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Lecture 24 - Capacitance
• Review
• Spherical Capacitor
• Capacitors in Parallel
• Capacitors in Series
• Storing Energy in the Electric Field
• Other Expressions for the Energy
• Energy Density
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Lecture 24: Capacitance and Energy Storage, Study notes of Physics
The concept of capacitance, the relationship between capacitance and geometry, and the calculation of capacitance for parallel plate, cylindrical, and spherical capacitors. It also discusses capacitors in parallel and series, and the energy stored in the electric field. Problem-solving examples and exercises.
Typology: Study notes
Pre 2010
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Lecture 24 - Capacitance
- Review • Spherical Capacitor • Capacitors in Parallel • Capacitors in Series • Storing Energy in the Electric Field • Other Expressions for the Energy • Energy Density
Summary
A Capacitor is an object with two spatially separated conductingsurfaces.
The definition of the capacitance of such an object is:
Q V
C
≡
The capacitance depends on the geometry :
d
A
+ + + + - - - - -
Parallel Plates
A d
C
∝
a b
L
r
+
Q
-
Q
Cylindrical
)
/
ln(
a
b L
C
∝
a
b
+
Q
-
Q
Spherical
a
b
ab
C
−
∝
Problem 25-
The plates of a spherical capacitor have radii 38.0 mm and 40.0 mm. (a)
Calculate the capacitance.
(b)
What must be the plate area of a
parallel-plate capacitor with the same plate separation and capacitance?
Capacitors in Parallel
Equivalent capacitor
: single capacitor
that has the same capacitance as acombination
Capacitors in parallel
when a potential
difference that is applied across theircombination results in the same potentialdifference across each capacitor
Figure 25-
(
)
∑
=
=
=
=
- = + + =
=
=
=
n j
j
eq eq
C
C
C
C
C
V
q
C
V C C C q q q q
V C q V C q V C q
1
3
2
1
3 2 1 3 2 1
3 3 2 2 1 1
Capacitors in Series
Capacitors in series
when a potential
difference that is applied across theircombination is the sum of the resultingpotential differences across each capacitor
Equivalent capacitor
: single capacitor
that has the same capacitance as acombination
Figure 25-
∑
=
=
=
=
⎞ ⎟⎟ ⎠
⎛ ⎜⎜ ⎝
- = + + =
=
=
=
n j
j
eq eq
C
C
C
C
C
q V
C
C C C q V V V V
C
q
V
C
q
V
C
q
V
1
3
2
1
3 2 1 3 2 1
3 3 2 2 1 1
1
1
1
1
1
1
1
1
1
Checkpoint 25-
A battery of potential V stores charge
q
on a combination of two
identical capacitors. What are the potential difference and the chargeon either capacitor if the capacitors are
(a)
in parallel and
(b)
in
series?
(a) V
1
= V
2
= V
q
1
= q
= q/
( q
1
- q
= q )
(b) V
1
= V
2
= V/
( V
1
- V
2
=V )
q
1
= q
= q
Example
(a)
V
= (2/3)
V
0
(b)
V
=
V
0
(c)
V
= (3/2)
V
0
What is the relationship between
V
0
and
V
in the
systems shown below?
d
(Area
A
)
V
0
+
Q
-
Q
conductor
(Area
A
)
V
+
Q
-
Q
d
/ d
/
3B
The arrangement on the right is equivalent to capacitors (each withseparation =
d
/3) in SERIES!!
C
C
eq
1 2
=
conductor
d
/
(Area
A
)
V
+
Q
-
Q
d
/
d
/
+
Q
-
Q
d
/
≡
(
)
0
0
0
3 2
(^32)
3 /
1 2
C
d A
A d
C
eq
=
=
=
ε
ε
(
)
0
0
2 3
2 /
3
V
C
Q
C
Q
V
eq
=
=
=
14
Sample Problem 25-
(a) Find the equivalent capacitance for the combination of capacitancesshown in Fig. 26-9a, across which the potential difference V is applied.Assume
C
1
= 12.
μ
F, C
2
= 5.
μ
F,
and
C
= 4.
μ
F.
F F F C C C
μ
μ
μ
- 7 3. 5 0
.
12
2
1
12
=
=
=
1
3
12
123
28 . 0
5 . 4
1
3 .
17
1
1
1
1
− = + = + =
F F F C C C
μ
μ
μ
F
F
C
57 .
3
)
28 .
0
/(
1
123
=
=
(b) The potential difference that is applied to the input terminal in Fig.26-9 is V = 12.5 V. What is the charge on C
1
? C V F V C q
μ
μ
6 .
44
)
5 .
12
)(
57 .
3 (
123
123
=
=
=
V
C F
q C
V
q
q
58 .
2
3 .
17
6 .
44
12 12
12
123
12
=
=
=
⇒
=
μ μ
C
V
F
V
C
V
C
q
μ
μ
0 .
31
)
58 .
2
)(
0 .
12 (
12
1
1
1
1
=
=
=
=
=
Energy Stored in an Electric Field
- Energy stored equals the work done by an
external agent to charge the capacitor
C
q
U
C
q
dq
q
C
dW
W
dq
q C
dq
V
dW
q
2
2
0
∫
∫
Where is the Energy Stored?
Claim: energy is stored in the electric field itself.Think of the energy needed to charge the capacitoras being the energy needed to create the field.
The electric field is given by:
A
Q
E
0
0
ε
σ ε
=
=
⇒
2
0
U
E
Ad
ε
The energy density
u
in the field is given
by:
2
0
E
W Ad
volume
W
u
ε
3
m
J
Units:
This is the energydensity,
u
, of the
electric field….
To calculate the energy density in the field, first consider theconstant field generated by a parallel plate capacitor, where
2
2
1
1
Q
Q
U
=
=
0
2
2 (
/
)
C
A
d
ε
++++++++ +++++++
+
Q
- - - - - - - - - - - - - - -
Q
Energy Density
- Energy stored in the volume between the
plates
2
0
2
0
0
2
1 2
1 2
2
E
u
d V
u
d
A
C
Ad
CV
U Ad
VOL
U
u
ε
ε
ε
=
⎞ ⎟ ⎠
⎛ ⎜ ⎝
=
⇒
=
=
=
=
Energy Density
Example
(and another exercise for the student!)
Consider
E
-field between surfaces of cylindrical
capacitor:
Calculate the energy in the field of the capacitor byintegrating the above energy density over the volume ofthe space between cylinders.
is general and is not restricted to the special case of theconstant field in a parallel plate capacitor.
Claim: the expression for the energy density of theelectrostatic field
2
0
1 2
E
u
ε
=
2
1 2
CV
W
=
2
2
0
0
1
1
2
.
2
2
U
E dV
E
rdrdl
etc
ε
ε
π
=
=
=
∫
∫ ∫
Compare this value with what you expect from thegeneral expression: