Capacitor and Capacitance
Capacitor (device):
• Two oppositely charged conductors separated by an insulator.
~ and a potential
• The charges +Q and −Q on conductors generate an electric field E
difference V (voltage).
• Only one conductor may be present. Then the relevant potential difference is between the
conductor and a point at infinity.
Capacitance (device property):
Q
V
• SI unit: 1F = 1C/V
• Definition: C =
(one Farad)
14/9/2015
[tsl103 – 1/21]
Parallel-Plate Capacitor
• A: area of each plate
• d: distance between plates
• Q: magnitude of charge on inside surface of each plate
• Charge per unit area (magnitude) on each plate: σ =
• Uniform electric field between plates:
Q
σ
=
E=
ǫ0
ǫ0 A
Q
A
−Q
+Q
• Voltage between plates:
V ≡ V+ − V− = Ed =
Qd
ǫ0 A
E
• Capacitance for parallel-plate geometry:
Q
ǫ0 A
C≡
=
V
d
V+
d
V−
14/9/2015
[tsl104 – 2/21]
Cylindrical Capacitor
Conducting cylinder of radius a and length L surrounded concentrically by conducting cylindrical
shell of inner radius b and equal length.
• Assumption: L ≫ b.
• λ: charge per unit length (magnitude) on each cylinder
• Q = λL: magnitude of charge on each cylinder
• Electric field between cylinders: use Gauss’ law
λL
λ
E[2πrL] =
⇒ E(r) =
ǫ0
2πǫ0 r
• Electric potential between cylinders: use V (a) = 0
Z r
Z r
dr
λ
r
λ
=−
ln
E(r)dr = −
V (r) = −
2πǫ0 a r
2πǫ0
a
a
• Voltage between cylinders:
V ≡ V+ − V−
−Q
+Q
b
b
Q
ln
= V (a) − V (b) =
2πǫ0 L
a
• Capacitance for cylindrical geometry:
2πǫ0 L
Q
=
C≡
V
ln(b/a)
a
E
14/9/2015
[tsl105 – 3/21]
Spherical Capacitor
Conducting sphere of radius a surrounded concentrically
by conducting spherical shell of inner radius b.
• Q: magnitude of charge on each sphere
• Electric field between spheres: use Gauss’ law
Q
Q
⇒ E(r) =
E[4πr2 ] =
ǫ0
4πǫ0 r2
• Electric potential between spheres: use V (a) = 0
»
–
Z r
Z r
1
1
dr
Q
Q
−
=
V (r) = −
E(r)dr = −
2
4πǫ
r
4πǫ
r
a
0
0
a
a
−Q
• Voltage between spheres:
V ≡ V+ − V− = V (a) − V (b) =
Q b−a
4πǫ0 ab
• Capacitance for spherical geometry:
ab
Q
= 4πǫ0
C≡
V
b−a
+Q
b
a
E
14/9/2015
[tsl106 – 4/21]
Energy Stored in Capacitor
Charging a capacitor requires work.
The work done is equal to the potential energy stored in the capacitor.
While charging, V increases linearly with q:
V (q) =
q
.
C
Increment of potential energy:
dU = V dq =
q
dq.
C
Potential energy of charged capacitor:
U=
Z
0
Q
1
V dq =
C
Z
0
Q
1
1
Q2
= CV 2 = QV.
qdq =
2C
2
2
Q: where is the potential energy stored?
A: in the electric field.
14/9/2015
[tsl107 – 5/21]
Energy Density Between Parallel Plates
Energy is stored in the electric field between the plates of a capacitor.
• Capacitance: C =
ǫ0 A
.
d
• Voltage: V = Ed.
1
1
• Potential energy: U = CV 2 = ǫ0 E 2 (Ad).
2
2
• Volume between the plates: Ad.
• Energy density of the electric field: uE =
−Q
+Q
U
1
= ǫ0 E 2
Ad
2
E
A
A
d
14/9/2015
[tsl108 – 6/21]
Integrating Energy Density in Spherical Capacitor
• Electric field: E(r) =
−Q
Q 1
4πǫ0 r2
»
Q
Q b−a
=
• Voltage: V =
4πǫ0 ab
4πǫ0
1
1
−
a
b
–
• Energy stored in capacitor: U =
b
b
a
1
• Energy density: uE (r) = ǫ0 E 2 (r)
2
Z
+Q
E
uE (r)(4πr2 )dr
a
b
Q2
1
1
2
ǫ0
(4πr
)dr
• ⇒ U =
2 r4
2
(4πǫ
)
0
a
»
–
Z b
1
1
1 Q2
1
1
1 Q2
−
QV
dr
=
=
• ⇒ U =
2 4πǫ0 a r2
2 4πǫ0 a
b
2
Z
14/9/2015
[tsl109 – 7/21]
Capacitor Problem (1)
Consider two oppositely charged parallel plates separated by a very small distance d.
What happens when the plates are pulled apart a fraction of d? Will the quantities listed below
increase or decrease in magnitude or stay unchanged?
~ between the plates.
(a) Electric field E
(b) Voltage V across the plates.
(c) Capacitance C of the device.
(d) Energy U stored in the device.
+
+
+
+
+
+
+
+
+
+
+
+
+
d
Ε
− − − − − − − − − − − − −
14/9/2015
[tsl110 – 8/21]
Capacitors Connected in Parallel
Find the equivalent capacitance of two capacitors connected in parallel:
• Charge on capacitors: Q1 + Q2 = Q
• Voltage across capacitors: V1 = V2 = V
• Equivalent capacitance:
Q1 + Q2
Q1
Q2
Q
=
=
+
C≡
V
V
V1
V2
• ⇒ C = C1 + C2
V1 = V
x
V0
C1
−Q 1
+Q 1
V0
V0 + V
−Q 2
V0
C2
+Q2
V2 = V
x
14/9/2015
[tsl111 – 9/21]
Capacitors Connected in Series
Find the equivalent capacitance of two capacitors connected in series:
• Charge on capacitors: Q1 = Q2 = Q
• Voltage across capacitors: V1 + V2 = V
• Equivalent capacitance:
• ⇒
1
V2
V
V1 + V2
V1
+
≡
=
=
C
Q
Q
Q1
Q2
1
1
1
+
=
C
C1
C2
V2
V1
V0
C1
V0
−Q
+Q
−Q
x
C2
+Q
V0 + V
.
14/9/2015
[tsl112 – 10/21]
Capacitor Problem (2)
Consider two equal capacitors connected in series.
(a) Find the voltages VA − VB , VB − VC , VA − VD .
(b) Find the charge QA on plate A.
(c) Find the electric field E between plates C and D.
2cm
2cm
D
B
24V
A
C
2µF
.
12V
2µ F
14/9/2015
[tsl113 – 11/21]
Capacitor Circuit (1)
Find the equivalent capacitances of the two capacitor networks.
All capacitors have a capacitance of 1µF .
(a)
(b)
14/9/2015
[tsl114 – 12/21]
Capacitor Circuit (2)
Consider the two capacitors connected in parallel.
(a) Which capacitor has the higher voltage?
12V
C1 = 1µF
C 2 = 2 µF
(b) Which capacitor has more charge?
(c) Which capacitor has more energy?
Consider the two capacitors connected in series.
(d) Which capacitor has the higher voltage?
(e) Which capacitor has more charge?
(f) Which capacitor has more energy?
C1 = 1µF
C 2 = 2µF
12V
14/9/2015
[tsl115 – 13/21]
Capacitor Circuit (3)
Connect the three capacitors in such a way that the equivalent capacitance is Ceq = 4µF. Draw
the circuit diagram.
2µF
2µ F
3µ F
4µ F
14/9/2015
[tsl116 – 14/21]
Capacitor Circuit (4)
Connect the three capacitors in such a way that the equivalent capacitance is Ceq = 2µF. Draw
the circuit diagram.
1µ F
3 µF
5µ F
2µ F
14/9/2015
[tsl117 – 15/21]
Capacitor Circuit (5)
Find the equivalent capacitances of the following circuits.
1µ F
2µF
(a)
3µF
1µF
2µ F
1µF
(b)
1µF
3µ F
14/9/2015
[tsl118 – 16/21]
Capacitor Circuit (6)
(a) Name two capacitors from the circuit on the left that are connected in series.
(b) Name two capacitors from the circuit on the right that are connected in parallel.
C1
C1
C3
C2
C2
C4
C5
C3
C4
C6
14/9/2015
[tsl119 – 17/21]
Capacitor Circuit (7)
(a) In the circuit shown the switch is first thrown to A. Find the charge Q0 and the energy UA on
the capacitor C1 once it is charged up.
(b) Then the switch is thrown to B, which charges up the capacitors C2 and C3 . The capacitor
C1 is partially discharged in the process. Find the charges Q1 , Q2 , Q3 on all three capacitors
and the voltages V1 , V2 , V3 across each capacitor once equilibrium has been reached again.
What is the energy UB now stored in the circuit?
A
B
S
C 2 = 3µ F
C1 = 4 µF
V0 = 120V
C3 = 6 µ F
14/9/2015
[tsl120 – 18/21]
Capacitor Circuit (8)
In the circuit shown find the charges Q1 , Q2 , Q3 , Q4 on each capacitor and the voltages
V1 , V2 , V3 , V4 across each capacitor
(a) when the switch S is open,
(b) when the switch S is closed.
C1 = 1 µF
C 2 = 2 µF
C 3 = 3µ F
S
C4 = 4 µ F
V = 12V
14/9/2015
[tsl122 – 19/21]
Capacitor Circuit (9)
The circuit of capacitors connected to a battery is at equilibrium.
(a) Find the equivalent capacitance Ceq .
(b) Find the total energy U stored in the circuit (excluding the battery).
(c) Find the the charge Q3 on capacitor C3 .
(d) Find the voltage V2 across capacitor C2 .
C1 = 3 µ F
12V
C3 = 4 µF
C2 = 6 µF
14/9/2015
[tsl335 – 20/21]
More Complex Capacitor Circuit
+ −
No two capacitors are in parallel or in series.
Solution requires different strategy:
C1
• zero charge on each conductor
(here color coded),
C2
• zero voltage around any closed loop.
+ −
Specifications: C1 , . . . , Q5 , V .
Five equations for unknowns Q1 , . . . , Q5 :
• Q1 + Q2 − Q4 − Q5 = 0
C
3
C4
+
−
V
C5
+ −
(a) Cm = 1pF, m = 1, . . . , 5 and V = 1V:
• Q3 + Q4 − Q1 = 0
Ceq = 1pF, Q3 = 0,
Q3
Q4
Q5
+
−
=0
•
C5
C3
C4
Q2
Q1
Q3
•
−
−
=0
C2
C1
C3
Q4
Q1
−
=0
• V −
C1
C4
Equivalent capacitance: Ceq =
+ −
Q1 = Q2 = Q4 = Q5 =
1
pC.
2
(b) Cm = m pF, m = 1, . . . , 5 and V = 1V:
Ceq =
Q1 + Q2
V
55
104
159
pF, Q1 =
pC, Q2 =
pC,
71
71
71
Q3 = −
9
64
95
pC, Q4 =
pC, Q5 =
pC.
71
71
71
14/9/2015
[tsl511 – 21/21]