Chapter 30
Capacitance
Capacitors
A cup can store water
charge
A device that stores
charge (and then energy
in electrostatic field) is
called a capacitor.
The capacitance of an isolated conductor
q
C  ( F ,  F , pF )
V
V 

a

E
q
rˆ
2
40 r
q
a
q
q
q
dr

C   4 0 a
2
40 r
40 a
V
Example What is the
capacitance of the Earth,
viewed as an isolated
conducting sphere of
radius R=6370km?
 4  3.14  (8.85 10
12
)
 (6.37 106 )
 7.1  10 F  710 μF
4
A capacitor consists of two conductors a and b of
arbitrary shape:
These two conductors are called plates (no matter
what their shapes).
Symbolically, a capacitor is represented as:
C stands for the capacitance
of the capacitor.
C
or
C
The
charge
The
charge q:A
q appears
is
always
Remarks
capacitor
on
the capacitor
plates
directly
proportional
to
is said
to
be charged
if
its plates
carry
equal and
the
potential
difference
There is a potential
opposite
charges
+q and
DV
between
the
plates
difference DV
-q. q is not the net charge
between
q capacitor,
Dthe
V  Vplates
V is
on the
which

zero.
q  C DV
capacitance
q
C
DV
Capacitors in Series and Parallel
1. Capacitors connected in Parallel :
DV
C1
q1  C1DV , q2  C2 DV
Ceq DV  q  q1  q2  (C1  C2 )DV
b
a
C2
Question: If we identify the above capacitors connected in
parallel as a single capacitor, DV
Ceqis its Ccapacitance?
what
1  C2
Ceq  i Ci
a
b
Ceq
2. Capacitors connected in Series :
DV1
a
DV2
q
q
q q
C1
DV
b

C2
a
q q
Ceq
q C1  q C2  DV1  DV2  DV  q Ceq
1
1
1


C1 C2
Ceq
1
1
 i
Ceq
Ci
b
The capacitance is a geometrical factor that
depends on the size, shape and separation of the
capacitor plates, as well as the material that
occupies the space between the plates.
The SI unit of
capacitance is farad :
1 farad = 1 F = 1 coulomb/volt
1 F = 10-6 F
1 pF = 10-12 F
Calculating the capacitance
Procedure:
1. Suppose that the capacitor is charged, with ±q on
the two plates respectively.
2. Find the electric field E in the region between the
plates.
3. Evaluate the potential difference between the
positive and negative plates, by using the

formula:
 
DV  V  V   E  ds

4.The expected capacitance is then:
C  q DV
A Parallel-plate Capacitor :
E  E  E

q
 
0 0 A
qd
DV  Ed 
0 A
q 0 A
C

DV d
A Cylindrical Capacitor :
 (q L)rˆ
E
, ar b
20 r

 
DV   E  ds


b
a
qdr
rˆ  rˆ
20 rL
b

ln  
20 L  a 
q
L
C  q DV  20
ln b a 
q
b
q
a
L
The capacitor has length L,
and L>>a, b.
A Spherical Capacitor :

E
qrˆ
40 r
2
, ar b

 
DV   E  ds


b
a
qdr
rˆ  rˆ
2
40 r
q 1 1

  
40  a b 
ab
C  q DV  40
ba
Capacitor with Dielectric
q
We now consider the effect of filling the
interior of a capacitor with a dielectric material
-q
The effect of the dielectric material is to
reduce the strength of the electric field in its
interior from the initial E0 in vacuum to E =E0/ke.
C  0 A / d
E0  q  0 A
E  E0 /  e  q  e 0 A
DV  Ed  qd  e 0 A
C '  q / DV   e 0 A / d
C '   eC
Capacitor with Dielectric
A⊕ ⊕ ⊕
d
⊕
⊕
⊕ q
 e1
e2
q
E  q 0 A
E1  E /  e1  q /  0 e1 A
E2  E /  e 2  q /  0 e 2 A
qd 1
1
DV  qd / 2 0 e1 A  qd / 2 0 e 2 A 
(  )
2 0 A  e1  e 2
q 2 0 A  e1 e 2
C

DV
d  e1   e 2
Capacitor with Dielectric
A⊕ ⊕ ⊕
q
d
 e1
-q
⊕
⊕ ⊕ q
q’
e2
-q’
q
E  2q  0 A
E1  E /  e1  2q /  0 e1 A
E2  E /  e 2  2q '/  0 e 2 A
DV1  2qd /  0 e1 A DV2  2q ' d /  0 e 2 A C  C  C
1
2
q  0 e1 A
0 A
q '  0 e 2 A
C1 


( e1   e 2 )
C2 

DV1
2d
2d
DV2
2d
Energy storage
During the
time
interval
[t, t+dt],
if an has
additional
Suppose
thatnext
at the
instant
t, the
capacitor
been
charge
is added
the increase
chargeddq'
with
chargeon
q', the
the plates,
voltagethen
between
its
plates
DV´.
of
the is
electrostatic
energy is,
dq′
q′
ΔV´
DV   q C
dU  dqDV   qdq C
If the process is continued until a total charge q
has been transferred, the total potential energy is:
U
q
2
1
1
q
2
 C (DV )  qDV
U   dU   q' dq' /C 
2
2C 2
0
0
Why do we say that the energy is stored in the
electric field between the capacitor plates?
Take the parallel-plate capacitor as an example.
C
C  0 A d C' 
2
charged with q, then:
E  q 0 A
q
q
q
q
q 2d
U

U   2U
2C 2 0 A
double the volume
2
2
1  q 
1
 Ad   0 E 2 Ω
  0 
2  0 A 
2
double the energy
q
Why do we say that the energy is stored in the
electric field between the capacitor plates?
Take the parallel-plate capacitor as an example.
C
C  0 A d C' 
2
charged with q, then:
E  q 0 A
2
2
q
q d
U

U   2U
2C 2 0 A
2
1  q 
1
 Ad   0 E 2 Ω
  0 
2  0 A 
2
q
q
energy density
U 1
2
u   0E
Ω 2
A' 2 A
Why do we say that the energy is stored in the
electric field between the capacitor plates?
Take the parallel-plate capacitor as an example.
C
q
C  0 A d C' 
2
charged with q, then:
q
E  q 0 A
2
2
q
q d
U

U   2U
2C 2 0 A
2
1  q 
1
 Ad   0 E 2 Ω
  0 
2  0 A 
2
q
2q
q
 2q
energy density
U 1
2
u   0E
Ω 2
UU' ' 2U
2U?/ ?2 ?
A' 2 A
U
Energy storage
Charge storage
Electric Potential
Charge storage
Electric field
Energy storage
Energy storage
1 2 1
1
2
U
q  C (DV )  qDV
2C
2
2
1
U   0 E 2Ω
2
Example
An isolated conducting sphere of
radius R carries a charge q.
How much energy is stored?
1 N
1
U   qiVi   Vdq
2 i 1
2 V
2
1
q
q
1 q
1 q
2

U 
dq '

ds

4

R

80 R
2 4 0 R
2 40 R
2 4 0 R
q
q'
0
4 0 R
U 
dq ' 
2
q'
q
8 0 R 0
U  q 2 2C  q 2 2(40 R) 

q2
80 R
q2
80 R
Example
An isolated conducting sphere of
radius R carries a charge q.
How much energy is stored?
1
U 1
2
2


E
d
U

ud

u   0E
0

2
Ω 2
E (r )  q 40 r 2 , R  r  
2

q
1
q2
2
U  0 
4

r
dr 
2
2
4
80 R
2 R 16  0 r
What is the radius b of an imaginary spherical surface
such as one thirds of the stored energy lie within it?
b
U 3  q 240 R   u (r )4r 2dr
2
b
R

  q

b
  q 2 32 2 0 r 4 4r 2dr
R
b
R
U
q
2
80 R
2

80 r 2 dr
 q 2 80  1 R   1 b 
1
1 1
 
3R R b
b  3R 2
qe q
DV
C    eC
q
2
U
q
U '

2C '  e
 e q
?
If the potential difference between the capacitor plates are the
same, the electric fields inside the capacitor are the same also.
E  q  0 A  q ( e 0 ) A
( e q)
q'
U'

  eU
2C ' 2 eC
2
2
q   e q
E '  E / e
q
DV '  DV /  e
C    eC
q
C
U '(DV ') 2
U'
e 2
q2 U
U '

2C '  e
If the potential difference between the capacitor plates are the
same, the electric fields inside the capacitor are the same also.
E  q  0 A  q ( e 0 ) A
( e q)
q'
U'

  eU
2C ' 2 eC
2
2
q   e q
Dielectrics and Gauss’ Law
 
 0  E  dA   0 E0 A  q
E0  q ( 0 A)
Example
C ?
 
 0  E0  ds  q
e
q
E0 
0 A
 
 e 0  E  ds  q
E
q
 e 0 A

DV   Eds  E0 (d  b)  Eb

0 A
q
q


C
DV E0 (d  b)  E0b /  e d  b(1 /  e  1)
Dielectrics and Gauss’ Law
 
 0  E  dA   0 E0 A  q
E0  q ( 0 A)
AGauss’
dielectric
law slab
should
is be
inserted,
amended as:
 
  

 0  E  dA  q  q  E

E
q

E
d
(
A


d

A
A
q
)


/

q
q
( 0 A)
00
e
0
e

q ’ is kinduced
surface
 instead
of  charge. .
e 0
0
1
E0 q qcontainedqwithin qthe Gauss
q
The charge
surface
q  q(  1)
E 

, 


is taken
the free
e
 e to
 ebe
 e 0 Acharge
 0 A only.
0 A
0A
Dielectrics and Gauss’ Law
 
 
 0  E  dA  q  q' p  dA

 
 0  ( E0  E ' )  dA  q  q'
 
 0  E0  dA  q
 
 
 0  E 'dA  q'    p  dA
p   e 0 E
Electric polarization vector
  
 ( 0 E0  p)  dA  q
D   0 E +p
 
 D  dA  q
Electric displacement vector
  0 E + e 0 E  (1+ e ) 0 E   0 E

E

D
0


E0



E0
e
Exercises
P695 27
Problems
P 696~699
6, 9, 20, 24