90Srisln2ln2101===7:5810s:t1=2(9:15108s)Thenumberofnucleipresentin400gof90Sris(6:021023=mol)N=(400g)=2:681024;(89:9g/mol)sotheoverallactivityofthe400gof90SrisR=N=(7:581010s1)(2:681024)=(3:71010=Ci
s)=5:49104Ci:Thisisspreadoutovera2000km2area,sotheactivitysurfacedensity"is(5:49104Ci)=2:74105Ci=m2:khdaw.com(20006m2)Iftheallowablelimitis0.002mCi,thentheareaoflandthatwouldcontainthisactivityis(0:002103Ci)=7:30102m2:(2:74105Ci=m2)P50-14(a)N=mNA=Mr,soN=(2:5103g)(6:021023/mol)=(239g/mol)=6:31018:(b)A=ln2N=t1=2,sothenumberthatdecayin12hoursis18www.khdaw.comln2(6:310)(12)(3600s)11=2:510:(24100)(3:15107s)(c)TheenergyabsorbedbythebodyisE=(2:51011)(5:2MeV)(1:61019J=eV)=0:20J:(d)Thedoseinradis(0:20J)=(87kg)=0:23rad:(e)Thebiologicaldoseinremis(0:23)(13)=3rem:P50-15(a)Theamountof课后答案网238Uperkilogramofgraniteis(4106kg)(6:021023/mol)N==1:011019:(0:238kg/mol)Theactivityisthenln2(1:011019)A==49:7=s:(4:47109y)(3:15107s/y)TheenergyreleasedinonesecondisE=(49:7=s)(51:7MeV)=4:11010J:Theamountof232Thperkilogramofgraniteis(13106kg)(6:021023/mol)N==3:371019:(0:232kg/mol)khdaw.com312若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comTheactivityisthenln2(3:371019)A==52:6=s:(1:411010y)(3:15107s/y)TheenergyreleasedinonesecondisE=(52:6=s)(42:7MeV)=3:61010J:Theamountof40Kperkilogramofgraniteis(4106kg)(6:021023/mol)N==6:021019:(0:040kg/mol)Theactivityisthenln2(6:021019)A==1030=s:(1:28109y)(3:15107s/y)Theenergyreleasedinonesecondiskhdaw.comE=(1030=s)(1:32MeV)=2:21010J:Thetotalofthethreeis9:91010Wperkilogramofgranite.(b)ThetotalfortheEarthis2:71013W.P50-16(a)Sinceonlyaismovingoriginallythenthevelocityofthecenterofmassismava+mX(0)maV==va:mX+mama+mXNo,sincemomentumisconserved.www.khdaw.com(b)MovingtothecenterofmassframegivesthevelocityofXasV,andthevelocityofaasvaV.Thekineticenergyisnow1
K=mV2+m(vV)2;cmXaa2v2m2m2=ama+mX;2X(m+m)2a(m+m)2aXaXmv2mm+m2=aaaXX;2(ma+mX)2mX课后答案网=Klab:ma+mXYes;kineticenergyisnotconserved.p(c)va=2K=m,sopva=2(15:9MeV)=(1876MeV)c=0:130c:Thecenterofmassvelocityis(2)3V=(0:130c)=2:8310c:(2)+(90)Finally,(90)Kcm=(15:9MeV)=15:6MeV:(2)+(90)khdaw.com313若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP50-17LetQ=KcmintheresultofProblem50-16,andinvert,solvingforKlab.P50-18(a)Removingaprotonfrom209Bi:E=(207:976636+1:007825208:980383)(931:5MeV)=3:80MeV:Removingaprotonfrom208Pb:E=(206:977408+1:007825207:976636)(931:5MeV)=8:01MeV:(b)Removinganeutronfrom209Pb:E=(207:976636+1:008665208:981075)(931:5MeV)=3:94MeV:Removinganeutronfrom208Pb:khdaw.comE=(206:975881+1:008665207:976636)(931:5MeV)=7:37MeV:www.khdaw.com课后答案网314khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE51-1(a)Forthecoal,m=(1109J)=(2:9107J=kg)=34kg:(b)Fortheuranium,m=(1109J)=(8:21013J=kg)=1:2105kg:E51-2(a)TheenergyfromthecoalisE=(100kg)(2:9107J=kg)=2:9109J:(b)TheenergyfromtheuraniumintheashisE=(3106)(100kg)(8:21013J)=2:51010J:E51-3(a)Thereare(1:00kg)(6:021023mol1)=2:561024khdaw.com(235g/mol)atomsin1.00kgof235U.(b)Ifeachatomreleases200MeV,then(200106eV)(1:61019J=eV)(2:561024)=8:191013Jofenergycouldbereleasedfrom1.00kgof235U.(c)Thisamountofenergywouldkeepa100-Wlamplitfor(8:191013J)t==8:191011s26;000y!(100W)19www.khdaw.comE51-42W=1:2510eV=s.Thisrequires(1:251019eV=s)=(200106eV)=6:251010=sasthessionrate.E51-5Thereare(1:00kg)(6:021023mol1)=2:561024(235g/mol)atomsin1.00kgof235U.Ifeachatomreleases200MeV,then课后答案网(200106eV)(1:61019J=eV)(2:561024)=8:191013Jofenergycouldbereleasedfrom1.00kgof235U.Thisamountofenergywouldkeepa100-Wlamplitfor(8:191013J)t==8:191011s30;000y!(100W)E51-6Thereare(1:00kg)(6:021023mol1)=2:521024(239g/mol)atomsin1.00kgof239Pu.Ifeachatomreleases180MeV,then(180106eV)(1:61019J=eV)(2:521024)=7:251013Jofenergycouldbereleasedfrom1.00kgof239Pu.khdaw.com315若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE51-7Whenthe233Unucleusabsorbsaneutronwearegivenatotalof92protonsand142neutrons.Galliumhas31protonsandaround39neutrons;chromiumhas24protonsandaround28neutrons.Therearethen37protonsandaround75neutronsleftover.Thiswouldberubidium,butthenumberofneutronsisverywrong.Althoughtheelementalidenticationiscorrect,becausewemustconserveprotonnumber,theisotopesarewronginourabovechoicesforneutronnumbers.E51-8Betadecayistheemissionofanelectronfromthenucleus;oneoftheneutronschangesintoaproton.Theatomnowneedsonemoreelectronintheelectronshells;byusingatomicmasses(asopposedtonuclearmasses)thenthebetaelectronisaccountedfor.Thisisonlytruefornegativebetadecay,notforpositivebetadecay.E51-9(a)Thereare(1:0g)(6:021023mol1)=2:561021(235g/mol)atomsin1.00gof235U.Thessionrateiskhdaw.comA=ln2N=t=ln2(2:561021)=(3:51017y)(365d/y)=13:9=d:1=2(b)Theratioistheinverseratioofthehalf-lives:(3:51017y)=(7:04108y)=4:97108:E51-10(a)TheatomicnumberofYmustbe9254=38,sotheelementisSr.Themassnumberis235+11401=95,soYis95Sr.(b)TheatomicnumberofYmustbe9253=39,sotheelementisY.Themassnumberis235+11392=95,soYis95Y.(c)TheatomicnumberofXmustbe9240=52,sotheelementisTe.Themassnumberis235+11002=134,soXis134Te.www.khdaw.com(d)Themassnumberdierenceis235+114192=3,sob=3.E51-11TheQvalueisQ=[51:940122(25:982593)](931:5MeV)=23MeV:Thenegativevalueimpliesthatthisssionreactionisnotpossible.E51-12TheQvalueis课后答案网Q=[97:9054082(48:950024)](931:5MeV)=4:99MeV:ThetwofragmentswouldhaveaverylargeCoulombbarriertoovercome.E51-13Theenergyreleasedis(235:043923140:92004491:91972621:008665)(931:5MeV)=174MeV:E51-14SinceEn>Ebssionispossiblebythermalneutrons.E51-15(a)Theuraniumstartswith92protons.Thetwoendproductshaveatotalof58+44=102.Thismeansthattheremusthavebeentenbetadecays.(b)TheQvalueforthisprocessisQ=(238:050783+1:008665139:90543498:905939)(931khdaw.com:5MeV)=231MeV:316若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE51-16(a)Theotherfragmenthas9232=60protonsand235+183=153neutrons.Thatelementis153Nd.(b)SinceK=p2=2mandmomentumisconserved,then2mK=2mK.Thismeansthat1122K2=(m1=m2)K1.ButK1+K2=Q,som2+m1K1=Q;m2orm2K1=Q;m1+m2withasimilarexpressionforK.Thenfor83Ge2(153)K=(170MeV)=110MeV;(83+153)whilefor153Nd(83)K=(170MeV)=60MeV;khdaw.com(83+153)(c)For83Ge,rs2K2(110MeV)v==c=0:053c;m(83)(931MeV)whilefor153Ndrs2K2(60MeV)v==c=0:029c:m(153)(931MeV)E51-17Since239Puisonenucleonheavierthanwww.khdaw.com238Uonlyoneneutroncaptureisrequired.TheatomicnumberofPuistwomorethanU,sotwobetadecayswillberequired.Thereactionseriesisthen238U+n!239U;239U!239Np++;239Np!239Pu++:E51-18Eachssionreleases200MeV.Thetotalenergyreleasedoverthethreeyearsis课后答案网(190106W)(3)(3:15107s)=1:81016J:That"s(1:81016J)=(1:61019J=eV)(200106eV)=5:61026ssionevents.Thatrequiresm=(5:61026)(0:235kg/mol)=(6:021023/mol)=218kg:Butthisisonlyhalftheoriginalamount,or437kg.E51-19AccordingtoSampleProblem51-3therateatwhichnon-ssionthermalneutroncaptureoccursisonequarterthatofssion.Hencethemasswhichundergoesnon-ssionthermalneutroncaptureisonequartertheanswerofEx.51-18.Thetotalisthen(437kg)(1+0:25)=546kg:khdaw.com317若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE51-20(a)Qe=E=
N,whereEisthetotalenergyreleasedand
Nisthenumberofdecays.ThiscanalsobewrittenasPPt1=2Pt1=2MrQe===;Aln2Nln2NAmwhereAistheactivityandPthepoweroutputfromthesample.Solving,(2:3W)(29y)(3:15107s)(90g/mol)Q==4:531013J=2:8MeV:eln2(6:021023/mol)(1g)(b)P=(0:05)m(2300W=kg),so(150W)m==1:3kg:(0:05)(2300W=kg)E51-21LettheenergyreleasedbyonessionbeE1.Iftheaveragetimetothenextssioneventistgen,thentheaverage"poweroutputfromtheonessionisP1=E1=tgen.Ifeveryssioneventresultsinthereleaseofkneutrons,eachofwhichcausealaterssionevent,thenaftereverytimeperiodkhdaw.comtgenthenumberofssionevents,andhencetheaveragepoweroutputfromallofthessionevents,willincreasebyafactorofk.ForlongenoughtimeswecanwriteP(t)=Pkt=tgen:0E51-22InverttheexpressionderivedinExercise51-21:t3gen=t(1:310s)=(2:6s)P(350)k===0:99938:P0(1200)E51-23Eachssionreleases200MeV.Thenthessionrateiswww.khdaw.com(500106W)=(200106eV)(1:61019J=eV)=1:61019=sThenumberofneutronsintransit"isthen(1:61019=s)(1:0103s)=1:61016:E51-24UsingtheresultsofExercise51-21:P=(400MW)(1:0003)(300s)=(0:03s)=8030MW:E51-25Thetimeconstantforthisdecayis课后答案网ln2101==2:5010s:(2:77109s)Thenumberofnucleipresentin1.00kgis(1:00kg)(6:021023mol1)N==2:531024:(238g/mol)ThedecayrateisthenR=N=(2:501010s1)(2:531024)=6:331014s1:Thepowergeneratedisthedecayratetimestheenergyreleasedperdecay,P=(6:331014s1)(5:59106eV)(1:61019J/eV)=566W:318khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE51-26Thedetectordetectsonlyafractionoftheemittedneutrons.ThisfractionisA(2:5m2)==1:62104:4R24(35m)2Thetotal uxoutofthewarheadisthen(4:0=s)=(1:62104)=2:47104=s:Thenumberof239PuatomsisA(2:47104=s)(1:341011y)(3:15107s/y)N===6:021022:ln2(2:5)That"sonetenthofamole,sothemassis(239)=10=24g.E51-27UsingtheresultsofSampleProblem51-4,khdaw.comln[R(0)=R(t)]t=;58soln[(0:03)=(0:0072)]9t==1:7210y:(0:9840:155)(1109/y)E51-28(a)(15109W
y)(2105y)=7:5104W:(b)Thenumberofssionsrequiredis(15109W
y)(3:15107s/y)N==1:51028:(200MeV)(1:610www.khdaw.com19J=eV)Themassof235Uconsumedism=(1:51028)(0:235kg/mol)=(6:021023/mol)=5:8103kg:E51-29If238Uabsorbsaneutronitbecomes239U,whichwilldecaybybetadecaytorst239Npandthen239Pu;welookedatthisinExercise51-17.Thiscandecaybyalphaemissionaccordingto239Pu!235U+:E51-30Thenumberofatomspresentinthesampleis课后答案网N=(6:021023/mol)(1000kg)=(2:014g/mol)=2:991026:Ittakestwotomakeafusion,sotheenergyreleasedis(3:27MeV)(2:991026)=2=4:891026MeV:That"s7:81013J,whichisenoughtoburnthelampfort=(7:81013J)=(100W)=7:81011s=24800y:E51-31Thepotentialenergyatclosestapproachis(1:61019C)2U==9105eV:4(8:851012F=m)(1:61015m)khdaw.com319若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE51-32Theratiocanbewrittenasrn(K1)=K1e(K2K1)=kT;n(K2)K2sotheratioiss(5000eV)(3100eV)=(8:62105eV=K)(1:5107K)e=0:15:(1900eV)E51-33(a)SeeSampleProblem51-5.E51-34AddupalloftheQvaluesinthecycleofFig.51-10.E51-35Theenergyreleasediskhdaw.com(34:00260312:0000000)(931:5MeV)=7:27MeV:E51-36(a)Thenumberofparticleofhydrogenin1m3isN=(0:35)(1:5105kg)(6:021023/mol)=(0:001kg/mol)=3:161031(b)ThedensityofparticlesisN=V=p=kT;theratiois(3:161031)(1:381023J=K)(298K)=1:2106:(1:01105Pa)E51-37(a)Thereare(1:00kg)(6:021023www.khdaw.commol1)=6:021026(1g/mol)atomsin1.00kgof1H.Iffouratomsfusetoreleases26.7MeV,then(26:7MeV)(6:021026)=4=4:01027MeVofenergycouldbereleasedfrom1.00kgof1H.(b)Thereare(1:00kg)(6:021023mol1)=2:561024(235g/mol)atomsin1.00kgof课后答案网235U.Ifeachatomreleases200MeV,then(200MeV)(2:561024)=5:11026MeVofenergycouldbereleasedfrom1.00kgof235U.E51-38(a)E=
mc2,so(3:91026J=s)
m==4:3109kg=s:(3:0108m=s)2(b)ThefractionoftheSun"smasslost"is(4:3109kg=s)(3:15107s/y)(4:5109y)=0:03%:(2:01030kg)khdaw.com320若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE51-39Therateofconsumptionis6:21011kg=s,thecorehas1/8themassbutonly35%ishydrogen,sothetimeremainingist=(0:35)(1=8)(2:01030kg)=(6:21011kg=s)=1:41017s;orabout4:5109years.E51-40Forthersttworeactionsintoone:Q=[2(1:007825)(2:014102)](931:5MeV)=1:44MeV:Forthesecond,Q=[(1:007825)+(2:014102)(3:016029)](931:5MeV)=5:49MeV:Forthelast,Q=[2(3:016029)(4:002603)2(1:007825)](931:5MeV)=12:86MeV:E51-41khdaw.com(a)UsemNA=Mr=N,so7(0:012kg/mol)1(3:310J=kg)=4:1eV:(6:021023/mol)(1:61019J=eV)(b)Forevery12gramsofcarbonwerequire32gramsofoxygen,thetotalis44grams.Thetotalmassrequiredisthen40=12thatofcarbonalone.Theenergyproductionisthen(3:3107J=kg)(12=44)=9106J=kg:(c)Thesunwouldburnfor(21030kg)(9106J=kg)www.khdaw.com=4:61010s:(3:91026W)That"sonly1500years!E51-42Therateoffusioneventsis(5:31030W)=4:561042=s:(7:27106eV)(1:61019J=eV)Thecarbonisthenproducedatarate(4:56课后答案网1042=s)(0:012kg/mol)=(6:021023/mol)=9:081016kg=s:Theprocesswillbecompletein(4:61032kg)=1:6108y:(9:081016kg=s)(3:15107s/y)E51-43(a)Forthereactiond-d,Q=[2(2:014102)(3:016029)(1:008665)](931:5MeV)=3:27MeV:(b)Forthereactiond-d,Q=[2(2:014102)(3:016029)(1:007825)](931:5MeV)=4:03MeV:(c)Forthereactiond-t,Q=[(2:014102)+(3:016049)(4:002603)(1:008665)](931khdaw.com:5MeV)=17:59MeV:321若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE51-44Oneliterofwaterhasamassofonekilogram.Thenumberofatomsof2His(6:021023/mol)(0:00015kg)=4:51022:(0:002kg/mol)Theenergyavailableis(3:27106eV)(1:61019J=eV)(4:51022)=2=1:181010J:Thepoweroutputisthen(1:181010J)=1:4105W(86400s)E51-45Assumemomentumconservation,thenp=pnorvn=v=m=mn:Theratioofthekineticenergiesisthenkhdaw.comKmv2mnnn==4:Kmv2mnThenKn=4Q=5=14:07MeVwhileK=Q=5=3:52MeV.E51-46TheQvalueisQ=(6:015122+1:0086653:0160494:002603)(931:5MeV)=4:78MeV:CombinethetworeactionstogetanetQ=22:37MeV.Theamountofwww.khdaw.com6LirequiredisN=(2:61028MeV)=(22:37MeV)=1:161027:ThemassofLiDrequiredis(1:161027)(0:008kg/mol)m==15:4kg:(6:021023/mol)P51-1(a)Equation50-1isR=RA1=3;0whereR0=1:2fm.Thedistancebetweenthetwonucleiwillbethesumoftheradii,or课后答案网R(140)1=3+(94)1=3:0Thepotentialenergywillbe1q1q2U=;40re2(54)(38)=
;40R0(140)1=3+(94)1=3(1:601019C)2=211;4(8:851012C2=Nm2)(1:2fm)=253MeV:(b)Theenergywilleventuallyappearasthermalenergy.khdaw.com322若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comp32=3P51-2(a)SinceR=R0A,thesurfaceareaaisproportionaltoA.Thefractionalchangeinsurfaceareais(a+a)a(140)2=3+(96)2=3(236)2=3120==25%:a0(236)2=3(b)Nucleihaveaconstantdensity,sothereisnochangeinvolume.p223(c)SinceU/Q=R,U/Q=A.ThefractionalchangeintheelectrostaticpotentialenergyisU+UU(54)2(140)1=3+(38)2(96)1=3(92)2(236)1=3120==36%:U0(92)2(236)1=3P51-3(a)Thereare(2:5kg)(6:021023mol1)=6:291024(239g/mol)atomsin2.5kgof239Pu.Ifeachatomreleases180MeV,thenkhdaw.com(180MeV)(6:291024)=(2:61028MeV/megaton)=44kilotonisthebombyield.P51-4(a)Inanelasticcollisionthenucleusmovesforwardwithaspeedof2mnv=v0;mn+msothekineticenergywhenitmovesforwardism4m2mm
K=v2nwww.khdaw.com=Kn;20(m+m)2(m+m)2nnwherewecanwrite
Kbecauseinanelasticcollisionwhateverenergykineticenergythenucleuscarriesohadtocomefromtheneutron.(b)Forhydrogen,
K4(1)(1)==1:00:K(1+1)2Fordeuterium,
K4(1)(2)==0:89:课后答案网K(1+2)2Forcarbon,
K4(1)(12)==0:28:K(1+12)2Forlead,
K4(1)(206)==0:019:K(1+206)2(c)Ifeachcollisionreducestheenergybyafactorof10:89=0:11,thenthenumberofcollisionsrequiredisthesolutionto(0:025eV)=(1106eV)(0:11)N;whichisN=8.323khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP51-5Theradiiofthenucleiarep3R=(1:2fm)7=2:3fm:TheusingthederivationofSampleProblem51-5,(3)2(1:61019C)2K==1:4106eV:16(8:851012F=m)(2:31015m)P51-6(a)Addupthesixequationstoget12C+1H+13N+13C+1H+14N+1H+15O+15N+1H!13N++13C+e+++14N++15O++15N+e+++12C+4He:Canceloutthingsthatoccuronbothsidesandgetkhdaw.com1H+1H+1H+1H!+e+++++e+++4He:(b)AdduptheQvalues,andthenaddon4(0:511MeVfortheannihilationofthetwopositrons.P51-7(a)Demonstratingtheconsistencyofthisexpressionisconsiderablyeasierthanderivingitfromrstprinciples.FromProblem50-4wehavethatauniformsphereofchargeQandradiusRhaspotentialenergy3Q2U=:200RThisexpressionwasderivedfromthefundamentalexpressionwww.khdaw.com1qdqdU=:40rForgravitythefundamentalexpressionisGmdmdU=;rsowereplace1=40withGandQwithM.Butlikechargesrepelwhileallmassesattract,sowepickupanegativesign.(b)TheinitialenergywouldbezeroifR=1,sotheenergyreleasedis课后答案网3GM23(6:71011N
m2=kg2)(2:01030kg)2U===2:31041J:5R5(7:0108m)Atthecurrentrate(seeSampleProblem51-6),thesunwouldbe(2:31041J)t==5:91014s;(3:91026W)or187millionyearsold.324khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP51-8(a)Therateoffusioneventsis(3:91026W)=9:31037=s:(26:2106eV)(1:61019J=eV)Eacheventproducestwoneutrinos,sotherateis1:861038=s:(b)TheratetheseneutrinosimpingeontheEarthisproportionaltothesolidanglesubtendedbytheEarthasseenfromtheSun:r2(6:37106m)2==4:51010;4R24(1:501011m)2sotherateofneutrinosimpingingontheEarthiskhdaw.com(1:861038=s)(4:51010)=8:41028=s:P51-9(a)ReactionAreleases,foreachd(1=2)(4:03MeV)=2:02MeV;ReactionBreleases,foreachd(1=3)(17:59MeV)+(1=3)(4:03MeV)=7:21MeV:ReactionBisbetter,andreleaseswww.khdaw.com(7:21MeV)(2=02MeV)=5:19MeVmoreforeachN.P51-10(a)Themassofthepelletis453312m=(2:010m)(200kg=m)=6:710kg:3Thenumberofd-tpairsis课后答案网(6:71012kg)(6:021023/mol)N==8:061014;(0:005kg/mol)andif10%fusethentheenergyreleaseis(17:59MeV)(0:1)(8:061014)(1:61019J=eV)=230J:(b)That"s(230J)=(4:6106J=kg)=0:05kgofTNT.(c)Thepowerreleasedwouldbe(230J)(100=s)=2:3104W.325khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE52-1(a)ThegravitationalforceisgivenbyGm2=r2,whiletheelectrostaticforceisgivenbyq2=4r2.Theratiois04Gm24(8:851012C2=Nm2)(6:671011Nm2=kg2)(9:111031kg)20=;q2(1:601019C)2=2:41043:Gravitationaleectswouldbeswampedbyelectrostaticeectsatanyseparation.(b)Theratiois4Gm24(8:851012C2=Nm2)(6:671011Nm2=kg2)(1:671027kg)20=;q2(1:601019C)2=8:11037:E52-2(a)Q=938:27MeV0:511MeV)=937:76MeV.khdaw.com(b)Q=938:27MeV135MeV)=803MeV.E52-3ThegravitationalforcefromtheleadsphereisGmeM4GmeR=:R23SettingthisequaltotheelectrostaticforcefromtheprotonandsolvingforR,3e2R=;1620Gmea20orwww.khdaw.com3(1:61019C)2162(8:851012F=m)(6:671011Nm2=kg2)(11350kg=m3)(9:111031kg)(5:291011m)2whichmeansR=2:851028m:E52-4Eachtakeshalftheenergyofthepion,so(1240MeV
fm)==18:4fm:(135MeV)=2课后答案网E52-5TheenergyofoneofthepionswillbeppE=(pc)2+(mc2)2=(358:3MeV)2+(140MeV)2=385MeV:Therearetwoofthesepions,sotherestmassenergyofthe0is770MeV.E52-6E= mc2,so=(1:5106eV)=(20eV)=7:5104:Thespeedisgivenbypv=c11= 2cc=22;wheretheapproximationistrueforlarge.Then
v=c=2(7:5104)2=2:7102m=s:326khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE52-7d=c
t=hc=2
E.Then(1240MeV
fm)3d==2:1610fm:2(91200MeV)E52-8(a)Electromagnetic.(b)Weak,sinceneutrinosarepresent.(c)Strong.(d)Weak,sincestrangenesschanges.E52-9(a)Baryonnumberisconservedbyhavingtwop"ononesideandap"anda
0ontheother.Chargewillonlybeconservediftheparticlexispositive.Strangenesswillonlybeconservedifxisstrange.Sinceitcan"tbeabaryonitmustbeameson.ThenxisK+.(b)Baryonnumberontheleftis0,soxmustbeananti-baryon.Chargeontheleftiszero,soxmustbeneutralbecausen"isneutral.Strangenessiseverywherezero,sotheparticlemustben.(c)Thereisonebaryonontheleftandoneontheright,soxhasbaryonnumber0.Thechargeontheleftaddstozero,sokhdaw.comxisneutral.Thestrangenessofxmustalsobe0,soitmustbea0.E52-10Therearetwopositiveontheleft,andtwoontheright.Theanti-neutronmustthenbeneutral.Thebaryonnumberontherightisone,thatontheleftwouldbetwo,unlesstheanti-neutronhasabaryonnumberofminusone.Thereisnostrangenessontherightorleft,exceptpossibletheanti-neutron,soitmustalsohavestrangenesszero.E52-11(a)Annihilationreactionsareelectromagnetic,andthisinvolvesss.(b)Thisisneitherweaknorelectromagnetic,soitmustbestrong.(c)Thisisstrangenesschanging,soitisweak.www.khdaw.com(d)Strangenessisconserved,sothisisneitherweaknorelectromagnetic,soitmustbestrong.E52-12(a)K0!e++,e(b)K0!++0,(c)K0!++++,(d)K0!++0+0,E52-13(a)
0!p++.(b)n!p+e++.e(c)+!+++.(d)K!+.课后答案网E52-14E52-15Fromtoptobottom,theyare
++,
+,
0,+,0,0,
,,,and.E52-16(a)Thisisnotpossible.(b)uuuworks.E52-17Astrangenessof+1correspondstotheexistenceofansanti-quark,whichhasachargeof+1/3.Theonlyquarksthatcancombinewiththisanti-quarktoformamesonwillhavechargesof-1/3or+2/3.Itisonlypossibletohaveanetchargeof0or+1.Thereverseistrueforstrangeness-1.327khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE52-18Putbarsovereverything.Fortheanti-proton,uudZ,fortheanti-neutron,udd.quarksQSCparticleuc00-1D0dc-10-1Dsc-1-1-1DE52-19We"llconstructatable:scc000ccu001D0cd101D+cs111D+sE52-20(a)Writethequarkcontentoutthencanceloutthepartswhicharethesameonbothsides:dds!udd+du;sothefundamentalprocessiskhdaw.coms!u+d+u:(b)Writethequarkcontentoutthencanceloutthepartswhicharethesameonbothsides:ds!ud+du;sothefundamentalprocessiss!u+d+u:(c)Writethequarkcontentoutthencanceloutthepartswhicharethesameonbothsides:ud+uud!uus+us;sothefundamentalprocessiswww.khdaw.comd+d!s+s:(d)Writethequarkcontentoutthencanceloutthepartswhicharethesameonbothsides:+udd!uud+du;sothefundamentalprocessis!u+u:E52-21Theslopeis(7000km/s)=70km/s
Mpc:课后答案网(100Mpc)E52-22c=Hd,sod=(3105km/s)=(72km/s
Mpc)=4300Mpc:E52-23ThequestionshouldreadWhatisthe..."Thespeedofthegalaxyisv=Hd=(72km/s
Mpc)(240Mpc)=1:72107m=s:Theredshiftofthiswouldthenbep1(1:72107m=s)2=(3108m=s)2=(656:3nm)=695nm:1(1:72107m=s)=(3108m=s)khdaw.com328若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comE52-24Wecanapproximatetheredshiftas=0=(1u=c);so0(590nm)u=c1=c1=0:02c:(602nm)Thedistanceisd=v=H=(0:02)(3108m=s)=(72km/s
Mpc)=83Mpc:E52-25Theminimumenergyrequiredtoproducethepairsisthroughthecollisionoftwo140MeVphotons.ThiscorrespondstoatemperatureofT=(140MeV)=(8:62105eV=K)=1:621012K:Thistemperatureexistedatatimekhdaw.com(1:51010s1=2K)2t==86s:(1:621012K)2E52-26(a)0:002m.(b)f=(3108m=s)=(0:002m)=1:51011Hz:(c)E=(1240eV
nm)=(2106nm)=6:2104eV:E52-27(a)UseEq.52-3:p(1:51010sK)2t==91012s:(5000K)www.khdaw.com2That"sabout280,000years.(b)kT=(8:62105eV=K)(5000K)=0:43eV:(c)Theratiois(109)(0:43eV)=0:457:(940106eV)P52-1Thetotalenergyofthepionis135+80=215MeV.Thegammafactorofrelativityis=E=mc2=(215MeV)=(135MeV)=1:59;课后答案网sothevelocityparameterisp=11= 2=0:777:Thelifetimeofthepionasmeasuredinthelaboratoryist=(8:41017s)(1:59)=1:341016s;sothedistancetraveledisd=vt=(0:777)(3:00108m=s)(1:341016s)=31nm:329khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.compP52-2(a)E=K+mc2andpc=E2(mc2)2,soppc=(2200MeV+1777MeV)2(1777MeV)2=3558MeV:That"sthesameas(3558106eV)p=(1:61019J=eV)=1:901018kg
m=s(3108m=s).(b)qvB=mv2=r,sop=qB=r.Then(1:901018kg
m=s)r==9:9m:(1:61019C)(1:2T)P52-3(a)ApplytheresultsofExercise45-1:(1240MeV
fm)10khdaw.comE==(4:2810MeV=K)T:(2898m
K)T(b)T=2(0:511MeV)=(4:281010MeV=K)=2:39109K:P52-4(a)Sincep12=0;1wehavep
12=1;01www.khdaw.comorp12+1z=:1Nowinvert,pz(1)+1=12;(z+1)2(1)2=12;(z2+2z+1)(12+2)=12;(z课后答案网2+2z+2)22(z2+2z+1)+(z2+2z)=0:Solvethisquadraticfor,andz2+2z=:z2+2z+2(b)Usingtheresult,(4:43)2+2(4:43)==0:934:(4:43)2+2(4:43)+2(c)Thedistanceisd=v=H=(0:934)(3108m=s)=(72km/s
Mpc)=3893Mpc:330khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comP52-5(a)UsingEq.48-19,n1
E=kTln:n2Heren1=0:23whilen2=10:23,then
E=(8:62105eV=K)(2:7K)ln(0:23=0:77)=2:8104eV:(b)ApplytheresultsofExercise45-1:(1240eV
nm)==4:4mm:(2:8104eV)P52-6(a)UnlimitedexpansionmeansthatvHr,soweareinterestedinv=Hr.ThenpHr=2GM=r;H2r3=2G(4r3=3);khdaw.com3H2=8G=:(b)Evaluating,3[72103m=s
(3:0841022m)]2(6:021023/mol)=5:9=m3:8(6:671011N
m2=kg2)(0:001kg/mol)P52-7(a)Theforceonaparticleinasphericaldistributionofmatterdependsonlyonthemattercontainedinasphereofradiussmallerthanthedistancetothecenterofthesphericaldistribution.Andthenwecantreatallthatrelevantmatterasbeingconcentratedatthecenter.IfMisthetotalmass,thenwww.khdaw.comr3M0=M;R3isthefractionofmattercontainedinthesphereofradiusr