工程电磁场第八版课后答案第06章

CHAPTER6.6.1.Consideracoaxialcapacitorhavinginnerradiusa,outerradiusb,unitlength,andfilledwithamaterialwithdielectricconstant,≤r.Comparethistoaparallel-platecapacitorhavingplatewidth,w,plateseparationd,filledwiththesamedielectric,andhavingunitlength.Expresstheratiob/aintermsoftheratiod/w,suchthatthetwostructureswillstorethesameenergyforagivenappliedvoltage.Storingthesameenergyforagivenappliedvoltagemeansthatthecapacitanceswillbeequal.Withbothstructureshavingunitlengthandcontainingthesamedielectric(permittivity≤),weequatethetwocapacitances:2π≤ln(b/a)=≤wd⇒ba=expµ2πdw∂6.2.LetS=100mm2,d=3mm,and≤r=12foraparallel-platecapacitor.a)Calculatethecapacitance:C=≤r≤0Ad=12≤0(100×10−6)3×10−3=0.4≤0=3.54pfb)Afterconnectinga6Vbatteryacrossthecapacitor,calculateE,D,Q,andthetotalstoredelectrostaticenergy:First,E=V0/d=6/(3×10−3)=2000V/m,thenD=≤r≤0E=2.4×104≤0=0.21µC/m2ThechargeinthiscaseisQ=D·n|s=DA=0.21×(100×10−6)=0.21×10−4µC=21pCFinally,We=(1/2)QV0=0.5(21)(6)=63pJ.c)Withthesourcestillconnected,thedielectriciscarefullywithdrawnfrombetweentheplates.Withthedielectricgone,re-calculateE,D,Q,andtheenergystoredinthecapacitor.E=V0/d=6/(3×10−3)=2000V/m,asbefore.D=≤0E=2000≤0=17.7nC/m2ThechargeisnowQ=DA=17.7×(100×10−6)nC=1.8pC.Finally,We=(1/2)QV0=0.5(1.8)(6)=5.4pJ.d)Ifthechargeandenergyfoundin(c)arelessthanthatfoundin(b)(whichyoushouldhavediscovered),whatbecameofthemissingchargeandenergy?Intheabsenceoffrictioninremovingthedielectric,thechargeandenergyhavereturnedtothebatterythatgaveit.766.3.Capacitorstendtobemoreexpensiveastheircapacitanceandmaximumvoltage,Vmax,increase.ThevoltageVmaxislimitedbythefieldstrengthatwhichthedielectricbreaksdown,EBD.WhichofthesedielectricswillgivethelargestCVmaxproductforequalplateareas:(a)air:≤r=1,EBD=3MV/m;(b)bariumtitanate:≤r=1200,EBD=3MV/m;(c)silicondioxide:≤r=3.78,EBD=16MV/m;(d)polyethylene:≤r=2.26,EBD=4.7MV/m?NotethatVmax=EBDd,wheredistheplateseparation.Also,C=≤r≤0A/d,andsoVmaxC=≤r≤0AEBD,whereAistheplatearea.ThemaximumCVmaxproductisfoundthroughthemaximum≤rEBDproduct.Tryingthiswiththegivenmaterialsyieldsthewinner,whichisbariumtitanate.6.4.Anair-filledparallel-platecapacitorwithplateseparationdandplateareaAisconnectedtoabatterywhichappliesavoltageV0betweenplates.Withthebatteryleftconnected,theplatesaremovedaparttoadistanceof10d.Determinebywhatfactoreachofthefollowingquantitieschanges:a)V0:Remainsthesame,sincethebatteryisleftconnected.b)C:AsC=≤0A/d,increasingdbyafactoroftendecreasesCbyafactorof0.1.c)E:WerequireE×d=V0,whereV0hasnotchanged.Therefore,Ehasdecreasedbyafactorof0.1.d)D:AsD=≤0E,andsinceEhasdecreasedby0.1,Ddecreasesby0.1.e)Q:SinceQ=CV0,andasCisdownby0.1,Qalsodecreasesby0.1.f)ρS:AsQisreducedby0.1,ρSreducesby0.1.ThisisalsoconsistentwithDhavingbeenreducedby0.1.g)We:UseWe=1/2CV20,toobserveitsreductionby0.1,sinceCisreducedbythatfactor.6.5.Aparallelplatecapacitorisfilledwithanonuniformdielectriccharacterizedby≤r=2+2×106x2,wherexisthedistancefromoneplate.IfS=0.02m2,andd=1mm,findC:Startbyassumingchargedensityρsonthetopplate.Dwill,asusual,bex-directed,originatingatthetopplateandterminatingonthebottomplate.ThekeyhereisthatDwillbeconstantoverthedistancebetweenplates.Thiscanbeunderstoodbyconsideringthex-varyingdielectricasconstructedofmanythinlayers,eachhavingconstantpermittivity.Thepermittivitychangesfromlayertolayertoapproximatethegivenfunctionofx.Theapproximationbecomesexactasthelayerthicknessesapproachzero.WeknowthatD,whichisnormaltothelayers,willbecontinuousacrosseachboundary,andsoDisconstantovertheplateseparationdistance,andwillbegiveninmagnitudebyρs.TheelectricfieldmagnitudeisnowE=D≤0≤r=ρs≤0(2+2×106x2)ThevoltagebeweenplatesisthenV0=Z10−30ρsdx≤0(2+2×106x2)=ρs≤01√4×106tan−1√x√4×1062!ØØØ10−30=ρs≤012×103≥π4¥NowQ=ρs(.02),andsoC=QV0=ρs(.02)≤0(2×103)(4)ρsπ=4.51×10−10F=451pF776.6.RepeatProblem6.4assumingthebatteryisdisconnectedbeforetheplateseparationisin-creased:TheorderingofparametersischangedoverthatinProblem6.4,astheprogressionofthoughtonthematterisdifferent.a)Q:Remainsthesame,sincewiththebatterydisconnected,thechargehasnowheretogo.b)ρS:AsQisunchanged,ρSisalsounchanged,sincetheplateareaisthesame.c)D:AsD=ρS,itwillremainthesamealso.d)E:SinceE=D/≤0,andasDisnotchanged,Ewillalsoremainthesame.e)V0:WerequireE×d=V0,whereEhasnotchanged.Therefore,V0hasincreasedbyafactorof10.f)C:AsC=≤0A/d,increasingdbyafactoroftendecreasesCbyafactorof0.1.ThesameresultoccursbecauseC=Q/V0,whereV0isincreasedby10,whereasQhasnotchanged.g)We:UseWe=1/2CV20=1/2QV0,toobserveitsincreasebyafactorof10.6.7.Let≤r1=2.5for0y1mm,≤r2=4for1y3mm,and≤r3for3y5mm.Conductingsurfacesarepresentaty=0andy=5mm.Calculatethecapacitancepersquaremeterofsurfaceareaif:a)≤r3isthatofair;b)≤r3=≤r1;c)≤r3=≤r2;d)region3issilver:Thecombinationwillbethreecapacitorsinseries,forwhich1C=1C1+1C2+1C3=d1≤r1≤0(1)+d2≤r2≤0(1)+d3≤r3≤0(1)=10−3≤0∑12.5+24+2≤r3∏SothatC=(5×10−3)≤0≤r310+4.5≤r3Evaluatingthisforthefourcases,wefinda)C=3.05nFfor≤r3=1,b)C=5.21nFfor≤r3=2.5,c)C=6.32nFfor≤r3=4,andd)C=9.83nFifsilver(takenasaperfectconductor)formsregion3;thishastheeffectofremovingtheterminvolving≤r3fromthe