gauss's law - ucf physics：高斯定律-物理

Gauss’sLawElectricFluxWedefinetheelectricflux,oftheelectricfieldE,throughthesurfaceA,as:F=E.AWhere:

Aisavectornormaltothesurface (magnitudeA,anddirectionnormaltothesurface).

istheanglebetweenEandAareaAEA=EAcos()ElectricFluxHerethefluxis

F=E

·

AYoucanthinkofthefluxthroughsomesurfaceasameasureofthenumberoffieldlineswhichpassthroughthatsurface.FluxdependsonthestrengthofE,onthesurfacearea,andontherelativeorientationofthefieldandsurface.Normaltosurface,magnitudeAareaAEAEA

ElectricFluxThefluxalsodependsonorientationareaAqAcos

qThenumberoffieldlinesthroughthetiltedsurfaceequalsthenumberthroughitsprojection.Hence,thefluxthroughthetiltedsurfaceissimplygivenbythefluxthroughitsprojection:E(Acosq).F=E.A=EAcos

qareaAqAcos

qEEAACalculatethefluxoftheelectricfieldE,throughthesurfaceA,ineachofthethreecasesshown:a)=b)=c)=qdAEWhatifthesurfaceiscurved,orthefieldvarieswithposition??1.WedividethesurfaceintosmallregionswithareadA2.ThefluxthroughdAis

dF=EdA

cos

q

dF=E.dA3.ToobtainthetotalfluxweneedtointegrateoverthesurfaceAA=

d

=

E.dA

=E.AInthecaseofaclosedsurfaceTheloopmeanstheintegralisoveraclosedsurface.qdAEForaclosedsurface:ThefluxispositiveforfieldlinesthatleavetheenclosedvolumeThefluxisnegativeforfieldlinesthatentertheenclosedvolumeIfachargeisoutsideaclosedsurface,thenetfluxiszero.Asmanylinesleavethesurface,aslinesenterit.Forwhichoftheseclosedsurfaces(a,b,c,d)thefluxoftheelectricfield,producedbythecharge+2q,iszero?SphericalsurfacewithpointchargeatcenterFluxofelectricfield:Gauss’sLawThisisalwaystrue.Occasionally,itprovidesaveryeasywaytofindtheelectricfield(forhighlysymmetriccases).Theelectricfluxthroughanyclosedsurfaceequalsenclosedcharge/

0Calculatethefluxoftheelectricfield

foreachoftheclosedsurfacesa,b,c,anddSurfacea,

a=Surfaceb,

b=Surfacec,

c=Surfaced,

d=CalculatetheelectricfieldproducedbyapointchargeusingGaussLawWechooseforthegaussiansurfaceasphereofradiusr,centeredonthechargeQ.Then,theelectricfieldE,hasthesamemagnitudeeverywhereonthesurface(radialsymmetry)Furthermore,ateachpointonthesurface,thefieldEandthesurfacenormaldAareparallel(bothpointradiallyoutward).E.dA=EdA[cos

=1]EQCoulomb’sLaw!

E.dA=Q/e0

E.dA=E

dA=EAA=4r2EA=E4r2=

Q/e0ElectricfieldproducedbyapointchargeEQk=1/4

0

0=

permittivity

0=8.85x10-12C2/Nm2IsGauss’sLawmorefundamentalthanCoulomb’sLaw?No!HerewederivedCoulomb’slawforapointchargefromGauss’slaw.OnecaninsteadderiveGauss’slawforageneral(evenverynasty)chargedistributionfromCoulomb’slaw.Thetwolawsareequivalent.Gauss’slawgivesusaneasywaytosolveafewverysymmetricproblemsinelectrostatics.Italsogivesusgreatinsightintotheelectricfieldsinandonconductorsandwithinvoidsinsidemetals.Gauss’sLawThetotalfluxwithinaclosedsurface……isproportionaltotheenclosedcharge.Gauss’sLawisalwaystrue,butisonlyusefulforcertainverysimpleproblemswithgreatsymmetry.ApplyingGauss’sLawGauss’slawisusefulonlywhentheelectricfieldisconstantonagivensurfaceInfinitesheetofcharge1.SelectGausssurfaceInthiscaseacylindricalpillbox2.CalculatethefluxoftheelectricfieldthroughtheGausssurface=2EA3.Equate=qencl/02EA=qencl/04.SolveforEE=qencl

/2A0=/20

(with=qencl

/A)GAUSSLAW–SPECIALSYMMETRIES

SPHERICAL(pointorsphere)CYLINDRICAL(lineorcylinder)

PLANAR(planeorsheet)CHARGEDENSITYDependsonlyonradialdistancefromcentralpointDependsonlyonperpendiculardistancefromlineDependsonlyonperpendiculardistancefromplaneGAUSSIANSURFACESpherecenteredatpointofsymmetryCylindercenteredataxisofsymmetryPillboxorcylinderwithaxisperpendiculartoplane

ELECTRICFIELDEEconstantatsurfaceE║A-cos

=1EconstantatcurvedsurfaceandE║AE┴Aatendsurfacecos

=0EconstantatendsurfacesandE║AE┴Aatcurvedsurfacecos

=0

FLUX

CylindricalgeometryPlanargeometrySphericalgeometryEAchargeQisuniformlydistributedthroughasphereofradiusR.Whatistheelectricfieldasafunctionofr?.FindEatr1andr2.Problem:SphereofChargeQr2r1RAchargeQisuniformlydistributedthroughasphereofradiusR.Whatistheelectricfieldasafunctionofr?.FindEatr1andr2.Problem:SphereofChargeQUsesymmetry!Thisissphericallysymmetric.ThatmeansthatE(r)isradiallyoutward,andthatallpoints,atagivenradius(|r|=r),havethesamemagnitudeoffield.r2r1RE(r1)E(r2)Problem:SphereofChargeQE&dArRWhatistheenclosedcharge?FirstfindE(r)atapointoutsidethechargedsphere.ApplyGauss’slaw,usingastheGaussiansurfacethesphereofradiusr

pictured.Problem:SphereofChargeQrRE&dAWhatistheenclosedcharge?QFirstfindE(r)atapointoutsidethechargedsphere.ApplyGauss’slaw,usingastheGaussiansurfacethesphereofradiusr

pictured.Problem:SphereofChargeQrRE&dAWhatistheenclosedcharge?QWhatisthefluxthroughthissurface?FirstfindE(r)atapointoutsidethechargedsphere.ApplyGauss’slaw,usingastheGaussiansurfacethesphereofradiusr

pictured.Problem:SphereofChargeQrRE&dAWhatistheenclosedcharge?QWhatisthefluxthroughthissurface?FirstfindE(r)atapointoutsidethechargedsphere.ApplyGauss’slaw,usingastheGaussiansurfacethesphereofradiusr

pictured.Problem:SphereofChargeQrRE&dAWhatistheenclosedcharge?QWhatisthefluxthroughthissurface?Gauss

FirstfindE(r)atapointoutsidethechargedsphere.ApplyGauss’slaw,usingastheGaussiansurfacethesphereofradiusr

pictured.Problem:SphereofChargeQrRE&dAWhatistheenclosedcharge?QWhatisthefluxthroughthissurface?Gauss:SoExactlyasthoughallthechargewereattheorigin!(forr>R)FirstfindE(r)atapointoutsidethechargedsphere.ApplyGauss’slaw,usingastheGaussiansurfacethesphereofradiusr

pictured.Problem:SphereofChargeQRrE(r)NextfindE(r)atapointinsidethesphere.ApplyGauss’slaw,usingalittlesphereofradiusrasaGaussiansurface.Whatistheenclosedcharge?Thattakesalittleeffort.Thelittlespherehassomefractionofthetotalcharge.Whatfraction?That’sgivenbyvolumeratio:Againthefluxis:SettinggivesForr<RProblem:SphereofChargeQProblem:SphereofChargeQLookcloserattheseresults.Theelectricfieldatcomesfromasumoverthecontributionsofallthelittlebits.It’sobviousthatthenetEatthispointwillbehorizontal.Butthemagnitudefromeachbitisdifferent;andit’scompletelynotobviousthatthemagnitudeEjustdependsonthedistancefromthesphere’scentertotheobservationpoint.DoingthisasavolumeintegralwouldbeHARD.Gauss’slawisEASY.RQrr>RsProblem:InfinitechargedplaneConsideraninfiniteplanewithaconstantsurfacechargedensitys

(whichissomenumberofCoulombspersquaremeter).WhatisEatapointlocatedadistancezabovetheplane?xyzsProblem:InfinitechargedplaneConsideraninfiniteplanewithaconstantsurfacechargedensitys

(whichissomenumberofCoulombspersquaremeter).WhatisEatapointlocatedadistancezabovetheplane?xyzUsesymmetry!Theelectricfieldmustpointstraightawayfromtheplane(ifs>0).MaybetheMagnitudeofEdependsonz,butthedirectionisfixed.AndEisindependentofxandy.EEEGaussian“pillbox”sProblem:InfinitechargedplaneSochooseaGaussiansurfacethatisa“pillbox”,whichhasitstopabovetheplane,anditsbottombelowtheplane,eachadistancezfromtheplane.Thatwaytheobservationpointliesinthetop.zzEEGaussian“pillbox”sProblem:InfinitechargedplanezzLettheareaofthetopandbottombeA.EEGaussian“pillbox”sProblem:InfinitechargedplanezzLettheareaofthetopandbottombeA.Totalchargeenclosedbybox=EEGaussian“pillbox”sProblem:InfinitechargedplanezzLettheareaofthetopandbottombeA.Totalchargeenclosedbybox=AsEEGaussian“pillbox”sProblem:InfinitechargedplanezzLettheareaofthetopandbottombeA.Outwardfluxthroughthetop:EEGaussian“pillbox”sProblem:InfinitechargedplanezzLettheareaofthetopandbottombeA.Outwardfluxthroughthetop:EAEEGaussian“pillbox”sProblem:InfinitechargedplanezzLettheareaofthetopandbottombeA.Outwardfluxthroughthetop:EAOutwardfluxthroughthebottom:EEGaussian“pillbox”sProblem:InfinitechargedplanezzLettheareaofthetopandbottombeA.Outwardfluxthroughthetop:EAOutwardfluxthroughthebottom:EAEEGaussian“pillbox”sProblem:InfinitechargedplanezzLettheareaofthetopandbottombeA.Outwardfluxthroughthetop:EAOutwardfluxthroughthebottom:EAOutwardfluxthroughthesides:EEGaussian“pillbox”sProblem:InfinitechargedplanezzLettheareaofthetopandbottombeA.Outwardfluxthroughthetop:EAOutwardfluxthroughthebottom:EAOutwardfluxthroughthesides:Ex(somearea)xcos(900)=0EEGaussian“pillbox”sProblem:InfinitechargedplanezzOutwardfluxthroughthetop:EAOutwardfluxthroughthebottom:EAOutwardfluxthroughthesides:Ex(somearea)xcos(900)=0Sothetotalfluxis:2EALettheareaofthetopandbottombeA.EEGaussian“pillbox”sProblem:InfinitechargedplanezzGauss’slawthensaysthatAs/e0=2EAsothatE=s/2e0,outward.Thisisconstanteverywhereineachhalf-space!LettheareaofthetopandbottombeA.NoticethattheareaAcanceled:thisistypical!Problem:InfinitechargedplaneImaginedoingthiswithanintegraloverthechargedistribution:breakthesurfaceintolittlebitsdA…sdEDoingthisasasurfaceintegralwouldbeHARD.Gauss’slawisEASY.ConsideralongcylindricalchargedistributionofradiusR,withchargedensity=a–br(withaandbpositive).Calculatetheelectricfieldfor:r<Rr=Rr>RAconductorisamaterialinwhichchargescanmoverelativelyfreely.Usuallythesearemetals(Au,Cu,Ag,Al).Excesscharges(ofthesamesign)placedonaconductorwillmoveasfarfromeachotheraspossible,sincetheyrepeleachother.Forachargedconductor,inastaticsituation,allthechargeresidesatthesurfaceofaconductor.Forachargedconductor,inastaticsituation,theelectricfieldiszeroeverywhereinsideaconductor,andperpendiculartothesurfacejustoutsideConductorsConductorsWhyisE=0insideaconductor?ConductorsWhyisE=0insideaconductor?Conductorsarefulloffreeelectrons,roughlyonepercubicAngstrom.Thesearefreetomove.IfEisnonzeroinsomeregion,thentheelectronstherefeelaforce-eEandstarttomove.ConductorsWhyisE=0insideaconductor?Conductorsarefulloffreeelectrons,roughlyonepercubicAngstrom.Thesearefreetomove.IfEisnonzeroinsomeregion,thentheelectronstherefeelaforce-eEandstarttomove.Inanelectrostaticsproblem,theelectronsadjusttheirpositionsuntiltheforceoneveryelectroniszero(orelseitwouldmove!).Thatmeanswhenequilibriumisreached,E=0everywhereinsideaconductor.ConductorsBecauseE=0inside,theinsideofaconductorisneutral.Supposethereisanextrachargeinside.Gauss’slawforthelittlesphericalsurfacesaystherewouldbeanonzeroEnearby.Buttherecan’tbe,withinametal!Consequentlytheinteriorofametalisneutral.Anyexcesschargeendsuponthesurface.ElectricfieldjustoutsideachargedconductorTheelectricfieldjustoutsideachargedconductorisperpendiculartothesurfaceandhasmagnitudeE=/0PropertiesofConductorsInaconductortherearelargenumberofelectronsfreetomove.ThisfacthasseveralinterestingconsequencesExcesschargeplacedonaconductormovestotheexteriorsurfaceoftheconductorTheelectricfieldinsideaconductoriszerowhenchargesareatrestAconductorshieldsacavitywithinitfromexternalelectricfieldsElectricfieldlinescontactconductorsurfacesatrightanglesAconductorcanbechargedbycontactorinductionConnectingaconductortogroundisreferredtoasgroundingThegroundcanacceptofgiveupanunlimitednumberofelectrons

abProblem:ChargedcoaxialcableThispictureisacrosssectionofaninfinitelylonglineofcharge,surroundedbyaninfinitelylongcylindricalconductor.FindE.Thisrepresentsthelineofcharge.Sayithasalinearchargedensityofl(somenumberofC/m).Thisisthecylindricalconductor.Ithasinnerradiusa,andouterradiusb.Usesymmetry!ClearlyEpointsstraightout,anditsamplitudedependsonlyonr.Problem:ChargedcoaxialcableFirstfindE