Physics 202, Lecture 6
Today’s Topics

Capacitance (Ch. 24.1-24.3)
Calculating capacitance
Combinations of capacitors: series and parallel
Energy stored in capacitors (24.4)
Capacitance
Capacitor: two (spatially separated) conductors, charged
to +Q and -Q, with constant potential difference ΔV
stores electrical energy (by storing charge)
Capacitance:
Our first example
of a circuit element:
C!
Q
"V
a
b
C
Units: Farad (F)
1 F= 1 C/V
1
Charging a Capacitor
Battery: source of ΔV
Charging
Uncharged
Charge flows
until Q=CΔV
C
flow of e-
C
ΔV=V+-V-
Charging/discharging capacitor:
Text problems: 24.2, 24.6, 24.8
Calculating Capacitance
Capacitance is independent of charge, voltage on
capacitor: it depends on geometry of the conductors.
Examples:
Parallel conducting plates
Concentric spherical conductors
Concentric cylindrical conductors
Procedure: determine
!V as a function of Q
Simple example: capacitance of sphere
!V = V =
kQ
R
C=
(imagine second
conductor at infinity)
Q
Q
R
=
= = 4"# 0 R
!V kQ R k
Text: 24.13
2
Example: Parallel Plate Capacitor
Consider two metallic parallel plates with area A, separation d:
A !d
Step 1. Use Gauss’s law to get E:
!"
!
! !
Eid
" A = Ein A! = # A! $ 0
Ein = ! " 0
! =Q A
+Q
A
++++
d -----Q
A!
!
E
Step 2. Get potential difference:
! ! $ d Qd
!V = " # E i dl =
=
%0 %0 A
Step 3. Get Capacitance:
C=
Q
" A
= 0
!V
d
Parallel Plate Capacitor
C=
!0 A
d
fringe field
Realistic case
Text: 24.15,…
3
Spherical Capacitor
Consider two concentric conducting spherical shells, radii a,b
a<b
-Q
Step 1. Get electric field:
Gauss’s Law:
! kQ
E = 2 r̂
r
(a < r < b)
b
Step 2. Get potential difference:
a
+Q
a
! !
kQ
$ 1 1 ' kQ(b " a)
!V = " # Eidl = " # 2 dr = kQ & " ) =
% a b(
r
ab
b
C=
Step 3:
ab
k(b ! a)
Cylindrical Capacitor
Consider two concentric cylindrical conducting shells, radii a,b
a<b
Q = !!
Step 1. Get electric field:
!
Gauss’s Law: E =
2k !
r̂
r
(a < r < b)
Step 2. Get potential difference:
a
! !
2k $
!V = " # E idl = " #
dr = 2k $ ln ( b a )
r
b
Step 3:
C=
!
2k ln(b / a)
4
Calculating Capacitance
+Q
• Given 4 concentric cylinders
of radii a,b,c,d and charges
+Q, -Q, +Q, -Q
-Q
b
• Note: E=0 between b and c! WHY??
A cylinder of radius r1: b < r1< c
encloses zero charge!
C=
a
d
Q
Q
Q $ $ b'
$ d''
#a 2!" 0 rL dr + 0 + #c 2!" 0 rL dr = 2!" 0 L &% ln &% a )( + ln &% c )( )(
b
Vad =
+Q
c
• Question: What is the
capacitance between a and d?
•
-Q
d
Q
2!" 0 L
=
Vad # # b &
# d&&
%$ ln %$ a (' + ln %$ c (' ('
Note: result of 2 cylindrical capacitors
“in series” (next slides)
Combinations of Capacitors
Parallel combination: ΔV same
ΔV1 Q
C1 1
C2
Q2
ΔV2
C1ΔV1 =Q1
C2ΔV2 =Q2
ΔV
CP
Q=Q1+Q2
effectively
ΔV
ΔV1 =ΔV2 =ΔV
(why?)
CP= C1+C2+C3+…
ΔV
Equivalent Capacitance
CP=Q/ΔV  C = C1+ C2
CP always > Ci!
5
Combinations of Capacitors
Series combination:
-Q1
+Q2
+Q1
C1ΔV1 =Q
C2ΔV2 =Q
-Q2
Charge conservation:
Q1=Q2(=Q)
C1
Q
C2
ΔV1 ΔV2
effectively
CS
ΔV=ΔV1+ΔV2
ΔV
Equivalent Capacitance
CS=Q/ΔV  1/CS = 1/C1+1/C2
1/CS= 1/C1+1/C2+1/C3+…
Cs =
C1C2
C1 + C2
CS always < Ci!
Example: Connection of Charged Capacitors
Two capacitors, C1=1µF and C2=2µF are initially
charged to ΔV1=1V and ΔV2=2V, respectively.
 What are the charges in each capacitor?
 If the capacitors are connected in parallel as shown by
the dashed lines, what is the charge in each capacitor
after the connection?
-
C1
-
+
+
C2
More examples: 24.29-31
6
Energy of a Capacitor
• How much energy is stored in a charged capacitor?
– Calculate the work provided (usually by a battery) to charge a
capacitor to +/- Q:
Incremental work dW needed to add charge dq to capacitor at voltage V:
-
+
! q$
dW = V (q)i dq = # & i dq
" C%
• The total work W to charge to Q is then given by:
Q
W=
1
1 Q2
qdq
=
C !0
2 C
• In terms of the voltage V:
Two ways to write W
W=
1
CV 2
2
Capacitor Variables
• The total work to charge capacitor to Q equals the energy U
stored in the capacitor:
Q
1
1 Q2
U = ! qdq =
C0
2 C
• In terms of the voltage V:
1
U = CV 2
2
You can do one of two things to a capacitor :
• hook it up to a battery
specify V and Q follows
• put some charge on it
specify Q and V follows
Q = CV
Q
V=
C
7
Example (I)
• Suppose the capacitor shown here is charged
to Q. The battery is then disconnected.
A
++++
d -----
• Now suppose the plates are pulled further apart to a final
separation d1.
• How do the quantities Q, C, E, V, U change?
•
•
•
•
•
Q:
C:
E:
V:
U:
remains the same.. no way for charge to leave.
decreases.. capacitance depends on geometry
remains the same... depends only on charge density
increases.. since C ↓, but Q remains same (or d ↑ but E the same)
increases.. add energy to system by separating
• How much do these quantities change?.. See board.
Answers:
C1 =
d
C
d1
V1 =
d1
V
d
U1 =
d1
U
d
Example (II)
• Suppose the battery (V) is kept
attached to the capacitor.
A
++++
d
-----
V
• Again pull the plates apart from d to d1.
• Now what changes?
•
•
•
•
•
C:
V:
Q:
E:
U:
decreases (capacitance depends only on geometry)
must stay the same - the battery forces it to be V
must decrease, Q=CV charge flows off the plate
must decrease ( E = V , E =
D
must decrease (U = 1 CV 2)
2
!
)
E0
• How much do these quantities change?.. See board.
Answers:
C1 =
d
C
d1
E1 =
d
E
d1
U1 =
d
U
d1
8
Where is the Energy stored?
• Claim: energy is stored in the electric field itself.
• Consider the example of a constant field generated by a parallel
plate capacitor:
-Q
-------- ------
++++++++ +++++++
+Q
1 Q2 1 Q2
U=
=
2 C 2 ( A! 0 / d)
• The electric field is given by:
E=
!
Q
=
"0 "0 A
⇒
1
U = ! 0 E 2 Ad
2
• The energy density u in the field is given by:
u=
U
U
1
=
= !0 E 2
volume Ad 2
Units:
J
m3
Examples (today or next lecture): 24.42, 24.48,…
9