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Electricity & MagnetismLecture 9: Conductors and Capacitance
Today’sConcept:
A)Conductors
B)Capacitance
(
Electricity&Magne7smLecture7,Slide1
Some of your comments
Thischaptermakesabsolute0sense.
Iwouldreallyloveanexplana7ononhowcapacitorsareusedastemporarybaHeriesforstoringenergyinthem.
Capacitors vs. Batteries
‣ AbaHeryusesanelectrochemicalreac7ontoseparatecharges.ΔVbetweenterminalsiscalled“EMF”,symbol:,E orE.
‣ Capacitorsstorechargeonitsplates.
Capacitance and Capacitors (30.5)
We have been using capacitors a lot without defining capacitanceor describing how to charge-up these devices.
A battery will create a potential difference across a capacitorwhich is equal to the potential difference in the battery.
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 4 / 14
E
Some capacitors
! 0
WE BELIEVE THERE ARE ONLY THREE THINGS YOU NEED TO KNOW TO DO ALL OF HOMEWORK!
1. E = 0 withinthematerialofaconductor:(atsta7cequilibrium)
Chargesmoveinsideaconductorinordertocanceloutthefieldsthatwouldbethereintheabsenceoftheconductor.Thisprincipledeterminestheinducedchargedensi7esonthesurfacesofconductors.
CONCEPTSDETERMINETHECALCULATION!
Our Comments
2.Gauss’Law:Ifchargedistribu7onshavesufficientsymmetry(spherical,
cylindrical,planar),thenGauss’lawcanbeusedtodeterminetheelectricfieldeverywhere.
Electricity&Magne7smLecture7,Slide3
I
surface
~E · d ~A = Qenclosed
✏0
�Va⇤b ⇥�Ua⇤b
q= �Z b
a~E · d~l
Conductors
" Chargesfreetomove" E=0inaconductor" Surface=Equipoten7al" Eatsurfaceperpendiculartosurface
TheMainPoints
Electricity&Magne7smLecture7,Slide4
CheckPoint: Two Spherical Conductors 1
Electricity&Magne7smLecture7,Slide5
Twosphericalconductorsareseparatedbyalargedistance.Theyeachcarrythesameposi7vechargeQ.ConductorAhasalargerradiusthanconductorB.
Comparethepoten7alatthesurfaceofconductorAwiththepoten7alatthesurfaceofconductorB.
A.VA>VB
B.VA=VB
C.VA<VB
“Thepoten7alisthesameasfromapointchargeatthecenterofAorB.So,followingV=kQ/r,Awouldhaveasmallerpoten7al“
CheckPoint: Two Spherical Conductors 2
Electricity&Magne7smLecture7,Slide6
Twosphericalconductorsareseparatedbyalargedistance.Theyeachcarrythesameposi7vechargeQ.ConductorAhasalargerradiusthanconductorB.
Thetwoconductorsarenowconnectedbyawire.Howdothepoten7alsattheconductorsurfacescomparenow?
A.VA>VB
B.VA=VB
C.VA<VB
“Thepoten7alswillbecomeequalsincethechargeswillwanttogotoplacesoflowerpoten7al,un7litbalancesout.“
CheckPoint: Two Spherical Conductors 3
Electricity&Magne7smLecture7,Slide7
Twosphericalconductorsareseparatedbyalargedistance.Theyeachcarrythesameposi7vechargeQ.ConductorAhasalargerradiusthanconductorB.
WhathappenstothechargeonconductorAaberitisconnectedtoconductorBbythewire?
A.QAincreasesB.QAdecreasesC.QAdoesnotchange
“Since B initially has a higher potential, charges move from B to A. “
786 | C H A P T E R 2 3 Electric Potential
Example 23-15 Two Charged Spherical Conductors
Two uncharged spherical conductors of radius and(Figure 23-26) and separated by a distance much greater than
are connected by a long, very thin conducting wire. A total chargeis placed on one of the spheres and the system is allowed to
reach electrostatic equilibrium. (a) What is the charge on each sphere?(b) What is the electric field strength at the surface of each sphere?(c) What is the electric potential of each sphere? (Assume that the chargeon the connecting wire is negligible.)
PICTURE The total charge will be distributed with on sphere 1 and on sphere 2 so that the spheres will be at the same potential. We can use
for the potential of each sphere.
SOLVE
V ! kQ>R Q2Q1
Q ! "80 nC6.0 cmR2 ! 2.0 cm
R1 ! 6.0 cm
(a) 1. Conservation of charge gives us one relation between thecharges and Q2:Q1
Q1 " Q2 ! Q
2. Equating the potential of the spheres gives us a secondrelation for the charges and Q2:Q1
kQ1
R1!kQ2
R2⇒ Q2 !
R2
R1Q1
3. Combine the results from steps 1 and 2 and solve for andQ2:
Q1 so
20 nCQ2 ! Q # Q1 !
60 nCQ1 !R1
R1 " R2Q !
6.0 cm8.0 cm
(80 nC) !
Q1 "R1
R2Q1 ! Q
(b) Use these results to calculate the electric field strengths at thesurface of the spheres:
450 kN>C!
E2 !kQ2
R22
!(8.99 $ 109 N # m2>C2)(20 $ 10#9 C)
(0.020 m)2
150 kN>C!
E1 !kQ1
R21
!(8.99 $ 109 N # m2>C2)(60 $ 10#9 C)
(0.060 m)2
(c) Calculate the common potential from for either sphere:kQ>R9.0 kV!
V1 !kQ1
R1!
(8.99 $ 109 N # m2>C2)(60 $ 10#9 C)0.060 m
CHECK If we use sphere 2 to calculate we obtain An additional check is available, be-
cause the electric field strength at the surface of each sphere is proportional to its chargedensity. The radius of sphere 1 is three times the radius of sphere 2, so its surface area isnine times the surface area of sphere 2. And because sphere 1 has three times the charge,its charge density is one-third the charge density of sphere 2. Therefore, the electric fieldstrength at the surface of sphere 1 should be one-third of the electric field strength at thesurface of sphere 2, which is what we found in Part (b).
TAKING IT FURTHER The presence of the long, very thin wire connecting the spheresmakes the result of this example only approximate because the potential function is valid for the region outside an isolated conducting sphere. With the wire in place thespheres cannot accurately be modeled as isolated spheres.
V ! kQ>r
109 N # m2>C2)(20 $ 10#9 C)>0.020 m ! 9.0 $ 103 V.V2 ! kQ2 >R2 ! (8.99 $V,
R1R2
F I G U R E 2 3 - 2 6
CheckPoint: Charged Parallel Plates 1
Electricity&Magne7smLecture7,Slide8
TwoparallelplatesofequalareacarryequalandoppositechargeQ0.Thepoten7aldifferencebetweenthetwoplatesismeasuredtobeV0.Anunchargedconduc7ngplate(thegreenthinginthepicturebelow)isslippedintothespacebetweentheplateswithouttouchingeitherone.ThechargeontheplatesisadjustedtoanewvalueQ1suchthatthepoten7aldifferencebetweentheplatesremainsthesame.
CompareQ1andQ0.oQ1<Q0
oQ1=Q0
oQ1>Q0
THECAPACITORQ
UESTIONSWERETOU
GH!
THEPLAN:
We’llwork
throughtheexam
pleinthe
PreLectureandth
endothepreflig
htques7ons.
Capacitanceisdefinedforanypairofspa7allyseparatedconductors.
Howdoweunderstandthisdefini7on?
+Q
−Qd
!Considertwoconductors,onewithexcesscharge=+Qandtheotherwithexcesscharge=−Q
!Thesechargescreateanelectricfieldinthespacebetweenthem
E
!Thispoten7aldifferenceshouldbepropor7onaltoQ!•Thera7oofQtothepoten7aldifferenceisthecapacitanceandonlydependsonthegeometryoftheconductors
V
!Wecanintegratetheelectricfieldbetweenthemtofindthepoten7aldifferencebetweentheconductor
Capacitance
Electricity&Magne7smLecture7,Slide9
First determine E field produced by charged conductors:
Second,integrateEtofindthepotenHaldifferenceV
x
y
Aspromised,VisproporHonaltoQ!
Cdeterminedbygeometry!
Example (done in Prelecture 7)
Whatisσ ?
A=areaofplate
+Q
−Qd E
Electricity&Magne7smLecture7,Slide10
V = �Z d
0~E · d~y
+Q0
−Q0
dIniHalchargeoncapacitor=Q0
PlatesnotconnectedtoanythingA)Q1<Q0
B)Q1=Q0
C)Q1>Q0
HowisQ1relatedtoQ0?
Clicker Question Related to CheckPoint
td
+Q1
−Q1
Insertunchargedconductor
Chargeoncapacitornow=Q1
CHARGECANNOTCHANGE!
Electricity&Magne7smLecture7,Slide11
td
+Q1
−Q1
A)+Q1
B)−Q1
C) 0D)PosiHvebutthemagnitudeunknownE)NegaHvebutthemagnitudeunknown
WhatisthetotalchargeinducedontheboHomsurfaceoftheconductor?
Where to Start ?
Electricity&Magne7smLecture7,Slide12
E
+Q0
−Q0
E
WHATDOWEKNOW?
E = 0
Emustbe= 0inconductor!
Why ?
+Q0
−Q0
ChargesinsideconductormovetocancelEfieldfromtop&bo8omplates.
Electricity&Magne7smLecture7,Slide13
NowcalculateVasafuncHonofdistancefromtheboSomconductor.
+Q0
−Q0
E = 0
y
V
y
d t
WhatisΔV=V(d)?A)ΔV=E0dB)ΔV=E0(d−t)C)ΔV=E0(d+t)
d
tEy
-E0
Theintegral=areaunderthecurve
Calculate V
Electricity&Magne7smLecture7,Slide14
V( y) = �Z y
0~E · d~y
CheckPoint Results: Charged Parallel Plates 1
Electricity&Magne7smLecture7,Slide15
TwoparallelplatesofequalareacarryequalandoppositechargeQ0.Thepoten7aldifferencebetweenthetwoplatesismeasuredtobeV0.Anunchargedconduc7ngplate(thegreenthinginthepicturebelow)isslippedintothespacebetweentheplateswithouttouchingeitherone.ThechargeontheplatesisadjustedtoanewvalueQ1suchthatthepoten7aldifferencebetweentheplatesremainsthesame.
CompareQ1andQ0.A.Q1<Q0
B.Q1=Q0
C.Q1>Q0
“ThedistanceforQ1issmallerthereforethechargemustdecreasetocompensateforthechangeindistance““Sincethepoten?alremainsthesame,thereisnochangeinthechargeofQ.““thefieldthroughtheconductoriszero,soithasconstantpoten?al.BecauseofthisitmusthavegreaterchargesothetotalVisthatsame.“
CheckPoint Results: Charged Parallel Plates 2
Electricity&Magne7smLecture7,Slide16
TwoparallelplatesofequalareacarryequalandoppositechargeQ0.Thepoten7aldifferencebetweenthetwoplatesismeasuredtobeV0.Anunchargedconduc7ngplate(thegreenthinginthepicturebelow)isslippedintothespacebetweentheplateswithouttouchingeitherone.ThechargeontheplatesisadjustedtoanewvalueQ1suchthatthepoten7aldifferencebetweentheplatesremainsthesame.
Comparethecapacitanceofthetwoconfigura7onsintheaboveproblem.A.C1>C0B.C1=C0C.C1<C0
“ThedistanceforQ1issmallerthereforethechargemustdecreasetocompensateforthechangeindistance““Sincethepoten?alremainsthesame,thereisnochangeinthechargeofQ.““thefieldthroughtheconductoriszero,soithasconstantpoten?al.BecauseofthisitmusthavegreaterchargesothetotalVisthatsame.“
CheckPoint Results: Charged Parallel Plates 2
Electricity&Magne7smLecture7,Slide17
TwoparallelplatesofequalareacarryequalandoppositechargeQ0.Thepoten7aldifferencebetweenthetwoplatesismeasuredtobeV0.Anunchargedconduc7ngplate(thegreenthinginthepicturebelow)isslippedintothespacebetweentheplateswithouttouchingeitherone.ThechargeontheplatesisadjustedtoanewvalueQ1suchthatthepoten7aldifferencebetweentheplatesremainsthesame.
Comparethecapacitanceofthetwoconfigura7onsintheaboveproblem.A.C1>C0B.C1=C0C.C1<C0
WecandetermineCfromeithercase sameV(preflight) sameQ(lecture)Cdependsonlyongeometry! E0 = Q0 /ε0A
C0 = Q0 /E0d
C1 = Q0 /[E0(d − t)]
C0 = ε0A /d
C1 = ε0A /(d − t)
BANG
Energy in Capacitors
Electricity&Magne7smLecture7,Slide18
!ConceptualAnalysis:
ButwhatisQ andwhatisV ?Theyarenotgiven?
!ImportantPoint:C isapropertyoftheobject!(concentriccylindershere)• AssumesomeQ (i.e.,+Q ononeconductorand−Q ontheother)• ThesechargescreateEfieldinregionbetweenconductors• This E fielddeterminesapoten7aldifferenceV betweentheconductors• V shouldbepropor7onaltoQ; thera7oQ/V isthecapacitance.
Calculation
metal
metal
a1
a2
a3a4
cross-sec7on Acapacitorisconstructedfromtwoconduc7ngcylindricalshellsofradiia1,a2,a3,anda4andlength
L(L>>ai).
WhatisthecapacitanceCofthiscapacitor?
Electricity&Magne7smLecture7,Slide19
!StrategicAnalysis:• Put+Qonoutershelland −Qoninnershell
• Cylindricalsymmetry:UseGauss’LawtocalculateEeverywhere• IntegrateEtogetV• Takera7oQ/V:shouldgetexpressiononlyusinggeometricparameters(ai,L)
Calculationcross-sec7on
metal
metal
a1
a2
a3a4 Acapacitorisconstructedfromtwoconduc7ng
cylindricalshellsofradiia1,a2,a3,anda4andlength
L(L>>ai).
WhatisthecapacitanceCofthiscapacitor?
Electricity&Magne7smLecture7,Slide20
Calculation
metal
metal
+Q
a1
a2
a3a4
−Q
Why?
Whereis+Qonouterconductorlocated?
A)atr=a4B)atr=a3C)bothsurfacesD)throughoutshell
WeknowthatE = 0 inconductor(betweena3anda4)
++
+
+
+++
+
+
++
+
+
+++
+
+
cross-sec7on Acapacitorisconstructedfromtwoconduc7ngcylindricalshellsofradiia1,a2,a3,anda4andlength
L(L>>ai).
WhatisthecapacitanceCofthiscapacitor?
+Q mustbeoninsidesurface(a3), sothat Qenclosed = + Q − Q = 0
Electricity&Magne7smLecture7,Slide21
Gauss’law:
I
surface
~E · d ~A = Qenclosed
✏0
Calculation
metal
+Q
a1
a2
a3a4
−Q
Why?
+ ++
+
+
++
+
+
+ ++
+
+
+++
+
+
We know that E = 0 in conductor (between a1 and a2)
−
metal
−−−−−
−−− −
−−−−
−
−−−
cross-sec7on Acapacitorisconstructedfromtwoconduc7ngcylindricalshellsofradiia1,a2,a3,anda4andlength
L(L>>ai).
WhatisthecapacitanceCofthiscapacitor?
Whereis−Qoninnerconductorlocated?
A)atr=a2B)atr= a1C)bothsurfacesD)throughoutshell
+Q mustbeonoutersurface(a2),
sothatQenclosed = 0
Electricity&Magne7smLecture7,Slide22
Gauss’law:
I
surface
~E · d ~A = Qenclosed
✏0
Calculation
metal
+Q
a1
a2
a3a4
−Q
+ ++
+
+
++
+
+
+ ++
+
+
++
+
+
−
metal
−−−−−
−−− −
−−−−
−
−−−
−Why?
cross-sec7on Acapacitorisconstructedfromtwoconduc7ngcylindricalshellsofradiia1,a2,a3,anda4andlength
L(L>>ai).
WhatisthecapacitanceCofthiscapacitor?
a2<r<a3:WhatisE(r)?
A) 0 B) C)D)E)
E(2⇡rL) =Q✏0
Direc7on:RadiallyIn
Electricity&Magne7smLecture7,Slide23
Gauss’law:
I
surface
~E · d ~A = Qenclosed
✏0
Calculation
r < a2: E(r) = 0
since Qenclosed = 0
a2<r<a3:
WhatisV ?Thepoten7aldifferencebetweentheconductors.
WhatisthesignofV=Vouter−Vinner?
A)Vouter − Vinner<0B)Vouter − Vinner= 0C)Vouter−Vinner>0
Q(2πε0a2L)
metal
+Q
a1
a2
a3a4
−Q
+ ++
+
+
++
+
+
+ ++
+
+
++
+
+
−
metal
−−−−−
−−− −
−−−
−
−−−
−
cross-sec7on Acapacitorisconstructedfromtwoconduc7ngcylindricalshellsofradiia1,a2,a3,anda4andlength
L(L>>ai).
WhatisthecapacitanceCofthiscapacitor?
Electricity&Magne7smLecture7,Slide24
Q(2πε0a2L)
WhatisV≡Vouter−Vinner?
(A)(B)(C)(D)
Calculation
V propor7onaltoQ,aspromised
metal
+Q
a1
a2
a3a4
−Q
+ ++
+
+
++
+
+
+ ++
+
+
++
+
+
−
metal
−−−−−
−−− −
−−−
−
−−−
−
cross-sec7on Acapacitorisconstructedfromtwoconduc7ngcylindricalshellsofradiia1,a2,a3,anda4andlength
L(L>>ai).
WhatisthecapacitanceCofthiscapacitor?
a2<r<a3:
Electricity&Magne7smLecture7,Slide25
C ⌘ QV=
2⇡✏0Lln(a3/a2)
The Potential of a Ring of Charge
The Potential of a Ring of Charge (Example 29.11)
A thin, uniformly charged ring of radius R
has total charge Q . Find the potential atdistance z on the axis of the ring.
The distance r
i
isr
i
=p
R
2 + z
2
The potential is then the sum:
V =NX
i=1
14⇤�0
�Q
r
i
=1
4⇤�01⌃
R
2 + z
2
NX
i=1
�Q =1
4⇤�0Q⌃
R
2 + z
2
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 5 / 19
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