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Physics 202, Lecture 6Today’s Topics
Capacitance (Ch. 24.1-24.3)Calculating capacitanceCombinations of capacitors: series and parallel
Energy stored in capacitors (24.4)
Capacitance
Capacitance:
stores electrical energy (by storing charge)
Capacitor: two (spatially separated) conductors, charged to +Q and -Q, with constant potential difference ΔV
C !Q
"V
Units: Farad (F)1 F= 1 C/VC
a bOur first exampleof a circuit element:
2
Charging a Capacitor
UnchargedCharging
flow
of e
-
ΔV=V+-V-
Battery: source of ΔV
CC
Text problems: 24.2, 24.6, 24.8
Charge flows until Q=CΔV
Charging/discharging capacitor:
Calculating CapacitanceCapacitance is independent of charge, voltage oncapacitor: it depends on geometry of the conductors.
Examples: Parallel conducting platesConcentric spherical conductors
Concentric cylindrical conductors
Procedure: determine as a function of !V Q
Simple example: capacitance of sphere (imagine second conductor at infinity)
!V = V =kQ
RC =
Q
!V=
Q
kQ R=R
k= 4"#
0R
Text: 24.13
3
Example: Parallel Plate CapacitorConsider two metallic parallel plates with area A, separation d:
d
A
- - - - -+ + + +!
+Q
-Q
Step 1. Use Gauss’s law to get E: A ! d
!"
! = Q A
!A
!Eid!A = E
in!A ="" # !A $
0
Ein= ! "
0
!E
Step 2. Get potential difference:
!V = "!Eid!l =#
$d
%0
=Qd
%0A
Step 3. Get Capacitance: C =Q
!V="0A
d
Parallel Plate Capacitor
Realistic case
C =!0A
d
fringe field
Text: 24.15,…
4
Spherical Capacitor
C =ab
k(b ! a)
-Q
+Q
Consider two concentric conducting spherical shells, radii a,ba < b
Step 1. Get electric field:
Gauss’s Law: ba
!E =
kQ
r2r̂
!V = "!Eid#!l = "
kQ
r2dr
b
a
# = kQ1
a"1
b
$%&
'()=kQ(b " a)
ab
(a < r < b)
Step 2. Get potential difference:
Step 3:
C =!
2k ln(b / a)
Cylindrical Capacitor
Consider two concentric cylindrical conducting shells, radii a,b
a < b
Step 1. Get electric field:
Gauss’s Law:
!E =
2k!
rr̂
!V = "!Eid#!l = "
2k$
rdr
b
a
# = 2k$ ln b a( )
(a < r < b)
Step 2. Get potential difference:
Step 3:
Q = !!
5
Calculating Capacitance• Given 4 concentric cylinders
of radii a,b,c,d and charges+Q, -Q, +Q, -Q
• Question: What is thecapacitance between a and d?
• Note: E=0 between b and c! WHY??• A cylinder of radius r1: b < r1< c
encloses zero charge!
-Q+Q
-Q+Qa
b
c
d
Note: result of 2 cylindrical capacitors“in series” (next slides)
Vad =Q
2!"0rL
a
b
# dr + 0 +Q
2!"0rL
c
d
# dr =Q
2!"0Lln
b
a
$%&
'()+ ln
d
c
$%&
'()
$%&
'()
C =Q
Vad=
2!"0L
lnb
a
#$%
&'(+ ln
d
c
#$%
&'(
#$%
&'(
Combinations of Capacitors
ΔV1
ΔV2
ΔV
C1
C2
Q1
Q2
ΔV
ΔV
CP
Q=Q1+Q2
Equivalent Capacitance CP=Q/ΔV C = C1+ C2
C1ΔV1 =Q1C2ΔV2 =Q2
ΔV1 =ΔV2 =ΔV(why?)
CP= C1+C2+C3+…
effectively
Parallel combination: ΔV same
CP always > Ci!
6
Combinations of Capacitors
effectivelyΔV1 ΔV2
+Q1
ΔV=ΔV1+ΔV2
Q
ΔV
Equivalent CapacitanceCS=Q/ΔV 1/CS = 1/C1+1/C2
C1C2
CS
Charge conservation: Q1=Q2(=Q)
-Q1 +Q2-Q2
1/CS= 1/C1+1/C2+1/C3+…
C1ΔV1 =QC2ΔV2 =Q
Series combination:
Cs=
C1C2
C1+ C
2
CS always < Ci!
Example: Connection of Charged Capacitors
Two capacitors, C1=1µF and C2=2µF are initiallycharged 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 capacitorafter the connection?
C1
C2
- +
- +
More examples: 24.29-31
7
Energy of a Capacitor• How much energy is stored in a charged capacitor?
– Calculate the work provided (usually by a battery) to charge acapacitor to +/- Q:
Incremental work dW needed to add charge dq to capacitor at voltage V:- +
• In terms of the voltage V:
• The total work W to charge to Q is then given by:
Two ways to write W
dW = V (q)idq =q
C
!"#
$%&idq
W =1
Cqdq =
0
Q
!1
2
Q2
C
W =1
2CV
2
Capacitor Variables
• In terms of the voltage V:
• The total work to charge capacitor to Q equals the energy Ustored in the capacitor:
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
U =1
2CV
2
U =1
Cqdq =
0
Q
!1
2
Q2
C
Q = CV
V =Q
C
8
Example (I)
d
A
- - - - -+ + + +• Suppose the capacitor shown here is charged
to Q. The battery is then disconnected.
• Now suppose the plates are pulled further apart to a finalseparation d1.
• How do the quantities Q, C, E, V, U change?
• How much do these quantities change?.. See board.
• Q:• C:• E:• V:• U:
remains the same.. no way for charge to leave.
increases.. add energy to system by separating
decreases.. capacitance depends on geometry
increases.. since C ↓, but Q remains same (or d ↑ but E the same)remains the same... depends only on charge density
Answers: C1=d
d1
C V1=d1
dV
U1=
d1
dU
• Suppose the battery (V) is keptattached to the capacitor.
• 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 Vmust decrease, Q=CV charge flows off the plate
d
A
- - - - -+ + + +V
• How much do these quantities change?.. See board.
Answers:
U1=
d
d1
U
C1=
d
d1
C
E1=
d
d1
E
must decrease ( , )
E =!
E0
E =V
D
must decrease ( ) U =
1
2CV
2
Example (II)
9
Where is the Energy stored?• Claim: energy is stored in the electric field itself.
• The electric field is given by:
⇒• The energy density u in the field is given by:
Units:
• Consider the example of a constant field generated by a parallelplate capacitor:
++++++++ +++++++
- - - - - - - - - - - - - -- Q
+Q
u =U
volume=
U
Ad=
1
2!
0E
2
E =!
"0
=Q
"0A
U =1
2!
0E
2Ad
U =1
2
Q2
C=
1
2
Q2
( A!0
/ d)
J
m3
Examples (today or next lecture): 24.42, 24.48,…