1040 Phys
Lecture 5
Capacitance and Dielectrics
Definition of Capacitance
Consider two conductors carrying charges of equal magnitude and opposite sign, as
shown in Figure 1. Such a combination of two conductors is called a capacitor.The
conductors are called plates. A potential difference V exists between the conductors due
to the presence of the charges.
Figure (1)
The capacitance C of a capacitor is defined as the ratio of the magnitude of the charge on
either conductor to the magnitude of the potential difference between the conductors:
From this Equation we see that capacitance has SI units of coulombs per volt. The SI
unit of capacitance is the farad (F), which was named in honor of Michael Faraday:
1 F = 1 C/V
Calculating Capacitance
We can derive an expression for the capacitance of a pair of oppositely charged
conductors in the following manner: assume a charge of magnitude Q, and calculate the
potential difference using the techniques described in the preceding chapter. We then use
the expression C = Q/V to evaluate the capacitance.
1040 Phys
Lecture 5
The electric potential of the sphere of radius R is simply ke Q/R, and setting V = 0 for the
infinitely large shell, we have
This expression shows that the capacitance of an isolated charged sphere is proportional
to its radius and is independent of both the charge on the sphere and the potential
difference.
Figure (2)
Parallel-Plate Capacitors
Two parallel metallic plates of equal area A are separated by a distance d, as shown in
Figure 2. One plate carries a charge Q , and the other carries a charge -Q . The surface
charge density on either plate is  = Q/A. If the plates are very close together (in
comparison with their length and width), we can assume that the electric field is uniform
1040 Phys
Lecture 5
between the plates and is zero elsewhere. The value of the electric field between the
plates is
Because the field between the plates is uniform, the magnitude of the potential difference
between the plates equals Ed ; therefore,
That is, the capacitance of a parallel-plate capacitor is proportional to the area of its plates
and inversely proportional to the plate separation.
Example 1 Parallel-Plate Capacitor
A parallel-plate capacitor with air between the plates has an area A = 2.00 x 10-4 m2 and a
plate separation d = 1.00 mm. Find its capacitance.
Solution
1040 Phys
Lecture 5
Figure (3)
Cylindrical and Spherical Capacitors
From the definition of capacitance, we can, in principle, find the capacitance of any
geometric arrangement of conductors. Assume a solid cylindrical conductor of radius a
and charge Q is coaxial with a cylindrical shell of negligible thickness, radius b , a, and
charge -Q (Fig. 3a). The capacitance of this cylindrical capacitor if its length is l.
We must first calculate the potential difference between the two cylinders, which
is given in general by
where E is the electric field in the region between the cylinders. According to Gauss’s
law the magnitude of the electric field of a cylindrical charge distribution having linear
charge density  is
E  2 Ke

r
1040 Phys
Lecture 5
The same result applies here because, according to Gauss’s law, the charge on the outer
cylinder does not contribute to the electric field inside it. Using this result and noting
from Figure 3b that E is along r, we find that
(Note that
)
Suppose b = 2.00 a for the cylindrical capacitor. We would like to increase the
capacitance, and we can do so by choosing to increase l by 10% or by increasing a
by 10%. Which choice is more effective at increasing the capacitance?
Answer C is proportional to l so increasing l by 10% results in a 10% increase in C.
For the result of the change in a, let us first evaluate C for b = 2.00a:
1040 Phys
Lecture 5
The Spherical Capacitor
Suppose a spherical capacitor consists of a spherical conducting shell of radius b and
charge -Q concentric with a smaller conducting sphere of radius a and charge Q (Fig. 4).
Find the capacitance of this device.
Figure (4)
The field outside a spherically symmetric charge distribution is radial and given by the
expression
E  Ke
Q
r2
1040 Phys
Lecture 5
In this case, this result applies to the field between the spheres (a < r < b). From Gauss’s
law we see that only the inner sphere contributes to this field. Thus, the potential
difference between the spheres is
The magnitude of the potential difference is
1040 Phys
Lecture 5
Combinations of Capacitors
(1) Parallel Combination
Two capacitors connected as shown in Figure 5a are known as a parallel combination of
capacitors. The individual potential differences across capacitors connected in parallel are
the same and are equal to the potential difference applied across the combination.
Figure (5)
Let us call the maximum charges on the two capacitors Q1 and Q2. The total charge Q
stored by the two capacitors is
That is, the total charge on capacitors connected in parallel is the sum of the charges on
the individual capacitors. Because the voltages across the capacitors are the same, the
charges that they carry are
1040 Phys
Lecture 5
Suppose that we wish to replace these two capacitors by one equivalent capacitor having
a capacitance Ceq, as in Figure 4c.
If we extend this treatment to three or more capacitors connected in parallel, we find the
equivalent capacitance to be
Thus, the equivalent capacitance of a parallel combination of capacitors is the
algebraic sum of the individual capacitances and is greater than any of the
individual capacitances.
(2) Series Combination
Two capacitors connected as shown in Figure 6a and the equivalent circuit diagram in
Figure 6b are known as a series combination of capacitors.
Figure (6)
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Lecture 5
The charges on capacitors connected in series are the same. From Figure 6a, we see that
the voltage V across the battery terminals is split between the two capacitors:
where V1 and V2 are the potential differences across capacitors C1 and C2,
respectively. In general, the total potential difference across any number of
capacitors connected in series is the sum of the potential differences across the
individual capacitors.
Because we can apply the expression Q = C V to each capacitor shown in Figure 5b, the
potential differences across them are
When this analysis is applied to three or more capacitors connected in series, the
relationship for the equivalent capacitance is
This shows that the inverse of the equivalent capacitance is the algebraic sum of the
inverses of the individual capacitances and the equivalent capacitance of a series
combination is always less than any individual capacitance in the combination.
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Lecture 5
Example Equivalent Capacitance
Find the equivalent capacitance between a and b for the combination of capacitors shown
in Figure 7a. All capacitances are in microfarads.
Figure (7)
Finally, the 2.0 µF and 4.0 µF capacitors in Figure 7c are in parallel and thus have an
equivalent capacitance of 6.0 µF.
Energy Stored in a Charged Capacitor
Suppose that q is the charge on the capacitor at some instant during the charging process.
At the same instant, the potential difference across the capacitor is V = q/C. We know
that the work necessary to transfer an increment of charge dq from the plate carrying
charge -q to the plate carrying charge q (which is at the higher electric potential) is
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Lecture 5
The total work required to charge the capacitor from q =0 to some final charge q = Q is
The work done in charging the capacitor appears as electric potential energy U stored in
the capacitor. We can express the potential energy stored in a charged capacitor in the
following forms:
Because the volume occupied by the electric field is Ad, the energy per unit volume uE =
U/Ad, known as the energy density, is
The energy density in any electric field is proportional to the square of the
magnitude of the electric field at a given point
References
This lecture is a part of chapter 26 from the following book
Physics for Scientists and Engineers (with Physics NOW and InfoTrac),
Raymond A. Serway - Emeritus, James Madison University , Thomson
Brooks/Cole © 2004, 6th Edition, 1296 pages
1040 Phys
Lecture 5
Problems
(1) An air-filled capacitor consists of two parallel plates, each with an area of 7.60
cm2, separated by a distance of 1.80 mm. A 20.0-V potential difference is applied
to these plates. Calculate (a) the electric field between the plates, (b) the surface
charge density, (c) the capacitance, and (d) the charge on each plate. ( problem
26.7)
(a) E 
E 
V

d
20
1.8 x 10 3
 11.1 x 10 3 volt / m

o
(b)   E  o  11.1 x 10 3 x 8.85 x 10 12
 98.235 x 10 9 C / m 2
(c) C 
o A
d

8.85 x 10 12 x 7.6 x 10 4
 37.367 x 10 13 C  3.7367 pF
1.8 x 10 3
(d) Q  C V  37.367 x10 13 x 20  74.73 x 10 12  74.73 pC
(2) When a potential difference of 150 V is applied to the plates of a parallel-plate
capacitor, the plates carry a surface charge density of 30.0 nC/cm2. What is the
spacing between the plates? ( problem26.9)
V  E d 
d 
V  o

 d
o
150 x 8.85 x 10 12

30 x 10 9
 44.25 x 10 3 m
1040 Phys
Lecture 5
(3) A 50.0-m length of coaxial cable has an inner conductor that has a diameter of
2.58 mm and carries a charge of 8.10 µC. The surrounding conductor has an inner
diameter of 7.27 mm and a charge of -8.10 µC. (a) What is the capacitance of this
cable? (b) What is the potential difference between the two conductors? Assume
the region between the conductors is air. ( problem26.11)
C 
Q

V
(a)
l
b
2 K e ln  
a
 2.688 x 10 9 F
(b) V 
50
 7.27 x 10 3 

2 x 9 x 10 9 ln 
3 
 2.58 x 10 
 2.688 p F

8.1 x 10 6
Q

 3.0134 x 10 3 volt  3.0134 kv
9
C
2.688 x 10
(4) An air-filled spherical capacitor is constructed with inner and outer shell radii of
7.00 and 14.0 cm, respectively. (a) Calculate the capacitance of the device. (b)
What potential difference between the spheres results in a charge of 4.00 µC on
the capacitor? ( problem26.13)
(a)
C 
ab
7 x 10 2 x 14 x 10 2
Q


V
Ke ( b  a )
9 x 10 9 x (14 x 10 2  74 x 10 2 )
 15.6 x 10 12 F  15.6 pF
4 x 10 6
Q

 0.25 x 10 6  250 kv
(b) V 
12
C 15.6 x 10
1040 Phys
Lecture 5
(5) Two capacitors, C1 = 5.00 µF and C2 = 12.0 µF, are connected in parallel, and
the resulting combination is connected to a 9.00-V battery. (a) What is the
equivalent capacitance of the combination? What are (b) the potential difference
across each capacitor and (c) the charge stored on each capacitor?
(a)
C eq  C1  C 2  5 x 10 6
 17 x 10 6
 12 x 10 6
F  17 F
(b) The individual potential differences across capacitors connected in parallel are
the same and are equal to the potential difference applied across the
combination, So V1  V2  V  9 volt
(c)
Q1  C1 x V  5 x 10 6 x 9  45 x 10 6 C  45 C
Q2  C 2 x V  12 x 10 6 x 9  108 x 10 6 C  108 C
1040 Phys
Lecture 5
(6) What If? The two capacitors of Problem 5 are now connected in series and to a
9.00-V battery. Find (a) the equivalent capacitance of the combination, (b) the
potential difference across each capacitor, and (c) the charge on each capacitor.
(a)
1
1
1
1
1 12  5 17






C eq
C1
C2 5
12
60
60
 C eq 
60
 3.529 F
17
(b)
Q  Q1  Q2
Q  C eq V  3.529 x 10 6 x 9  31.765 x 10 6 C  31.765 C
V1 
V2 
Q
31.765

 6.353 volt
C1
5
Q
31.765

 2.647 volt
C2
12
(c)
Q  Q1  Q2
Q  Ceq V  3.529 x 106 x 9  31.765 x 106 C  31.765 C
1040 Phys
Lecture 5
(7) Two capacitors when connected in parallel give an equivalent capacitance of 9.00
pF and give an equivalent capacitance of 2.00 pF when connected in series. What
is the capacitance of each capacitor?
C1  C 2  9
 C1  9  C 2
(1)
1
1
1


C1
C2
2
1
1
1


9  C2
C2
2
C2  9  C2 1

C 2 9  C 2 
2
9 C 2  C 22  18
C 22  9 C  18  0
C 2
 6  C 2  3  0
C 2  3 pF
or C 2  6 pF
and from Eq. (1)
C1  6 pF or C1  3 pF
 C1  3 pF and C 2  6 pF
1040 Phys
Lecture 5
(8) Four capacitors are connected as shown in Figure 8. (a) Find the equivalent
capacitance between points a and b. (b) Calculate the charge on each capacitor if
Vab = 15.0 V.
Figure (8)
(a)
1
C eq1
C eq1 

1
1
6
 
15
3 15
15
 2.5 F
6
C eq 2  2.5  6  8.5 F
8.5  20 28.5
1
1
1




C eq
8.5
20 8.5 x 20 170
 C eq 
170
 5.965 F
28.5
1040 Phys
Lecture 5
1040 Phys
Lecture 5
(9) Find the equivalent capacitance between points a and b for the group of capacitors
connected as shown in the Figure . Take C1 = 5.00 µF, C2 = 10.0 µF, and C3 =
2.00 µF.
1040 Phys
Lecture 5
References
This lecture is a part of chapter 26 from the following book
Physics for Scientists and Engineers (with Physics NOW and InfoTrac),
Raymond A. Serway - Emeritus, James Madison University , Thomson
Brooks/Cole © 2004, 6th Edition, 1296 pages