1
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: V
QC =
Unit: farad (F): 1 F = 1 C / 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
0
2εQ
AE = , 02
2εA
QE = , d
A
QEdV
0ε==
d
A
V
QC 0ε==
2
Cylindrical Capacitor
0
2ε
π QrLE = -->
02 επrL
QER =
=−=−= ∫ a
b
L
Qdr
rL
QVVV
a
b
ba ln22 00 πεεπ
==
a
bL
V
QC
ln
2 0πε
Spherical Capacitor
0
24ε
π QEr = ,
−=−= ∫ ba
Qdr
r
QV
a
b
1144 0
20 πεπε
ab
ab
V
QC
−== 04πε
Self-Capacitance
The potential of a spherical conductor of radius R carrying a charge Q is R
kQV = .
The self-capacitance of a spherical conductor is:
Rk
R
R
kQQ
V
QC 04πε====
26.3 Combination of Capacitors
Capacitors Connected in Parallel
Obtain V and Q to calculate C.
VVV == 21 & 111 /VQC =
3
( )VCCVCVCQQQ 21221121 +=+=+=
21/ CCVQC +==
Capacitors connected in parallel:
...4321 ++++= CCCCCeq
Capacitors connected in series
QQQ == 21 & 111 /VQC =
+=+=+=
212
2
1
121
11
CCQ
C
Q
C
QVVV
21
111
CCV
QC
+==
series --> sum voltage & parallel --> sum charges
Capacitors connected in series:
...11111
4321
++++=CCCCC
Example: Two capacitors are removed from the battery and
carefully connected from each other.
21 VV = & 1
11 V
QC =
2
2
1
1
C
Q
C
Q = & CQQ µ9621 =+
Capacitors in Series and in Parallel
321
111CCCC
++
=
Q = ? V = ?
213 QQQQ +== , 31 VVV += , 21 VV =
2
2
1
1
C
Q
C
Q = & QQQ =+ 21 --> QCC
CQ
21
11 +
= & QCC
CQ
21
22 +
=
C1
C2
C3
4
QC
QCCC
Q
C
QVVV
3213
3
1
131
11 ++
=+=+= -->
321
111
CCCV
QC
++
==
26.4 Energy Stored in a Charged
Capacitor
dqC
qVdqdU ==
22
0 21
21
2CVQV
C
Qdq
C
qdUU
Q
===== ∫∫
Electrostatic Field Energy (derived from energy
stored in a capacitor)
d
A
V
QC 0ε== , EdV =
( ) VEAdEEdd
ACVU
=
=== 20
20
202
21
21
21
21 εεε
Electrostatic Energy Density: 202
1E
V
Uue ε== (energy per unit volume)
Example: Calculate the energy stored in the conductor
carrying a charge Q.
Rr < : 0=E
Rr > : 2r
kQE =
( ) ( )drrr
QkdrrEdVudU e
24
22
022
0 421
421 πεπε
=
==
QVR
kQQdr
rQkdUU
RR 21
211
22
220 ==== ∫∫
∞∞
πε
+
+
+
-
-
- +
+Q R
5
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 ( 1>κ ) that is characteristic of the
dielectric.
If the dielectric field is 0E before the dielectric slab is inserted, after the dielectric
slab is inserted between the plates the field is
κ0E
E = --> the potential is κκ
00 VdEEdV ===
If 0
0 V
QC = , the capacitor is 0
0 /' C
V
Q
V
QC κ
κ=== .
The capacitance of a parallel-plate capacitor filled with a dielectric of constant κ is
d
A
d
A
d
A
dE
A
V
A
V
QC
εκεκ
εσ
σκσκ
σ ====== 0
0
00 / --> 0κεε = is called the
permittivity of the dielectric.
Vacuum: κv = 1
the Dielectric: κ > 1
material: κm > 0
6
Energy Stored in The Presence of a Dielectric
The energy stored in a capacitor is:
VdqdU = --> dqC
qdU = --> ∫∫ =
QU
dqC
qdU
00
--> 22
21
2CV
C
QU ==
The energy of a capacitor with the dielectric is
( ) ( ) ( )AdEEdd
AEd
d
ACVU
==== 2222
21
21
21
21 εεε
20
2
21
21
EEue κεε ==
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 2=κ 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) d
AC 0ε= , (b)
d
AC
ε=
(c) series connection:
4/3
1
4/
1
4/3
1
4/
11
000
d
A
d
A
d
A
d
AC κεεεε +=+=
Example: The parallel plates of a given capacitor are square with 2aA = and
separation distance d. If the plates are maintained at a constant potential V and a
Vacuum: κv = 1
the Dielectric: κ > 1
material: κm > 0
- + E0
E
7
square of dielectric slab of constant κ , area 2aA = , thickness d is inserted between the capacitor plates to a distance x as shown in the following figure. Let 0σ
be the free charge density at the conductor-air surface. (a) Calculate the free charge
density κσ 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?
(a) In air: 0
0
εσ=E & dV
0
0
εσ= , in dielectric:
0εσK
E K= & dK
V K
0εσ=
� 0σσ KK =
(b) ( ) ( )( )xKa
d
a
d
xaa
d
axKCCC 1000
21 −+=−+=+= εεε
(c) 2
21
CVU = �
( )121
21
21 02222 −=+
−=+
−= Kd
aV
dx
dCV
dx
dCV
dx
dQV
dx
dCVF
ε
The first term is due to charge redistribution and the second is due to the additional
charges supplied by the constant voltage.
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 ( )
L
xLt − while that of the
k2 material is of L
tx
For a short stripe of dx, ( )
( ) LxLt
Wdx
d
AC
/01
1
11 −
== εκε,
( )Ltx
Wdx
d
AC
/02
1
12
εκε ==
The series connected capacitance 21
111
CCC+= ,
−+
=
L
xt
dxWC
121
0
111κκκ
ε
The total parallel connected capacitance
−
=
−+
= ∫2
1
12
0
0
121
0 ln11111 κ
κ
κκ
ε
κκκ
ε
t
LW
L
xt
dxWC
L
total
k2 k1 L W
8
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.
( ) EpEdqEqrEqrrrrrrrrrr ×=×=−×+×= −+τ
θτ sinpEEp =×=rrv
The potential energy: ( ) θθθτ dpEdFdxdU sin−−=−=−=
� ( )if
f
i
pEdpEU θθθθθ
θ
coscossin −−== ∫
EpUrr ⋅−=
26.7 An Atomic Description of
Dielectrics
- +
-q
+q
E
9
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
charge –e. Show that when such an atom is placed in a uniform external field Er
, the
induced dipole moment is proportional to Er
; that is, Eprr α= , where α is called
Qtotal = 0
Qtotal < 0 Qtotal > 0
bound charge
E0 E
bound charge
10
the polarizability.
rR
e
r
Re
r
A
QE efrominside 3
002
33
0__ 44
3
434
πεεπ
ππ
ε−=
−
==−
304 R
erEerp
πεαα === --> 3
04 Rπεα =
Magnitude of The Bound Charge
0εσ b
bE = & 0
0 εσ fE =
bEEE
E −== 00
κ --> 0
11 EEb
−=κ
--> fb σκ
σ
−= 11
0
1 σσκ
σσσ ==−= fbfeffective � 0κσσ =f
+
+
+
+
-
-
-
-
+
+
+
+
-
-
-
-
-
-
+
+