[SHIVOK SP212]
January 8, 2016 CH 24 ElectricPotential
I.
ElectricPotentialEnergy
A. Experimentally,physicistsandengineersdiscoveredthatthe
electricforceisconservativeandthushasanassociatedelectric
potentialenergy.
B. Themotivationforassociatingapotentialenergywithaforceis
thatwecanthenapplytheprincipleoftheconservationofmechanical
energytoclosedsystemsinvolvingtheforce.
C. Whenanelectrostaticforceactsbetweentwoormorecharged
particleswithinasystemofparticles,wecanassignanelectric
potentialenergyUtothesystem.
D. Ifthesystemchangesitsconfigurationfromaninitialstateitoa
differentfinalstatef,theelectrostaticforcedoesworkWonthe
particles.IftheresultingchangeisU,then
E. Aswithotherconservativeforces,theworkdonebythe
electrostaticforceispathindependent.
F. Usuallythereferenceconfigurationofasystemofcharged
particlesistakentobethatinwhichtheparticlesareallinfinitely
separatedfromoneanother.Thecorrespondingreferencepotential
energyisusuallysetbezero.Therefore,
Page1
[SHIVOK SP212]
January 8, 2016 II.
ElectricPotential:
A. Thepotentialenergyperunitchargeatapointinanelectricfield
iscalledtheelectricpotentialV(orsimplythepotential)atthatpoint.
Thisisascalarquantity.
Thus,
B. TheelectricpotentialdifferenceVbetweenanytwopointsiandfin
anelectricfieldisequaltothedifferenceinpotentialenergyperunit
chargebetweenthetwopoints.Thus,
C. Thepotentialdifferencebetweentwopointsisthusthenegativeof
theworkdonebytheelectrostaticforcetomoveaunitchargefrom
onepointtotheother.
D. IfwesetUi=0atinfinityasourreferencepotentialenergy,then
theelectricpotentialVmustalsobezerothere.Therefore,theelectric
potentialatanypointinanelectricfieldcanbedefinedtobe
1.
HereW istheworkdonebytheelectricfieldonachargedparticleas
∞
thatparticlemovesinfrominfinitytopointf.
E. TheSIunitforpotentialisthejoulepercoulomb.This
combinationiscalledthevolt(abbreviatedV).
1.
Thisunitofvoltallowsustoadoptamoreconventionalunitforthe
electricfield,E,whichisexpressedinnewtonspercoulomb.
Page2
[SHIVOK SP212]
January 8, 2016 F. Wecannowdefineanenergyunitthatisaconvenientonefor
energymeasurementsintheatomic/subatomicdomain:Oneelectron‐
volt(eV)istheenergyequaltotheworkrequiredtomoveasingle
elementarychargee,suchasthatoftheelectronortheproton,
throughapotentialdifferenceofexactlyonevolt.Themagnitudeof
thisworkisqV,and
G. ElectricPotential:WorkdonebyanAppliedForce
1.
Ifaparticleofchargeqismovedfrompointitopointfinanelectric
fieldbyapplyingaforcetoit,theappliedforcedoesworkW onthecharge
app
whiletheelectricfielddoesworkWonit.ThechangeKinthekineticenergy
oftheparticleis
2.
Iftheparticleisstationarybeforeandafterthemove;ThenK andK f
i
arebothzero.
3.
Relatingtheworkdonebyourappliedforcetothechangeinthe
potentialenergyoftheparticleduringthemove,onehas
4.
WecanalsorelateW
app
totheelectricpotentialdifferenceVbetween
theinitialandfinallocationsoftheparticle:
Page3
[SHIVOK SP212]
January 8, 2016 III.
EquipotentialSurfaces:
A. Adjacentpointsthathavethesameelectricpotentialforman
equipotentialsurface,whichcanbeeitheranimaginarysurfaceora
real,physicalsurface.
B. NonetworkWisdoneonachargedparticlebyanelectricfield
whentheparticlemovesbetweentwopointsiandfonthesame
equipotentialsurface.
C. Lookatandensureyouunderstandtheotherexamplesonpage
691.
Page4
[SHIVOK SP212]
January 8, 2016 IV.
CalculatingthePotentialfromtheField:
A. Giventhebelowsituation,weneedtocalculatethepotential
differencebetweenanypointsiandfinanelectricfieldifweknow
thefieldvectorallalonganypathconnectingthepoints.
B. Remember,
C. Thusfortheabovesituation,
1.
Totalwork:
2.
3.
Thus,thepotentialdifferenceVf‐Vibetweenanytwopointsiandfinan
electricfieldisequaltothenegativeofthelineintegralfromitof.Sincethe
electrostaticforceisconservative,allpathsyieldthesameresult.
4.
IfwesetpotentialVi=0,then
5.
ThisisthepotentialVatanypointfintheelectricfieldrelativetothe
zeropotentialatpointi.Ifpointiisatinfinity,thenthisisthepotentialVat
anypointfrelativetothezeropotentialatinfinity.
Page5
[SHIVOK SP212]
January 8, 2016 V.
PotentialDuetoaPointCharge:
A. Apositivelychargedparticleproducesapositiveelectricpotential.
Anegativelychargedparticleproducesanegativeelectricpotential.
B. ConsiderapointPatdistanceRfromafixedparticleofpositive
chargeq.Imaginethatwemoveapositivetestchargeq0frompointP
toinfinity.Thepathchosencanbethesimplestone—alinethat
extendsradiallyfromthefixedparticlethroughPtoinfinity.
C. IfVf=0(at∞)andVi=V(atR).Then,forthemagnitudeofthe
electricfieldatthesiteofthetestcharge,
D. Thatgives:
1.
SwitchingRtor,
Page6
[SHIVOK SP212]
January 8, 2016 E. PotentialDuetoaGroupofPointCharges
1.
Thenetpotentialatapointduetoagroupofpointchargescanbe
foundwiththehelpofthesuperpositionprinciple.Firsttheindividual
potentialresultingfromeachchargeisconsideredatthegivenpoint.Thenwe
sumthepotentials.
2.
Forncharges,thenetpotentialis
F. SampleProblem:
1.
TheelectricfieldinaregionofspacehasthecomponentsEy=Ez=0
andEx=(4.00N/C)x.PointAisontheyaxisaty=3.00m,andpointBisonthe
xaxisatx=4.00m.WhatisthepotentialdifferenceVB‐VA?
+Z Page7
[SHIVOK SP212]
January 8, 2016 VI.
PotentialDuetoanElectricDipole:
A. Considerthefollowingdiagramofadipole
B. AtP,thepositivepointcharge(atdistancer(+))setsuppotential
V(+)andthenegativepointcharge(atdistancer(‐))setsuppotential
V(‐).ThenthenetpotentialatPis:
C. Naturallyoccurringdipolesarequitesmall;soweareusually
interestedonlyinpointsthatarerelativelyfarfromthedipole,such
thatd«r,wheredisthedistancebetweenthecharges.Ifp=qd,
Page8
[SHIVOK SP212]
January 8, 2016 D. InducedDipoleMoment:
Page9
[SHIVOK SP212]
January 8, 2016 VII.
PotentialDuetoaContinuousChargeDistribution:LineofCharge
A. Fig.below(a)Athin,uniformlychargedrodproducesanelectric
potentialVatpointP.(b)Anelementcanbetreatedasaparticle.(c)
ThepotentialatPduetotheelementdependsonthedistancer.We
needtosumthepotentialsduetoalltheelements,fromtheleftside
(d)totherightside(e).
B. Ifisthechargeperunitlength,thenthechargeonlengthdxis:
1.
Step1: dq  2.
Step2: dV  3.
Step3: dV  4.
Step4: r   dV  5.
Step5: V  dV  
6.
Step6:#17inintegraltableAppendixE,pageA‐11
Remember_______________________ 7.
Step7:ReplaceLambawithQ/L
Page
10
[SHIVOK SP212]
January 8, 2016 VIII. PotentialDuetoaContinuousChargeDistribution:ChargedDisk
A. InFig.below,consideradifferentialelementconsistingofaflat
ringofradiusR’andradialwidthdR’.
B. Itschargehasmagnitude
C. ThecontributionofthisringtotheelectricpotentialatPis:
D. ThenetpotentialatPcanbefoundbyadding(viaintegration)the
contributionsofalltheringsfromR’=0toR’=R:
Note: Integral process is U substitution (See CH‐22 page 644) Let U=Z2+R’2 dU= 2R’dR’ Page
11
[SHIVOK SP212]
January 8, 2016 IX.
Sampleproblems:
A. TheFigurebelowshowsathinplasticrodoflengthL=14.9cm
anduniformpositivechargeQ=54.8fClyingonanxaxis.WithV=0
atinfinity,findtheelectricpotentialatpointP1ontheaxis,atdistance
d=2.59cmfromoneendoftherod.
1.
Solutionsetup:
Page
12
[SHIVOK SP212]
January 8, 2016 B. InFig.below,aplasticrodhavingauniformlydistributedcharge
Q=‐25.6pChasbeenbentintoacirculararcofradiusR=3.71cmand
centralangle=120°.WithV=0atinfinity,whatistheelectric
potentialatP,thecenterofcurvatureoftherod?
1.
Solution:
Page
13
[SHIVOK SP212]
January 8, 2016 C. InFig.below,aparticleofelementarycharge+eisinitiallyat
coordinatez=20nmonthedipoleaxis(hereazaxis)throughan
electricdipole,onthepositivesideofthedipole.(Theoriginofzisat
thecenterofthedipole.)Theparticleisthenmovedalongacircular
patharoundthedipolecenteruntilitisatcoordinatez=‐20nm,on
thenegativesideofthedipoleaxis.FigurebgivestheworkWadone
bytheforcemovingtheparticleversustheangleθthatlocatesthe
particlerelativetothepositivedirectionofthezaxis.Thescaleofthe
verticalaxisissetbyWas=4.0×10‐30J.Whatisthemagnitudeofthe
dipolemoment?
1.
Solutionsetup:
Page
14
[SHIVOK SP212]
January 8, 2016 X.
CalculatingtheFieldfromthePotential:
A. Supposethatapositivetestchargeq0movesthrougha
displacementfromoneequipotentialsurfacetotheadjacentsurface.
Theworktheelectricfielddoesonthetestchargeduringthemoveis
‐q0dV.
B. Theworkdonebytheelectricfieldmayalsobewrittenasthe
scalarproductor
=q0E(cos)ds.
1.
Therefore,
2.
Thatis,
3.
SinceEcosisthecomponentofEinthedirectionofds,
4.
Ifwetakethesaxistobe,inturn,thex,y,andzaxes,thex,y,andz
componentsofEatanypointare
5.
Therefore,thecomponentofEinanydirectionisthenegativeofthe
rateatwhichtheelectricpotentialchangeswithdistanceinthatdirection.
Page
15
[SHIVOK SP212]
January 8, 2016 C. Sampleproblem:
1.
Inaregionofspace,theelectricpotential,involts,isgivenby
V  2 xy 2 z  3 x 2 yz .Whatisthemagnitudeofelectricfieldvector,inN/C,
atthepoint{3,6,9}?
Page
16
[SHIVOK SP212]
January 8, 2016 XI.
ElectricPotentialEnergyofaSystemofPointCharges:
A. Theelectricpotentialenergyofasystemoffixedpointchargesis
equaltotheworkthatmustbedonebyanexternalagenttoassemble
thesystem,bringingeachchargeinfromaninfinitedistance.
1.
Theworkdoneisq2V,whereVisthepotentialthathasbeensetupby
q1atthepointwhereweputq2.
B. Sampleproblem:
1. Aparticleofcharge+7.5μCisreleasedfromrestatthepoint
x=60cmonanxaxis.Theparticlebeginstomoveduetothe
presenceofachargeQthatremainsfixedattheorigin.Whatisthe
kineticenergyoftheparticleattheinstantithasmoved40cmif(a)Q=
+20μCand(b)Q=‐20μC?
Page
17
[SHIVOK SP212]
January 8, 2016 XII.
PotentialofaChargedIsolatedConductor:
A. Remember,anexcesschargeplacedonanisolatedconductorwill
distributeitselfonthesurfaceofthatconductorsothatallpointsof
theconductor–whetheronthesurfaceorinside–cometothesame
potential.Thisistrueeveniftheconductorhasaninternalcavityand
evenifthatcavitycontainsanetcharge.
B. Weknowthat
C. SinceforallpointsE=0withinaconductor,itfollowsdirectlythat
Vf=Viforallpossiblepairsofpointsiandfintheconductor.
Page
18
[SHIVOK SP212]
January 8, 2016 D. Sampleproblem:
1.
Sphere1withradiusR1haspositivechargeq.Sphere2withradius
2.00R1isfarfromsphere1andinitiallyuncharged.Aftertheseparated
spheresareconnectedwithawirethinenoughtoretainonlynegligible
charge,(a)ispotentialV1ofsphere1greaterthan,lessthan,orequalto
potentialV2ofsphere2?Whatfractionofqendsupon(b)sphere1and(c)
sphere2?(d)Whatistheratioσ1/σ2ofthesurfacechargedensitiesofthe
spheres?
a)
Solutionsetup:
Page
19
[SHIVOK SP212]
January 8, 2016 XIII. IsolatedConductorinanIsolatedElectricField:
A. Ifanisolatedconductorisplacedinanexternalelectricfield,all
pointsoftheconductorstillcometoasinglepotentialregardlessof
whethertheconductorhasanexcesscharge.
B. Thefreeconductionelectronsdistributethemselvesonthe
surfaceinsuchawaythattheelectricfieldtheyproduceatinterior
pointscancelstheexternalelectricfieldthatwouldotherwisebe
there.
C. Furthermore,theelectrondistributioncausesthenetelectricfield
atallpointsonthesurfacetobeperpendiculartothesurface.Ifthe
conductorinFig.24‐20couldbesomehowremoved,leavingthe
surfacechargesfrozeninplace,theinternalandexternalelectricfield
wouldremainabsolutelyunchanged.
Page
20