Chapter 26 Capacitance and
Dielectrics
We must do work, qV, to bring a point charge q from far away (V = 0 at infinity) to a
region where other charges are present. The work done is stored in electrostatic field
energy.
26.1 Definition of Capacitance
26.2 Calculating Capacitance
Measure of the capacity to store charge: C =
Unit: farad (F): 1 F = 1 C / V;
Q
V
1 µF = 10-6 F;
1 pF = 10-12 F
The ratio of charge Q to the potential depends on the size and the shape of the
conductor.
Capacitors
A device consisting of two conductors carrying equal but
opposite charges is called a capacitor.
Parallel Plate Capacitor
E2A =
C=
Q
ε0
, E=2
Q
Q
d
, V = Ed =
Aε 0
2 Aε 0
Q ε0 A
=
V
d
1
Cylindrical Capacitor
2πrLE =
Q
ε0
--> E R =
Q
2πrLε 0
a
Q
Q
b
dr =
ln 
2πrLε 0
2πε 0 L  a 
b
V = Va − Vb = − ∫
C=
Q 2πε 0 L
=
V
b
ln 
a
Spherical Capacitor
4πr E =
2
C=
Q
ε0
a
, V = −∫
b
Q
4πε 0 r
2
dr =
Q 1 1
 − 
4πε 0  a b 
Q
ab
= 4πε 0
V
b−a
Self-Capacitance
The potential of a spherical conductor of radius R carrying a charge Q is V =
kQ
.
R
The self-capacitance of a spherical conductor is:
C=
Q
Q
R
=
= = 4πε 0 R
V kQ k
R
26.3 Combination of Capacitors
Capacitors Connected in Parallel
Obtain V and Q to calculate C.
V1 = V2 = V & C1 = Q1 /V1
2
Q = Q1 + Q2 = C1V1 + C2V2 = (C1 + C2 )V
C = Q / V = C1 + C2
Capacitors connected in parallel:
Ceq = C1 + C2 + C3 + C4 + ...
Capacitors connected in series
Q1 = Q2 = Q & C1 = Q1 /V1
V = V1 + V2 =
C=
 1
Q1 Q2
1 

+
= Q +
C1 C 2
C
C
 1
2 
Q
1
=
1
1
V
+
C1 C 2
series --> sum voltage & parallel --> sum charges
Capacitors connected in series:
1 1
1
1
1
=
+
+
+
+ ...
C C1 C 2 C3 C 4
Example: Two capacitors are removed from the battery and
carefully connected from each other.
Q
V1 = V2 & C1 = 1
V1
Q1 Q2
=
& Q1 + Q2 = 96 µC
C1 C 2
Capacitors in Series and in Parallel
1
1
1
=
+
C C1 + C2 C3
C1
C3
C2
Q=?V=?
Q = Q3 = Q1 + Q2 , V = V1 + V3 , V1 = V2
Q1 Q2
C1
C2
=
& Q1 + Q2 = Q --> Q1 =
Q & Q2 =
Q
C1 C2
C1 + C2
C1 + C2
3
Q1 Q3
Q
1
1
Q + Q --> C = =
+
=
V
C1 C3 C1 + C2
C3
V = V1 + V3 =
1
1
1
+
C1 + C2 C3
26.4 Energy Stored in a Charged
Capacitor
dU = Vdq =
+
-
+
+
-
+
q
dq
C
Q
q
Q2 1
1
dq =
= QV = CV 2
C
2C 2
2
0
U = ∫ dU = ∫
Electrostatic Field Energy (derived from energy
stored in a capacitor)
C=
Q ε0 A
=
, V = Ed
V
d
U=
1
1 ε0 A
(Ed )2 =  1 ε 0 E 2  Ad =  1 ε 0 E 2 V
CV 2 =
2
2 d
2

2

U 1
= ε 0 E 2 (energy per unit volume)
V 2
R
Example: Calculate the energy stored in the conductor
Electrostatic Energy Density: ue =
+Q
carrying a charge Q.
r < R: E =0
r > R: E =
kQ
r2
(
)
(
 1 k 2Q 2 
1

dU = ue dV =  ε 0 E 2  4πr 2 dr =  ε 0 4  4πr 2 dr
r 
2

2
∞
U = ∫ dU = 2πε 0 k Q
2
R
∞
2
1
∫r
R
2
dr =
)
1 kQ 1
Q
= QV
2 R
2
4
26.5 Capacitors and Dielectrics
When the space between the two conductors of a capacitor is occupied by a dielectric,
the capacitance is increased by a factor κ ( κ > 1 ) that is characteristic of the
dielectric.
If the dielectric field is E0 before the dielectric slab is inserted, after the dielectric
slab is inserted between the plates the field is
E=
E0
κ
If C0 =
--> the potential is V = Ed =
E0 d
κ
=
V0
κ
Q
Q
Q
, the capacitor is C ' = =
= κC0 .
V0
V V0 / κ
The capacitance of a parallel-plate capacitor filled with a dielectric of constant κ is
C=
Q
Aσ
Aσ
Aσ
Aκε 0 Aε
--> ε = κε 0 is called the
=
=
κ=
κ=
=
σ
V V0 / κ E0 d
d
d
d
ε0
material: κm > 0
permittivity of the dielectric.
Vacuum: κv = 1
the Dielectric: κ > 1
5
Energy Stored in The Presence of a Dielectric
The energy stored in a capacitor is:
q
dU = Vdq --> dU = dq -->
C
U
Q
q
Q2 1
dq --> U =
= CV 2
C
2
C
2
0
∫ dU = ∫
0
The energy of a capacitor with the dielectric is
U=
1
1 Aε
(Ed )2 = 1 Aε (Ed )2 =  1 εE 2 ( Ad )
CV 2 =
2
2 d
2 d
2

ue =
1 2 1
εE = κε 0 E 2
2
2
material: κm > 0
Vacuum: κv = 1
E0
E
-
+
the Dielectric: κ > 1
1.
You lose electric force to separate the charge.
2.
You enlarge the charging capacity as you know the dielectric will breakdown in a
high electric field. (If the same charge --> you lose some electric field,)
3.
You increase the energy per unit volume.
Combination of Capacitors
Example: A parallel-plate capacitor has square plates of edge length 10 cm and a
separation of d = 4 mm. A dielectric slab of constant κ = 2 has dimensions 10 cm X
10 cm X 4 mm. (a) What is the capacitance without the dielectric? (b) What is the
capacitance with the dielectric? (c) What is the capacitance if a dielectric slab with
dimensions 10 cm X 10 cm X 3 mm is inserted into the 4-mm gap?
(a) C =
Aε 0
Aε
, (b) C =
d
d
(c) series connection:
1
1
1
1
1
=
+
=
+
Aε
Aε 0 Aκε 0
C Aε 0
d / 4 3d / 4 d / 4 3d / 4
Example: The parallel plates of a given capacitor are square with A = a 2 and
separation distance d. If the plates are maintained at a constant potential V and a
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square of dielectric slab of constant κ , area A = a 2 , thickness d is inserted
between the capacitor plates to a distance x as shown in the following figure. Let σ 0
be the free charge density at the conductor-air surface. (a) Calculate the free charge
density σ κ at the capacitor-dielectric surface. (b) What is the effective capacitance?
(c) What is the magnitude of the required force to prevent the dielectric slab from
sliding into the plates?
σ0
σ
σ
σ
& V = 0 d , in dielectric: E = K & V = K d
ε0
ε0
Kε 0
Kε 0
σ K = Kσ 0
(a) In air: E =
(b) C = C1 + C2 =
axKε 0 a (a − x )ε 0 aε 0
+
=
(a + (K − 1)x )
d
d
d
1
(c) U = CV 2
2
dQ
dC 1 2 aε 0
 1  dC
 1  dC
(K − 1)
F = − V 2 
+V
= − V 2 
+V 2
= V
dx
dx 2
d
 2  dx
 2  dx
The first term is due to charge redistribution and the second is due to the additional
charges supplied by the constant voltage.
k1
L
W
k2
Example: A parallel plate capacitor with plates of area LW and separation t has the
region between its plates filled with wedges of two dielectric materials. Assume t is
much less than both W and L. (a) Please determine its capacitance.
The thickness of the k1 material decrease as a function of
k2 material is of
t (L − x )
while that of the
L
tx
L
A1ε Wdx (κ1ε 0 )
A ε Wdx (κ 2ε 0 )
=
, C2 = 1 =
d1 t (L − x ) / L
d1
tx / L
1 1
1
Wε 0 dx
The series connected capacitance
= +
, C=
C C1 C2
 1  1 1  x
t  +  −  
 κ1  κ 2 κ1  L 
The total parallel connected capacitance
For a short stripe of dx, C1 =
L
Ctotal = ∫
0
Wε 0dx
 1  1 1 x
t  +  −  
 κ1  κ 2 κ 1  L 
=
κ 
Wε 0 L
ln 1 
 1 1
κ
t  −   2 
 κ 2 κ1 
7
26.6 Electric Dipole in an Electric Field
Inside the material
to make sure that the electric field lines are from the positive
charge to the negative charge
-
+
If the field is uniform, it can rotate the dipole.
r r r r
r r
r
r r
τ = r+ × qE + r− × − qE = qd × E = p × E
(
v
r
+q
)
r
τ = p × E = pE sin θ
E
-q
The potential energy: dU = − Fdx = −τdθ = −(− pE sin θ )dθ
θf
U = pE ∫ sin θdθ = − pE (cos θ f − cos θ i )
θi
r r
U = −p⋅E
26.7 An Atomic Description of
Dielectrics
8
bound charge
Qtotal < 0
Qtotal > 0
Qtotal = 0
E0
E
bound charge
Example: A hydrogen atom consists of a proton nucleus of charge +e and an electron
of charge –e. The charge distribution of the atom is spherically symmetric, so the
atom is nonpolar. Consider a model in which the hydrogen atom consists of a positive
charge +e at the center of a uniformly charged spherical cloud of radius R and total
r
charge –e. Show that when such an atom is placed in a uniform external field E , the
r
r
r
induced dipole moment is proportional to E ; that is, p = αE , where α is called
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the polarizability.
Einside _ from _ −e
4π 3 − e
r
4πR 3
3
Q
3 = −e r
=
=
2
4πr ε 0
4πε 0 R 3
Aε 0
p = er = αE = α
er
4πε 0 R 3
--> α = 4πε 0 R 3
Magnitude of The Bound Charge
σ
σ
Eb = b & E0 = f
ε0
ε0
+
-
+-
+
-
+
+
++
+
E0
 1
 1
E=
= E0 − Eb --> Eb = 1 −  E0 --> σ b = 1 − σ f
κ
 κ
 κ
σ effective = σ f − σ b =
1
κ
σ f = σ0
++-
σ f = κσ 0
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