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January 4, 2016 CH 23 Gauss’Law
I.
IntroductiontoGauss’Law
A. Gauss’lawrelatestheelectricfieldsatpointsona(closed)
Gaussiansurfacetothenetchargeenclosedbythatsurface.
1.
Gauss’lawconsidersahypothetical(imaginary)closedsurface
enclosingthechargedistribution.
2.
ThisGaussiansurface,asitiscalled,canhaveanyshape,buttheshape
thatminimizesourcalculationsoftheelectricfieldisonethatmimicsthe
symmetryofthechargedistribution.
3.
WecanalsouseGauss’lawinreverse;ifweknowtheelectricfieldona
Gaussiansurface,wecanfindthenetchargeenclosedbythesurface.
4.
However,inordertocalculatehowmuchchargeisenclosedweneeda
wayofcalculatinghowmuchelectricfieldisinterceptedbytheGaussian
surface.Thismeasurementofinterceptedfieldiscalled__________________________.
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January 4, 2016 B. Flux–symbol(F)
1.
ThewordfluxcomesfromtheLatinwordmeaningto“Flow”.
2.
Let’ssupposethatasshowninthefigurebelow,weaimawide
airstreamofuniformvelocityatasmallsquareloopofareaA.
3.
Auniformairstreamofvelocityisperpendiculartotheplaneofa
squareloopofareaA.(b)Thecomponentofperpendiculartotheplaneofthe
loopisvcos,whereistheanglebetweenvandanormaltotheplane.(c)
TheareavectorAisperpendiculartotheplaneoftheloopandmakesanangle
withv.(d)Thevelocityfieldinterceptedbytheareaoftheloop.Therateof
volumeflowthroughtheloopis=(vcos)A.
4.
Thisrateofflowthroughanareaisanexampleofaflux—avolumeflux
inthissituation.Note:Thebelowequationonlyappliestothisspecificsample.
Don’tmemorizeit!
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January 4, 2016 II.
ElectricFlux
A. TheelectricfluxthroughaGaussiansurfaceisproportionaltothe
netnumberofelectricfieldlinespassingthroughthatsurface.
B. Diagram
C. Asshownabove,AGaussiansurfaceofarbitraryshapeimmersed
inanelectricfield.Thesurfaceisdividedintosmallsquaresofarea
A.TheareavectorsarealwaysperpendiculartoGaussiansurface!
TheelectricfieldvectorsEandtheareavectorsAforthree
representativesquares,marked1,2,and3,areshown.
D. Theexactdefinitionofthefluxoftheelectricfieldthrougha
closedsurfaceisfoundbyallowingtheareaofthesquaresshownin
Fig.23‐3tobecomesmallerandsmaller,approachingadifferential
limitdA.TheareavectorsthenapproachadifferentiallimitdA.The
sumoftheaboveequationthenbecomesanintegral:
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January 4, 2016 E. Example,Fluxthroughaclosedcube,Non‐uniformfield:
Left Face: Top Face: Page4
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January 4, 2016 III.
Gauss’sLaw
A. Gauss’lawrelatesthenetfluxofanelectricfieldthroughaclosed
surface(aGaussiansurface)tothenetchargeq thatisenclosedby
enc
thatsurface.
1.
netchargeq isthealgebraicsumofalltheenclosedpositiveand
enc
negativecharges,anditcanbepositive,negative,orzero.
B. Ifqencispositive,thenetfluxisoutward;ifqencisnegative,thenet
fluxisinward.
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January 4, 2016 C. Example,Enclosedchargeinanon‐uniformfield:
1.
Enclosedcharge:Onceweknowfluxitisrelativelyeasytofindqenc:
qenc = 2.
Thus,thecubeenclosesanet______________________________charge.
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January 4, 2016 IV.
Gauss’sLawandCoulomb’sLaw
A. Figurebelowshowsapositivepointchargeq,aroundwhicha
concentricsphericalGaussiansurfaceofradiusrisdrawn.Dividethis
surfaceintodifferentialareasdA.
B. TheareavectordAatanypointisperpendiculartothesurfaceand
directedoutwardfromtheinterior.
C. Fromthesymmetryofthesituation,atanypointtheelectricfield,
E,isalsoperpendiculartothesurfaceanddirectedoutwardfromthe
interior.
D. Thus,sincetheanglebetweenEanddAiszero,wecanrewrite
Gauss’lawas
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January 4, 2016 V.
AChargedIsolatedConductor
A. Ifanexcesschargeisplacedonanisolatedconductor,thatamount
ofchargewillmoveentirelytothesurfaceoftheconductor.Noneof
theexcesschargewillbefoundwithinthebodyoftheconductor.
1.
Figure23‐9ashows,incrosssection,anisolatedlumpofcopper
hangingfromaninsulatingthreadandhavinganexcesschargeq.The
Gaussiansurfaceisplacedjustinsidetheactualsurfaceoftheconductor.The
electricfieldinsidethisconductormustbezero.Sincetheexcesschargeisnot
insidetheGaussiansurface,itmustbeoutsidethatsurface,whichmeansit
mustlieontheactualsurfaceoftheconductor.
2.
Figure23‐9bshowsthesamehangingconductor,butnowwithacavity
thatistotallywithintheconductor.AGaussiansurfaceisdrawnsurrounding
thecavity,closetoitssurfacebutinsidetheconductingbody.Insidethe
conductor,therecanbenofluxthroughthisnewGaussiansurface.Therefore,
thereisnonetchargeonthecavitywalls;alltheexcesschargeremainsonthe
outersurfaceoftheconductor.
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January 4, 2016 3.
SampleProblem:Theshadedareabelowrepresentsthecrosssection
ofathick,solid,sphericalconductingshellthathasatotalchargeof+10C.The
conductorishollowanda+3Cchargeexistsatthecenterofthecavity.The
solidlineatthemiddleoftheconductorrepresentsthecrosssectionofa
sphericalgaussiansurface.Thenet(total)electricfluxthroughthegaussian
surfaceis
4.
SampleProblem#2:Theregionfromr=ator=bisasolid,spherical
conductorwithanet(total)chargeof+6.0C.Thereisa‐2.0Cchargeinsidethe
cavity(r<a).Thenet(total)chargethatcollectsontheoutsideofthe
conductor(r=b)isclosestto
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January 4, 2016 B. TheExternalElectricField:
1.
Theelectricfieldjustoutsidethesurfaceofaconductoriseasyto
determineusingGauss’law.
2.
Considerasectionofthesurfacethatissmallenoughtoneglectany
curvatureandthusthesectionisconsideredflat.
3.
AtinycylindricalGaussiansurfaceisembeddedinthesectionasinFig.
23‐10:Oneendcapisfullyinsidetheconductor,theotherisfullyoutside,and
thecylinderisperpendiculartotheconductor’ssurface.
4.
TheelectricfieldEatandjustoutsidetheconductor’ssurfacemust
alsobeperpendiculartothatsurface.
5.
WeassumethatthecapareaAissmallenoughthatthefieldmagnitude
Eisconstantoverthecap.ThenthefluxthroughthecapisEA,andthatisthe
netfluxthroughtheGaussiansurface.
6.
ThechargeqencenclosedbytheGaussiansurfaceliesontheconductor’s
surfaceinanareaA.Ifisthechargeperunitarea,thenqencisequaltoA.
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January 4, 2016 VI.
ApplyingGauss’sLaw:CylindricalSymmetry
A. Figurebelowshowsasectionofaninfinitelylongcylindrical
plasticrodwithauniformpositivelinearchargedensity.
LetusfindanexpressionforthemagnitudeoftheelectricfieldEata
distancerfromtheaxisoftherod.
1.
AteverypointonthecylindricalpartoftheGaussiansurface,must
havethesamemagnitudeEand(forapositivelychargedrod)mustbe
directedradiallyoutward.
2.
ThefluxofEthroughthiscylindricalsurfaceis
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January 4, 2016 VII.
ApplyingGauss’sLaw:PlanarSymmetry,Non‐conductingSheet
A. Figurebelowshowsaportionofathin,infinite,nonconducting
sheetwithauniform(positive)surfacechargedensity.Asheetof
thinplasticwrap,uniformlychargedononeside,canserveasasimple
model.
We need to find the electric field a distance r in front of the sheet.
1.
AusefulGaussiansurfaceisaclosedcylinderwithendcapsofareaA,
arrangedtopiercethesheetperpendicularlyasshown.Fromsymmetry,E
mustbeperpendiculartothesheetandhencetotheendcaps.
2.
Sincethechargeispositive,Eisdirectedawayfromthesheet.Thereis
nofluxthroughthisportionoftheGaussiansurface.ThusE.dAissimplyEdA,
and
HereAisthechargeenclosedbytheGaussiansurface.
3.
Therefore,
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January 4, 2016 B. TwoConductingPlates
1.
Figurebelowshowsacrosssectionofathin,infiniteconductingplate
withexcesspositivecharge.Theplateisthinandverylarge,andessentiallyall
theexcesschargeisonthetwolargefacesoftheplate.
2.
Ifthereisnoexternalelectricfieldtoforcethepositivechargeinto
someparticulardistribution,itwillspreadoutonthetwofaceswithauniform
surfacechargedensityofmagnitude1.
3.
Justoutsidetheplatethischargesetsupanelectricfieldofmagnitude
E=.
4.
Figure23‐16bshowsanidenticalplatewithexcessnegativecharge
havingthesamemagnitudeofsurfacechargedensity.Now,theelectricfieldis
directedtowardtheplate.
5.
IfwearrangefortheplatesofFigs.23‐16aandbtobeclosetoeach
otherandparallel(Fig.23‐16c),theexcesschargeononeplateattractsthe
excesschargeontheotherplate,andalltheexcesschargemovesontothe
innerfacesoftheplatesasinFig.23‐16c.
6.
Withtwiceasmuchchargenowoneachinnerface,thenewsurface
chargedensity,,oneachinnerfaceistwice.Thus,theelectricfieldatany
pointbetweentheplateshasthemagnitude
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January 4, 2016 VIII. ApplyingGauss’sLaw:SphericalSymmetry
A. Ashellofuniformchargeattractsorrepelsachargedparticlethat
isoutsidetheshellasifalltheshell’schargewereconcentratedatthe
centeroftheshell.
B. Ifachargedparticleislocatedinsideashellofuniformcharge,
thereisnoelectrostaticforceontheparticlefromtheshell.
C. Diagram
1.
ApplyingGauss’LawtoGaussiansurfaceS2weget:
2.
ApplyingGauss’LawtoGaussiansurfaceS1weget:
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January 4, 2016 D. TheElectricFieldinsideauniformsphereofchargeisdirected
radiallyandhasmagnitudeof:
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January 4, 2016 E. SampleProblem:
1.
Aconductingsphericalshellofradiusr0=0.070mhasauniform
surfacechargedensityof=2.0µC/m2onitsoutersurface.Themagnitudeof
theelectricfieldatpoints0.050mand0.100m(fromthecenteroftheshell),
respectivelyareclosestto
Row E at 0.050 m from center A B C D E E at 0.100 m from center 0 1.1x105 N/C
1.8x106 N/C 0
0 1.8x106 N/C
1.1x105 N/C 0
0 0
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