Electricity & Magnetism
Lecture 6: Electric Potential
Today’sConcept:
ElectricPoten4al
(DefinedintermsofPathIntegralofElectricField)
Electricity&Magne4smLecture6,Slide1
Stuff you asked about:
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“ExplainmorewhyEisnega4veofdeltaV”
“Idon'tlikethatweweretoldtousegradientswhenwehaven'tevendone
theminmathyet.”
“Sowejustneedtosuperimposetheradialfieldlineswhicharefoundbytaking
thenega4veofthegradientoftheelectricpoten4alin3dcartesian/spherical/
cylindricalcoordinatesystemandareperpendiculartoequipoten4als,thelocus
pointofallpointwiththesamepoten4aldifference.Simpleenough...We'lljust
dothat!....Ohhhhhwait...WHAT?.”
“soelectricpoten4alisthean4deriva4veofelectricfieldisthatcorrect?inother
wordstheareaunderelectricfieldfunc4ongivestheelectricpoten4al?
Whatisthedifferencebetweenanintegralofadotproductandaintegralofa
simpleproduct?
Can you please explain the Gradient in different Coordinate systems?
Electricity&Magne4smLecture6,Slide2
Big Idea
Last4mewedefinedtheelectricpoten4alenergyofchargeqinanelectricfield:
r
r r
b r
b
ΔU a→b = − ∫ F ⋅ dl = − ∫ qE ⋅ dl
a
a
Theonlymen4onofthepar4clewasthroughitschargeq.
Wecanobtainanewquan4ty,theelectricpoten4al,whichisaPROPERTYOF
€
THESPACE,asthepoten4alenergyperunitcharge.
b
r
r
Δ U a →b
ΔVa → b ≡
= − ∫ E ⋅ dl
q
a
Notethesimilaritytothedefini4onofanotherquan4tywhichisalsoaPROPERTY
r
OFTHESPACE,theelectricfield.
v F
E≡
q
Electricity&Magne4smLecture6,Slide3
Electric Potential from E field
ConsiderthethreepointsA,B,andClocatedinaregionofconstantelectric
fieldasshown.
D
Δx
WhatisthesignofΔVAC = VC-VA?
A)ΔV
AC < 0B)ΔVAC = 0C)ΔVAC > 0
C r
r
Rememberthedefini4on: ΔVA→C = − ∫ E ⋅ dl
Chooseapath(anywilldo!)
D r
r C r
Δ V A → C = − ∫ E ⋅ dl −
A
∫
D
r
E ⋅ dl
A
C
Δ V A →C = 0 −
∫
r r
E ⋅ dl = − E Δx < 0
D
Electricity&Magne4smLecture6,Slide4
CheckPoint: Zero Electric Field
Supposetheelectricfieldiszeroinacertainregionofspace.Whichofthefollowing
statementsbestdescribestheelectricpoten4alinthisregion?
A.  Theelectricpoten4aliszeroeverywhereinthisregion.
B.  Theelectricpoten4aliszeroatleastonepointinthisregion.
C.  Theelectricpoten4alisconstanteverywhereinthisregion.
D.  Thereisnotenoughinforma4ongiventodis4nguishwhichoftheabove
answersiscorrect.
Rememberthedefini4on
r r
= − ∫ E ⋅ dl
B
Δ V A →B
A
r
E=0
ΔVA→B = 0
Visconstant!
Electricity&Magne4smLecture6,Slide5
E from V
Ifwecangetthepoten4albyintegra4ngtheelectricfield:
b r
r
ΔVa → b = − E ⋅ dl
a
∫
Weshouldbeabletogettheelectricfieldbydifferen4a4ngthepoten4al?
r
v
E = −∇V
InCartesiancoordinates:
dV
Ex = −
Ey = −
dx
dV
dy
dV
Ez = −
dz
Electricity&Magne4smLecture6,Slide6
CheckPoint: Spatial Dependence of Potential 1
Theelectricpoten4alinacertainregionisplodedinthefollowinggraph
Atwhichpointisthemagnitudeofthe
E-FIELDgreatest?
o A
o B
o C
o D
HowdowegetEfrom V ?
r
v
E = −∇V
Ex = −
∂V
dx
Lookatslopes!
Electricity&Magne4smLecture6,Slide7
CheckPoint: Spatial Dependence of Potential 2
Theelectricpoten4alinacertainregionisplodedinthefollowinggraph
Atwhichpointisthedirec4onoftheEfieldalongthenega4vex-axis?
o A
o B
o C
o D
“At B, the slope is decreasing (-) so the direction of the E field is negative “
“E is negative when the slope of V is positive (E=-dV/dx). Therefore E is directed along the
x-axis at point C. “
HowdowegetEfrom V ?
r
v
E = −∇V
Ex = −
dV
dx
Lookatslopes!
Electricity&Magne4smLecture6,Slide8
Equipotentials
Equipoten4alsarethelocusofpointshavingthesamepoten4al.
Equipotentials produced
by a point charge
Equipoten4alsare
ALWAYS
perpendiculartotheelectricfieldlines.
TheSPACINGoftheequipoten4alsindicates
TheSTRENGTHoftheelectricfield.
Electricity&Magne4smLecture6,Slide9
Contour Lines on Topographic Maps
Electricity&Magne4smLecture6,Slide10
Visualizing the Potential of a Point Charge
Electricity&Magne4smLecture6,Slide11
CheckPoint: Electric Field Lines 1
Thefield-linerepresenta4onoftheE-fieldinacertainregioninspaceisshownbelow.
Thedashedlinesrepresentequipoten4allines.
AtwhichpointinspaceistheE-fieldthe
weakest?
o A
o B
o C
o D
“The electric field lines are the least dense at D”
“ From what I know, the answer should be D”
“ D is where the electric field lines are the least dense “
“ I’m pretty sure the electric field lines are the least dense at D”
“ I’d guess D “
Electricity&Magne4smLecture6,Slide12
CheckPoint: Electric Field Lines 2
Thefield-linerepresenta4onoftheE-fieldinacertainregioninspaceisshownbelow.
Thedashedlinesrepresentequipoten4allines.
Comparetheworkdonemovinganega4vecharge
fromAtoBandfromCtoD.Whichonerequires
morework?
A.  Moreworkisrequiredtomoveanega4ve
chargefromAtoBthanfromCtoD
B.  Moreworkisrequiredtomoveanega4ve
chargefromCtoDthanfromAtoB
C.  Thesameamountofworkisrequiredto
moveanega4vechargefromAtoBasto
moveitfromCtoD
D.  Cannotdeterminewithoutperformingthe
calcula4on
Electricity&Magne4smLecture6,Slide13
Clicker Question: Electronic Field 2
Whatarethese?
ELECTRICFIELDLINES!
Whatarethese?
EQUIPOTENTIALS!
WhatisthesignofWAC = workdonebyEfieldtomovenega4vechargefromAtoC?
A)WAC<0B)WAC=0C)WAC>0
AandCareonthesameequipoten4al
WAC = 0
Equipoten4alsareperpendiculartotheEfield:Noworkisdonealonganequipoten4al
Electricity&Magne4smLecture6,Slide14
CheckPoint Results: Electric Field Lines 2
Thefield-linerepresenta4onoftheE-fieldinacertainregioninspaceisshownbelow.
Thedashedlinesrepresentequipoten4allines.
Comparetheworkdonemovinganega4vecharge
fromAtoBandfromCtoD.Whichonerequires
morework?
A.  Moreworkisrequiredtomoveanega4ve
chargefromAtoBthanfromCtoD
B.  Moreworkisrequiredtomoveanega4ve
chargefromCtoDthanfromAtoB
C.  Thesameamountofworkisrequiredto
moveanega4vechargefromAtoBasto
moveitfromCtoD
D.  Cannotdeterminewithoutperformingthe
calcula4on
! AandCareonthesameequipoten4al
! BandDareonthesameequipoten4al
! Thereforethepoten4aldifferencebetweenAandBisthe
SAMEasthepoten4albetweenCandD
Electricity&Magne4smLecture6,Slide15
CheckPoint: Electric Field Lines 3
Thefield-linerepresenta4onoftheE-fieldinacertainregioninspaceisshownbelow.
Thedashedlinesrepresentequipoten4allines.
Comparetheworkdonemovinganega4vecharge
fromAtoBandfromAtoD.Whichonerequires
morework?
A.  Moreworkisrequiredtomoveanega4ve
chargefromAtoBthanfromAtoD
B.  Moreworkisrequiredtomoveanega4ve
chargefromAtoDthanfromAtoB
C.  Thesameamountofworkisrequiredto
moveanega4vechargefromAtoBasto
moveitfromAtoD
D.  Cannotdeterminewithoutperformingthe
calcula4on
Electricity&Magne4smLecture6,Slide16
Calculation for Potential
cross-sec4on
a4
a3
+Q
a2
a1
+q
metal
Pointchargeqatcenterofconcentric
conduc4ngsphericalshellsofradiia1,a2,a3,and
a4.Theinnershellisuncharged,buttheouter
shellcarrieschargeQ.
WhatisVasafunc4onof r?
metal
ConceptualAnalysis:
!  ChargesqandQwillcreateanEfieldthroughoutspace
r r
!  V ( r ) = − ∫ E ⋅ d l
r
r0
StrategicAnalysis:
!  Sphericalsymmetry:UseGauss’LawtocalculateEeverywhere
!  Integrate E togetV
Electricity&Magne4smLecture6,Slide17
Calculation: Quantitative Analysis
cross-sec4on
a4
a3
a2
a1
r > a4 : WhatisE(r)?
1 Q+q
1 Q
A)0B)C)
2
+Q
4πε 0 r
50
+q
metal
r
2πε 0
r
1 Q−q
1 Q+q
D)E)
4πε r 2
4πε 0 r 2
0
metal
Why?
Gauss’law:
r r Qe nc losed
∫ E ⋅ dA = ε
0
Q+q
E 4π r =
ε0
1 Q+q
E=
4πε 0 r 2
2
Electricity&Magne4smLecture6,Slide18
Calculation: Quantitative Analysis
cross-sec4on
a4
a3
a3< r<a4:WhatisE(r)?
+Q
1 q
1 q
4πε 0 r 2
2πε 0 r
A)0B)C)
a2
a1
+q
metal
1 −q
1 Q−q
D)E)
4πε 0 r 2
4πε 0 r 2
r
metal
ApplyingGauss’law,whatisQenclosedforredsphereshown?
A)qB)-qC)0
Howisthispossible?
-qmustbeinducedatr=a3surface
σ3 =
−q
4π a32
chargeatr=a4surface=Q+q
σ4 =
Q+q
4π a42
Electricity&Magne4smLecture6,Slide19
Calculation: Quantitative Analysis
cross-sec4on
a4
a3
+Q
a2
a1
Con4nueonin…
+q
metal
r
a2 < r < a3 : E = 1 q2
4πε 0 r
a1 < r < a2 :
r < a1 :
E=0
E=
1 q
4πε 0 r 2
metal
TofindV:
1) Chooser0suchthatV(r0)=0(usual:r0=infinity)
2) Integrate!
1 Q+q
V
=
r > a4 :
4π ε 0 r
a3 < r < a4 : A) V = 0
1 Q+q
V
=
= ΔV (∞ → a4 )+ 0
B)
4π ε 0 a4
C)
V=
1 Q+q
4π ε 0 a3
Electricity&Magne4smLecture6,Slide20
Calculation: Quantitative Analysis
cross-sec4on
a4
a3
r > a4 :
+Q
a2
a1
V=
1 Q+q
4π ε 0 r
a3 < r < a4 : V = 1 Q + q
4π ε 0
a4
+q
metal
metal
a2 < r < a3 :
a1 < r < a2 :
V (r ) = ΔV (∞ → a4 ) + 0 + ΔV (a3 → r )
V (r ) =
Q+q
q &1 1 #
$$ − !!
+0+
4πε 0 a4
4πε 0 % r a3 "
V (r ) =
1 &Q+q q q #
$$
+ − !!
4πε 0 % a4
a 2 a3 "
1 $Q +q q q '
V (r) =
+ − )
&
4πε 0 % a4
r a3 (
1 &Q+q q q q q #
$$
V
(
r
)
=
+ − + − !!
0 < r < a1 :
4πε 0 % a4
a2 a3 r a1 "
Electricity&Magne4smLecture6,Slide21