Chapter 24: Capacitance and Dielectrics
Capacitor: two conductors (separated by an insulator)
usually oppositely charged
a
+Q
b
-Q
Vab proportional to charge Q
C = Q/ Vab (defines capacitance)
units: 1F = 1 C/V
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The parallel plate capacitor
σ
Q A
E=
=
ε0
ε0
+Q
d
-Q
A
A
Q
Vab = Ed =
d
Aε 0
Q
A
C≡
= ε0
Vab
d
Capacitance does not depend upon Q, V!
=> C depends upon geometric factors only
How big is 1 Farad? (parallel plate example)
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Typical Capacitances ~ µF, nF, pF
Example: A parallel plate capacitor has plates 2.00 m2 in area, separated by a distance
of 5.00 mm. A potential difference of 10,000 V is applied across the capacitor.
Determine
the capacitance
the charge on each plate
the magnitude of the electric field in the region between the plates.
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A long cylindrical capacitor
rb
λ
Vab =
ln
2π ε 0 ra
Q= λL
L
Q 2π ε 0 L
C=
=
V ln rb ra
C
2π ε 0
=
L ln rb ra
rb
ra
coax :C ≈ 70 pF m
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A long cylindrical capacitor, small distance between cylinder walls
2π ε 0 L
C=
ln rb ra
rb = ra + d rb ≈ ra = R > > d
C=
[
2π ε 0 L
ln ( ra + d) ra
2
]
2π ε 0 L
=

d
ln 1+
 R
3
x
x
ln(1+ x) = x −
+

2
3
2π ε 0 L 2π RLε 0 A
C≈
=
= ε0
d R
d
d

L
rb
ra
Capacitor looks approximately like parallel plates, in appropriate limit.
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Capacitors in circuits
symbols
analysis follow from
conservation of energy (in terms of electric potential)
conservation of charge
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Q1 = Q2 = Q
V1 + V2 = V
Capacitors in series
a
+Q1
C1
V=Vab
C2
b
c
−Q1
+Q2
−Q2
Q1 = C1V1 Q2 = C2V2
1 V V1 + V2 V1 V2
≡ =
=
+
Ceq Q
Q
Q Q
=
V1 V2
+
Q1 Q2
1
1 1
=
+
Ceq C1 C2
Q1 = Q2 = Q
V1 + V2 = V
A 3 µF capacitor and a 6 µF capacitor are connected in series across an 18 V
battery. Determine the equivalent capacitance, the charge on each capacitor and
the potential difference across each capacitor.
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Capacitors in parallel
a
C1
V=Vab
+Q1 C2
+Q2
−Q1
−Q2
Q1 + Q2 = Q
V1 = V2 = V
Q1 = C1V1 Q2 = C2V2
Ceq ≡
b
Q Q1 + Q2 Q1 Q2
=
=
+
V
V
V V
Q1 Q2
=
+
V1 V2
Ceq = C1 + C2
Q1 + Q2 = Q
V1 = V2 = V
A 3 µF capacitor and a 6 µF capacitor are connected in parallel across an 18 V
battery. Determine the equivalent capacitance, the potential difference across each
capacitor and the charge on each capacitor.
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Combinations of combinations can be analyzed piecewise
C2
C1
C3
Some configurations are not combinations that can be treated as
combinations that can be analyzed as serial/parallel
C1
C2
C5
C3
C4
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Energy stored in a capacitor
When charged: Q = CV
Charging q = Cv
q
C
q= Q
q Q2
W = ∫ dq =
= U
C 2C
q= 0
dW = dq v = dq
dq
dq
q
-q
-dq
v = q/C
q
-q
dq
Q2 1
1 2
U=
= QV = CV
2C 2
2
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Electric Field Energy
Uniform field: parallel plate capacitor
+
+
−
−
−
−
+
−
+
+
−
−
+
−
+
+
−
−
+
−
+
+
1 2
Q = CV U = CV
2
ε 0A
C=
volume= Ad V = Ed
d
u ≡ U / volume= energy density
1ε 0A
=
(Ed)2 /(Ad)
2 d
1
u = ε 0 E2
2
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In the circuit shown V = 48V, C1 = 9µF, C2 = 4µF and
C3 = 8µF.
(a)Determine the equivalent capacitance of the
circuit,
(b) determine the energy stored in the combination
by calculating the energy stored in the equivalent
capacitance,
(c) calculate the charge on and potential difference
across each capacitor and
(d) calculate the energy stored in each individual
physical capacitor.
C1
V
C2
C3
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Dielectrics: insulating materials with other interesting properties
In parallel plate capacitors
For a charged, isolated capacitor
−Q
+Q
−Q
+Q
V0
V
potential difference decreases
same charge => capacitance increases
C = Q/V > C0 = Q/V0
Dielectric Constant: K = C/C0
material property
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Effect of dielectric on Electric field
parallel plates, constant charge
Q = CV = C0V0 => V = V0 /V (reduced)
=> E = E0/K
Material is polarized
Effective surface charge distribution
σ
σ net σ − σ i

1
E0 =
E=
=
σ i = σ  1− 

ε0
ε0
ε0
K
ε = Kε 0
permittivity of dielectric
+σ
−σi
−σ
+σi
σ
E=
ε
A
A
C = Kε 0 = ε
d
d
1
1 2
2
u = Kε 0 E = ε E
2
2
+σnet =σ −σi
−σnet =−(σ −σi)
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Example 25-8:
Take a parallel plate capacitor whose plates have an area of 2000 cm2 and are
separated by a distance of 1cm. The capacitor is charged to an initial voltage of 3 kV
and then disconnected from the charging source. An insulating material is placed
between the plates, completely filling the space, resulting in a decrease in the
capacitors voltage to 1kV. Determine the original and new capacitance, the charge
on the capacitor, the dielectric constant of the material, the permittivity of the
dielectric, the original and new electric fields, the energy stored in the capacitor with
and without the dielectric.
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−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
How does an insulating dielectric material reduce electric fields
by producing effective surface charge densities?
Reorientation of polar molecules
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
Induced polarization of non-polar molecules
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
Dielectric Breakdown: breaking of molecular bonds/ionization of
molecules.
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Polarization (approximately) proportional to applied Electric Field
beyond linear approximation: nonlinear optics...
Dielectric materials and Gauss’s Law
  Qenclosed
∫ KE ⋅ dA = ε 0
 
 
∫ ε E ⋅ dA = Qenclosed = ∫ D ⋅ dA
Qenclosed = enclosed free charge


D = ε E = Electric Displacement (trivia)
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