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Electricity & Magnetism
Lecture 9: Conductors and
Capacitance
Today’sConcept:
A)Conductors B)Capacitance
(
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Stuff you asked about:!
“Theyneversaidwhatcapacitanceisorwhatitsusedfor”!
Pleasegoovertheques
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Capacitors vs. Batteries
‣ AbaPeryusesanelectrochemicalreac
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Some capacitors" 0
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WE BELIEVE THERE ARE ONLY THREE THINGS YOU NEED TO KNOW TO DO
ALL OF HOMEWORK!
1. E = 0 withinthematerialofaconductor:(atsta
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Conductors
! Chargesfreetomove! E=0inaconductor! Surface=Equipoten
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CheckPoint: Two Spherical Conductors 1
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CheckPoint: Two Spherical Conductors 2
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CheckPoint: Two Spherical Conductors 3
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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
kQ1R1
!kQ2R2
⇒ Q2 !R2R1Q1
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 "R1R2Q1 ! Q
(b) Use these results to calculate the electric field strengths
at thesurface of the spheres:
450 kN>C!E2 !
kQ2R22
!(8.99 $ 109 N # m2>C2)(20 $ 10#9 C)
(0.020 m)2
150 kN>C!E1 !
kQ1R21
!(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 !kQ1R1
!(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
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CheckPoint: Charged Parallel Plates 1
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Capacitanceisdefinedforanypairofspa
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First determine E field produced by charged conductors:
Second,integrateEtofindthepoten
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+Q0
−Q0
dIniQ0
HowisQ1relatedtoQ0?
Clicker Question Related to CheckPoint
td
+Q1
−Q1
Insertunchargedconductor
Chargeoncapacitornow=Q1
CHARGECANNOTCHANGE!
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td
+Q1
−Q1
A)+Q1 B)−Q1 C) 0 D)Posi
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E
+Q0
−Q0
E
WHATDOWEKNOW?
E = 0
Emustbe= 0inconductor!
Why ?
+Q0
−Q0
ChargesinsideconductormovetocancelEfieldfromtop&boPomplates.
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NowcalculateVasafunc
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CheckPoint Results: Charged Parallel Plates 1
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CheckPoint Results: Charged Parallel Plates 2
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CheckPoint Results: Charged Parallel Plates 2
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Demo:BANG
Energy in Capacitors
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!ConceptualAnalysis:
ButwhatisQ andwhatisV ?Theyarenotgiven?
!ImportantPoint:C
isapropertyoftheobject!(concentriccylindershere)• AssumesomeQ
(i.e.,+Q ononeconductorand−Q ontheother)•
ThesechargescreateEfieldinregionbetweenconductors• This E
fielddeterminesapoten
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!StrategicAnalysis:• Put+Qonoutershelland −Qoninnershell•
Cylindricalsymmetry:UseGauss’LawtocalculateEeverywhere•
IntegrateEtogetV • Takera
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Calculation
metal
metal
+Q
a1a2
a3a4
−Q
Why?
Whereis+Qonouterconductorlocated?A)atr=a4B)atr=a3C)bothsurfacesD)throughoutshell
WeknowthatE = 0 inconductor(betweena3anda4)
++
+
+
+++
+
+
++
+
+
+++
+
+
cross-secai).
WhatisthecapacitanceCofthiscapacitor?
+Q mustbeoninsidesurface(a3), sothat Qenclosed = + Q − Q = 0
Electricity&Magne
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Calculation
metal
+Q
a1a2
a3a4
−Q
Why?
+ ++
+
+
++
+
+
+ ++
+
+
+++
+
+
We know that E = 0 in conductor (between a1 and a2)
−
metal
−−−−−
−−− −
−−−−
−
−−−
cross-secai).
WhatisthecapacitanceCofthiscapacitor?
Whereis−Qoninnerconductorlocated?A)atr=a2B)atr=
a1C)bothsurfacesD)throughoutshell
+Q mustbeonoutersurface(a2),sothatQenclosed = 0
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Calculation
metal
+Q
a1a2
a3a4
−Q
+ ++
+
+
++
+
+
+ ++
+
+
++
+
+
−
metal
−−−−−
−−− −
−−−−
−
−−−
−Why?
cross-secai).
WhatisthecapacitanceCofthiscapacitor?
a2<r<a3:WhatisE(r)?
A) 0 B) C)D)E)
E(2⇡rL) =Q✏0
Direc
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Calculation
r < a2: E(r) = 0
since Qenclosed = 0
a2<r<a3:
WhatisV ?Thepoten0
Q (2πε0a2L)
metal
+Q
a1a2
a3a4
−Q
+ ++
+
+
++
+
+
+ ++
+
+
++
+
+
−
metal
−−−−−
−−− −
−−−
−
−−−
−
cross-secai).
WhatisthecapacitanceCofthiscapacitor?
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Q (2πε0a2L)
WhatisV≡Vouter−Vinner?
(A)(B)(C)(D)
Calculation
V propor
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The Potential of a Ring of Charge
The Potential of a Ring of Charge (Example 29.11)
A thin, uniformly charged ring of radius Rhas total charge Q .
Find the potential atdistance z on the axis of the ring.
The distance ri isri =
pR2 + z2
The potential is then the sum:
V =NX
i=1
1
4⇤�0
�Qri
=1
4⇤�0
1⌃R2 + z2
NX
i=1
�Q =1
4⇤�0
Q⌃R2 + z2
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity
& Magnetism Spring 2010 5 / 19
