Download AP Physics Review Sheet #3 Emily Dickinson Electric Flux and

APPhysicsReviewSheet#3
EmilyDickinson
ElectricFluxandGauss’sLawCh.24
Equations
!
Φ! = 𝐸 ∗ 𝐴TheunitsforΦis𝑁 ∗ 𝑚 𝐶 Electricfluxcanbethoughtofasalloftheenergyenteringandexitinganobject.
TheelectricfluxisequaltotheEnergycrossedwiththearea.
ThisisdependentontheangleatwhichtheEnergyfieldenterstheareainwhose
fluxisbeingcalculated.Soitcanalsobewrittenas
Φ! = 𝐸 ∗ 𝐴 cos (𝜃)orΦ! = ∫ 𝐸 𝑑𝐴
sidenote:boththeEnergyandAreaarevectorsandshouldhaveatinyarrowontop
oftheEandAineachequation
Gauss’sLawisdefinedas:
𝑞!"
Φ! = ∮ 𝐸 𝑑𝐴 =
𝜀!
𝑞!" orqinternalreferstothenetchargeinsideofthesurface
!
𝜀! = !!"Epsilonknotreferstothepermittivityoffreespace
∮ Thisintegralsignrefersisforclosedsurfaces,sotheElectricfluxcalculatedby
takingthesurfaceintegraloftheenergyandtheareareferstothenetelectricflux
determinedatanypointonthesurface
ImportantNotes:
1.Ifthechargeistripled,thefluxwilltriplebecausethefluxisequaltotheinternal
chargeoverthepermittivityoffreespace,𝜀! 2.Ifthevolumeofthesphereisdoubled,thefluxremainsthesame
3.Ifthesurfaceischangedtoacubeshpe,thefluxwillremainthesamebecausethe
numberofelectricfieldlinesexitingandenteringwillremainthesamerelatively
basedontheshapeandareaoftheobject.
4.Ifthechargeismovedtoadifferentlocationinsidethesurface,thefluxwillstill
remainthesamebecausethenetfluxwillstillbethesame.
5.Ifthechargeismovedoutsideofthesurface,thefluxwilldroptozerobecause
thereisnointernalcharge.
Conductors(usuallymetal)havetheirownspecialrules.
1.Everywhereinsideoftheconductortheelectricfieldis0.
2.Allchargelocatedonsurfaceofconductor
3.Electricfieldjustoutsideofthechargedconductorthatisperpendiculartothe
!
surfacehasamagnitudeofE=! !
4.(Thisoneisn’tsuperimportant,butirregularlyshapedconductorstendto
accumulatechargeatlocationswheretheradiusofthecurvatureisthesmallest,so
atthesharpestpoints.
Problem1
Gauss’Law
UseGauss’Lawtodeterminetheelectricfieldofaspherewithauniformly
distributedcharge+Qatpositionr
a.Wherer<R
br>R
(ThedashedlinerepresentsaGaussian
spherevisualizedatthe2Rline)
a.r<R
+Q
Theinternalchargeisnotsimply+Q
becausesomeofthechargeisnot
R includedintheimaginedGaussiansphere,
2R
soitshouldnotcontributetowardsthe
Electricfield.Toaccountforthisweneed
tomakearatioofthevolumethatthe
chargeinhabits.
4
𝜋 𝑟!
𝑉!"#$%"&'
3
𝑄!"#$%"&' = +𝑄
=𝑄
4 !
𝑉!"!#$
𝜋𝑅
3
𝑟!
=𝑄 ! 𝑅
TheElectricfieldwillbeconstanteverywherealongtheimaginedGaussiansurface.
Soitcanbetakenoutoftheintegral
𝑟!
𝑄
𝑞!"
𝑅!
𝐸∮ 𝑑𝐴 =
= 𝐸 4𝜋𝑟 ! =
1
𝜀!
4𝜋𝑘
!
𝑟
𝑟!
𝐸 4𝜋𝑟 ! = 𝑄 ! 4𝜋𝑘 ; 𝐸 𝑟 ! = 𝑄 ! 𝑘
𝑅
𝑅
𝑟
𝐸 = 𝑄 ! 𝑘
𝑅
b.r>R
𝑄!" = +𝑄
EonceagainisconstanteverywherealongtheimaginedGaussiansphere
𝑞!"
𝑄
∮ 𝐸𝑑𝐴 =
; 𝐸 4𝜋𝑟 ! = = 𝐸 4𝜋𝑟 ! = 𝑄 4𝜋𝑘 𝜀!
𝜀!
𝑘𝑄
𝐸 = !
𝑟
Problem2:Cylinder
Findtheelectricfielddistancerawayfromanon-conductingcylinderofuniform
chargewhosechargeperunitlengthis𝜆andalengthofL
Solution:CreateacylindricalGaussiansurfacethatsurroundsthenon-conducting
cylinderwithuniformchargedensity.
TheGaussiansurfacedoesnotencompassthetoporbottomofthecylinder.
r
𝑞!"
∮ 𝐸𝑑𝐴 =
𝜀!
Gauss’Law
𝑞
𝐸 2𝜋𝑟𝐿 =
𝜀!
Theenergyremainsthesamesoonlytheareaneedstobeintegrated.Theareaof
thecylinderisjustthe2𝜋𝑟𝐿piecebecausetheelectricfielddoesnotgooutofthe
endsofthecylinder.
𝑞
𝜆 = ; 𝑞!" = 𝜆𝐿
𝐿
Theinternalchargecanbeexpressedas
𝑞!" = 𝜆𝐿
!
because𝜆 representsthechargeperunitlengthor! 𝜆𝐿
𝐸 2𝜋𝑟𝐿 =
;
𝜀
!
𝜀! = !!"soitcanbemanipulatedsothat
4𝜋𝑘𝜆𝐿
𝐸=
2𝜋𝑟𝐿
ThepiandLcancelout,the4andthe2leaveone2behindandwhat’sleftisequalto
theelectricfieldatsomedistancer.
2𝑘𝜆
𝐸=
𝑟
Problem3:Conductorwithnonconductorinside
Thereisaconductingshellwithacharge2Qsurroundinganon-conductingcylinder
oflengthLandacharge–2Qandachargedensityof𝜆. Whatistheelectricfield
when:
Aistheradiusofthenonconductingcylinder
Bistheradiusoftheinnerpartoftheconductingshell
Cistheouterradiusoftheconductingshell
c
a
b
a.r=a
Thisisjustlikethequestionabove.
𝑞!"
∮ 𝐸𝑑𝐴 =
𝜀!
𝜆𝐿
𝐸 2𝜋𝑟𝐿 =
;
𝜀
4𝜋𝑘𝜆𝐿
𝐸=
2𝜋𝑟𝐿
2𝑘𝜆
𝐸=
𝑟
b.whereb<r<c
Theelectricfieldis0!Becauseitisinsideaconductor
c.wherer>c
Theelectricfieldisalso0!Becausethechargeoftheconductingshellandthenon
conductioncylindercancel.