Chapter 26
Capacitance, dielectrics and
electric energy storage
• Basic circuit devices
–
–
–
–
Resistors
Capacitors
Inductors
Power supply (Battery, Generator)
• Use our knowledge of electric
fields, potentials, and energy to
describe how capacitors work.
Parallel Plate Capacitor
Fig 26-4b, p.800
Active Figure 26.4
(SLIDESHOW MODE ONLY)
Potential difference and electric
fields in a uniform electric field
+Q
b
d
a
Qenclosed
 E   E dA 
o
A
EA
o

E 
o
-Q
d
  ˆ


ˆ
ˆ
Vba     j  dxi  dyj   dy  d
o 

o
0
0 o
d


Potential difference and electric
fields in a uniform electric field

Qd
Vba  d 
o
Ao
+Q
b
d
a
-Q
Q  Vba
Q  CVba
The constant of proportionality is called “capacitance.”
For a parallel plate capacitor, the capacitance is:
A 0
C
d
Factors affecting capacitance
• Size of the capacitor (A, d)
• Geometric arrangement
– Plates
– Cylinders
• Material between conductors
– Air
– Paper
– Wax
Units of capacitance
Q
C
Vba
C
 Coulomb 
 Volt   Farad  V  F
A Farad is a lot of capacitance. Typical capacitors are
“micro, nano, pico-Farad
Capacitance – Isolated Sphere
• Assume a spherical charged conductor
• Assume V = 0 at infinity
Q
Q
R
C


 4πεoR
V keQ / R ke
• Note, this is independent of the charge and
the potential difference
Cylindrical capacitor
L
b
Qenclosed
 E   E dA 
o
a
Q
E 2rL 
o
+Q on center conducting cylinder
Q
E 
2o rL
-Q on outer conducting cylinder
b
 Q

Q
dr
Q
b

Vba    
rˆ   drrˆ   

ln  

2o rL 
2o L a r
2o L  a 
a
b
Capacitance of a cylindrical capacitor
L
b
a
Q  CVab
L
Q
C

Vab
2o L

Q
b
b
ln   ln  
2o L  a 
a
Q
Example
• How strong is the electric field between the
plates of a 0.80 mF air gap capacitor if they
are 2.0 mm apart and each has a charge of
72 mC?
Capacitors in Parallel
C3
Q3
Q2

a
Qtotal  Q1  Q2  Q3
V  V1  V2  V3
C2
C1
Q1
Q total  Ceq V
b
Q1  C1V
Q2  C2V
Q3  C3V
Ceq V  C1V  C2 V  C3V
V  Vab
Ceq  C1  C2  C3
Capacitors in Series
C1
C2
C3
Q1
Q2
Q3
a
Q
Q  Ceq V  V 
Ceq
V  V1  V2  V3
Q

V1 
b
C1
V  Vab
Qtotal  Q1  Q2  Q3
Q
V2 
C2
Q
V3 
C3
Q
Q Q Q
 

Ceq C1 C2 C3
1
1
1
1
 

Ceq C1 C2 C3
Capacitor circuit example
2 mF
3V
3 mF
4 mF
2 mF
What single capacitor can replace the four shown here?
How much charge can the system hold?
How much charge is on one of the 2 mF capacitors?
Energy Storage in Capacitors
dU  Vdq
Ub
dU
Vb 
V
q
dq
(Like problem 23-50)
q Q
Change in potential energy
while charging capacitor

Vdq
q 0
Qd
Vba 
A o
Parallel Plates
Concentric Cylinders
In General
U
Q
a
Vba 
ln  
2o L  b 
Q
V
C
q Q
2 Q
q
1q
U   dq 
C
C 2
q 0
0
Q2

2C
Alternate Energy Expressions
2
Q
U
2C
Q
V
C
2
2
CV
1
U
 CV 2
2C
2
1Q 2 1
U
V  QV
2V
2
Energy Density
Energy per unit volume:
Consider a Parallel Plate Capacitor:
U
u
Volume
1
2
U  CV
2
A 0
C
d
1 A0 2 2 1
2
U
E d   Ad   0 E
2 d
2
U
U 1
2
u

 0 E
Volume Ad 2
V  Ed
Dielectrics
•A dielectric is a nonconducting material that, when placed
between the plates of a capacitor, increases the capacitance
•Materials with Dipoles that can align with an external
electric Field. Dielectrics include rubber, plastic, and waxed
paper
- + +
+
-
E ind
-
+ +
-- +
E Dielectric  E o  E ind
Eo
Eo

K
K is the Dielectric Constant
Measure of the degree of dipole alignment in the material
Dielectrics
Example values of dielectric constant
“Dielectric strength” is the
maximum field in the
dielectric before breakdown.
(a spark or flow of charge)
E max  Vmax / d
Effect of a dielectric on capacitance
E Dielectric
Eo

K
VDielectric
Vo

K
Q
Q
C K
 KCo
V
Vo
E Dielectric
Eo
d 
d
K
Potential difference with a dielectric
is less than the potential difference
across free space
Results in a higher capacitance.
Allows more charge to be stored before breakdown voltage.
Effect of the dielectric constant
Parallel Plate Capacitor
o A
Ko A
Co 
 C  KCo 
d
d
Material permittivity measures
degree to which the material
permits induced dipoles
to align with an external field
  Ko
A
C
d
Example modifications
using permittivity
1
1
1 2
2
2
u o  0 E  u  K0 E  E
2
2
2
Example – Parallel Plate Capacitor
+Q
d
3
d
-Q
o A
C
d
A
What is new capacitance?
Dipoles
The combination of two equal charges of opposite sign,
+Q and –Q, separated by a distance l
-Q
p  Q2a
p
2a
p1
p2
p  p1  p2
+Q
Dipoles in a Uniform Electric Field
E

2a
p
-Q
F  QE
ˆi
  r F  x
Fx
F  QE
+Q
ˆj
y
Fy
kˆ
z
Fz
    r F sin 
  F a sin   Fa sin   QEa sin   QEa sin   Q2aEsin   pEsin 
  p E
Work Rotating a Dipole in an
Uniform Electric Field
E
F  QE
+Q

2a
p
F  QE
-Q
dW  d  pE sin d  dU
U  pE  sin d  pE cos   U 0
Let: U(=90o )  0
U0  0
U  p E
U  pE cos 
Example P26.9
When a potential difference of 150 V
is applied to the plates of a parallelplate capacitor, the plates carry a
surface charge density of 30.0
nC/cm2. What is the spacing between
the plates?
Q 
d
0  V 

0 A
 V 
d
8.85 10
12

 30.0  10
9
C cm
C2 N m
2
2
 150 V 
1.00  10
4
cm
2
m
2

 4.42 mm
Example P26.21
Four capacitors are connected as
shown in Figure P26.21.
(a) Find the equivalent capacitance
between points a and b.
(b) Calculate the charge on each
capacitor if ΔVab = 15.0 V.
1
1
1


C s 15.0 3.00
Q  C V   5.96 mF 15.0 V   89.5 mC
Q 89.5 m C

 4.47 V
C 20.0 mF
15.0  4.47  10.53 V
V 
C s  2.50 mF
C p  2.50  6.00  8.50 m F

1
1 
C eq  

 8.50 mF 20.0 mF 
1
 5.96 mF
Q  C V   6.00 mF  10.53 V   63.2 m C
89.5  63.2  26.3 mC
on 6.00 m F
Example P26.27
Find the equivalent capacitance
between points a and b for the
group of capacitors connected as
shown in Figure P26.27. Take C1 =
5.00 μF, C2 = 10.0 μF, and C3 =
2.00 μF.
1 
 1
Cs  

 5.00 10.0
1
 3.33 mF
C p1  2 3.33  2.00  8.66 mF
C p2  2 10.0  20.0 mF
1 
 1
C eq  

 8.66 20.0
1
 6.04 mF
Example P26.35
A parallel-plate capacitor is charged and then
disconnected from a battery. By what fraction
does the stored energy change (increase or
decrease) when the plate separation is doubled?
d2  2d1
,
C2 
1
C1
2
. Therefore, the
stored energy doubles
Example P26.43
Determine (a) the capacitance and (b)
the maximum potential difference that
can be applied to a Teflon-filled
parallel-plate capacitor having a plate
area of 1.75 cm2 and plate separation of
0.040 0 mm.
C
 0 A
d


1.75  10
2.10 8.85  1012 F m
4
5
4.00  10

m
Vm ax  Em axd  60.0  106 V m
m
2
  8.13 10
11
 4.00  10
5

F  81.3 pF
m  2.40 kV
Example P26.59
A parallel-plate capacitor is constructed
using a dielectric material whose dielectric
constant is 3.00 and whose dielectric strength
is 2.00 × 108 V/m. The desired capacitance
is 0.250 μF, and the capacitor must withstand
a maximum potential difference of 4 000 V.
Find the minimum area of the capacitor
plates.
  3.00
Em ax  2.00  108 V m 
C
A
 0 A
d


Vm ax
d
 0.250  106 F


0.250  106  4 000
C Vm ax
Cd


 0.188 m
 0  0 Em ax 3.00 8.85  1012 2.00  108

2