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APPhysicsReviewSheet#3 EmilyDickinson ElectricFluxandGaussโsLawCh.24 Equations ! ฮฆ! = ๐ธ โ ๐ดTheunitsforฮฆis๐ โ ๐ ๐ถ Electricfluxcanbethoughtofasalloftheenergyenteringandexitinganobject. TheelectricfluxisequaltotheEnergycrossedwiththearea. ThisisdependentontheangleatwhichtheEnergyfieldenterstheareainwhose fluxisbeingcalculated.Soitcanalsobewrittenas ฮฆ! = ๐ธ โ ๐ด cos (๐)orฮฆ! = โซ ๐ธ ๐๐ด sidenote:boththeEnergyandAreaarevectorsandshouldhaveatinyarrowontop oftheEandAineachequation GaussโsLawisdefinedas: ๐!" ฮฆ! = โฎ ๐ธ ๐๐ด = ๐! ๐!" orqinternalreferstothenetchargeinsideofthesurface ! ๐! = !!"Epsilonknotreferstothepermittivityoffreespace โฎ Thisintegralsignrefersisforclosedsurfaces,sotheElectricfluxcalculatedby takingthesurfaceintegraloftheenergyandtheareareferstothenetelectricflux determinedatanypointonthesurface ImportantNotes: 1.Ifthechargeistripled,thefluxwilltriplebecausethefluxisequaltotheinternal chargeoverthepermittivityoffreespace,๐! 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.(Thisoneisnโtsuperimportant,butirregularlyshapedconductorstendto accumulatechargeatlocationswheretheradiusofthecurvatureisthesmallest,so atthesharpestpoints. Problem1 GaussโLaw UseGaussโLawtodeterminetheelectricfieldofaspherewithauniformly 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 ๐ ๐! ๐!"#$%"&' 3 ๐!"#$%"&' = +๐ =๐ 4 ! ๐!"!#$ ๐๐ 3 ๐! =๐ ! ๐ TheElectricfieldwillbeconstanteverywherealongtheimaginedGaussiansurface. Soitcanbetakenoutoftheintegral ๐! ๐ ๐!" ๐ ! ๐ธโฎ ๐๐ด = = ๐ธ 4๐๐ ! = 1 ๐! 4๐๐ ! ๐ ๐! ๐ธ 4๐๐ ! = ๐ ! 4๐๐ ; ๐ธ ๐ ! = ๐ ! ๐ ๐ ๐ ๐ ๐ธ = ๐ ! ๐ ๐ b.r>R ๐!" = +๐ EonceagainisconstanteverywherealongtheimaginedGaussiansphere ๐!" ๐ โฎ ๐ธ๐๐ด = ; ๐ธ 4๐๐ ! = = ๐ธ 4๐๐ ! = ๐ 4๐๐ ๐! ๐! ๐๐ ๐ธ = ! ๐ Problem2:Cylinder Findtheelectricfielddistancerawayfromanon-conductingcylinderofuniform chargewhosechargeperunitlengthis๐andalengthofL Solution:CreateacylindricalGaussiansurfacethatsurroundsthenon-conducting cylinderwithuniformchargedensity. TheGaussiansurfacedoesnotencompassthetoporbottomofthecylinder. r ๐!" โฎ ๐ธ๐๐ด = ๐! GaussโLaw ๐ ๐ธ 2๐๐๐ฟ = ๐! Theenergyremainsthesamesoonlytheareaneedstobeintegrated.Theareaof thecylinderisjustthe2๐๐๐ฟpiecebecausetheelectricfielddoesnotgooutofthe endsofthecylinder. ๐ ๐ = ; ๐!" = ๐๐ฟ ๐ฟ Theinternalchargecanbeexpressedas ๐!" = ๐๐ฟ ! because๐ representsthechargeperunitlengthor! ๐๐ฟ ๐ธ 2๐๐๐ฟ = ; ๐ ! ๐! = !!"soitcanbemanipulatedsothat 4๐๐๐๐ฟ ๐ธ= 2๐๐๐ฟ ThepiandLcancelout,the4andthe2leaveone2behindandwhatโsleftisequalto theelectricfieldatsomedistancer. 2๐๐ ๐ธ= ๐ Problem3:Conductorwithnonconductorinside Thereisaconductingshellwithacharge2Qsurroundinganon-conductingcylinder oflengthLandachargeโ2Qandachargedensityof๐. Whatistheelectricfield when: Aistheradiusofthenonconductingcylinder Bistheradiusoftheinnerpartoftheconductingshell Cistheouterradiusoftheconductingshell c a b a.r=a Thisisjustlikethequestionabove. ๐!" โฎ ๐ธ๐๐ด = ๐! ๐๐ฟ ๐ธ 2๐๐๐ฟ = ; ๐ 4๐๐๐๐ฟ ๐ธ= 2๐๐๐ฟ 2๐๐ ๐ธ= ๐ b.whereb<r<c Theelectricfieldis0!Becauseitisinsideaconductor c.wherer>c Theelectricfieldisalso0!Becausethechargeoftheconductingshellandthenon conductioncylindercancel.