Capacitance

CAPACITANCE
A capacitor is a system that stores energy in the form of an electric field. In its simplest form, a capacitor consists
of a pair of parallel metal plates separated by air or other insulating material.
The potential difference V between the plates of a capacitor is directly proportional to the charge Q on either
of them, so the ratio Q/V is always the same for a particular capacitor. This ratio is called the capacitance C of
the capacitor:
Q
C =
V
charge on either plate
Capacitance =
potential difference between plates
The unit of capacitance is the farad (F), where 1 farad = 1 coulomb/volt. Since the farad is too large for
practical purposes, the microfarad and picofarad are commonly used, where
1 microfarad = 1 F = 10−6 F
1 picofarad = 1 pF = 10−12 F
A charge of 10−6 C on each plate of 1-F capacitor will produce a potential difference of V = Q/C = 1 V
between the plates.

PARALLEL-PLATE CAPACITOR
A capacitor that consists of parallel plates each of area A separated by the distance d (Fig. -1) has a capaci-
tance of
A
C = Kε 0
d
d

+Q −Q
A

Fig. 1. (From Modern Technical Physics, 6th Ed., Arthur Beiser, c⃝ 1992. Reprinted by permission of Pearson
Education, Inc.)
The constant ε0 is the permittivity of free space mentioned in Chapter 23; its value is

ε0 = 8.85 × 10−12 C2/(N·m2) = 8.85 × 10−12 F/m

The quantity K is the dielectric constant of the material between the capacitor plates; the greater K is, the more
effective the material is in diminishing an electric field. For free space, K = 1; for air, K = 1.0006; a typical
value for glass is K = 6; and for water, K = 80.

SOLVED PROBLEM 1
A 200-pF capacitor is connected to a 100-V battery. Find the charge on the capacitor’s plates.

Q = CV = (200 × 10−12 F)(100 V) = 2 × 10−8 C

SOLVED PROBLEM 2
A capacitor has a charge of 5 × 10−4 C when the potential difference across its plates is 300 V. Find its
capacitance.

= 5 × 10 C
Q −4
C= −6

V = 1.67 × 10 F = 1.67 F
300 V

SOLVED PROBLEM 3
A parallel-plate capacitor has plates 5 cm square and 0.1 mm apart. Find its capacitance (a) in air and
(b) with mica of K = 6 between the plates.
(a) In air K is very nearly 1, and so
A (0.05 m)2 = 2.21 × 10−10 F = 221 pF
C = Kε0 = (1) 8.85 × 10−12 F
d m (10−4 m)

(b) With mica between the plates the capacitance will be K = 6 times greater, or

C = (16)(221 pF) = 1326 pF

SOLVED PROBLEM 4
A parallel-plate capacitor has a capacitance of 2 F in air and 4.6 F when it is immersed in benzene.
What is the dielectric constant of benzene?
Since C is proportional to K , in general
C1 C2
=
K1 K2
for the same capacitor. Here, with K1 = Kair = 1, the dielectric constant K2 of benzene is
C2 4.6 F
K =K = (1) = 2.3
2 1
C1 2 F

SOLVED PROBLEM 26.5

A 10-F capacitor with air between its plates is connected to a 50-V source and then disconnected.
(a) What is the charge on the capacitor and the potential difference across it? (b) The space between the
plates of the charged capacitor is filled with Teflon (K = 2.1). What is the charge on the capacitor and
the potential difference across it now?
 (a) The capacitor’s charge is

Q = CV = (10 × 10−6 F)(50 V) = 5 × 10−4 C

The potential difference across it remains 50 V after it is disconnected.

(b) The presence of another dielectric does not change the charge on the capacitor. Since its capacitance is now
K2
C2 = C1
K1
and V = Q/C, the new potential difference is
1
V = Q = K1 Q = K1 V = (50 V) = 23.8 V
2 1
C2 K2 C1 K2 2.1

CAPACITORS IN COMBINATION
The equivalent capacitance of a set of connected capacitors is the capacitance of the single capacitor that can
replace the set without changing the properties of any circuit it is part of. The equivalent capacitance of a set of
capacitors joined in series (Fig. 26-2) is
1 1 1 1
= + + + ··· capacitors in series
C C1 C2 C3
When using a calculator with a reciprocal (1/ X) key, it is easiest to proceed in the way described in
Chapter 25 in the case of the similar formula for the equivalent resistance of a set of resistors in parallel. The
key sequence here would be
[C1][1/ X ][+][C2][1/ X ][+][C3][1/ X ][+] · · · [=][1/ X ]
If there are only two capacitors in series,
1 1 1 C1 + C2 C1C2
= + = and so C=
C C1 C2 C1C2 C 1 + C2
In a parallel set of capacitors (Fig. 26-3), the equivalent capacitance is the sum of the individual capacitances:

C = C1 + C2 + C3 + · · · capacitors in parallel

Fig. 2 Fig. 3

SOLVED PROBLEM 6
Show that the equivalent capacitance C of three capacitors connected in series is given by 1/C =
1/C1 + 1/C2 + 1/C3.
Each capacitor in a series connection has charges of the same magnitude Q on its plates, so the voltages across
them are, respectively,
Q Q Q
V1 = V2 = V3 =
C1 C2 C3
 If C is the equivalent capacitance of the set, then

V =V +V +V Q Q Q Q
1 2 3
= + +
C C1 C2 C3
Dividing through by the charge Q gives
1 1 1 1
= + +
C C1 C2 C3

SOLVED PROBLEM 7
Show that the equivalent capacitance C of three capacitors connected in parallel is given by C =
C1 + C2 + C3.
Now the same voltage V is across all the capacitors, and their respective charges are

Q1 = C1V Q2 = C2V Q3 = C3V

The total charge Q1 + Q2 + Q3 on either the + or − plates of the capacitors is equal to the charge Q on the
corresponding plate of the equivalent capacitor C , and so

Q = Q1 + Q2 + Q3 CV = C1V = C2V + C3V

Dividing through by V gives

C = C1 + C2 + C3

SOLVED PROBLEM 8
Three capacitors whose capacitances are 1, 2, and 3 F are connected in series. Find the equivalent
capacitance of the combination.

1 1 1 1 1 1 1 11
= = + + =
C C1 + C2 + C3 1 F 2 F 3 F 6 F

Hence,
6
C= F = 0.545 F
11

SOLVED PROBLEM 9
The three capacitors of Prob. 8 are connected in parallel. Find the equivalent capacitance of the
combination.

C = C1 + C2 + C3 = 1 F + 2 F + 3 F = 6 F

SOLVED PROBLEM 10
A 2- and 3-F capacitor are connected in series. (a) What is their equivalent capacitance? (b) A potential
difference of 500 V is applied to the combination. Find the charge on each capacitor and the potential
difference across it.
C1C2 (2 F)(3 F)
(a) C= = = 1.2 F
C1 + C2 2 F + 3 F
(b) The charge on the combination is

Q = CV = (1.2 × 10−6 F)(500 V) = 6 × 10−4 C

 The same charge is present on each capacitor (Fig. -4). Hence the potential difference across the 2-F
capacitor is
Q 6 × 10−4 C
V1 = = = 300 V
C1 2 × 10−6 F
and that across the 3-F capacitor is
Q 6 × 10−4 C
V2 = = = 200 V
C2 3 × 10−6 F
As a check we note that V1 + V2 = 500 V.

Fig. -4 Fig. -5

SOLVED PROBLEM 11
A 5- and 10-pF capacitor are connected in parallel. (a) What is their equivalent capacitance? (b) A
potential difference of 1000 V is applied to the combination. Find the charge on each capacitor and the
potential difference across it.
(a) C = C1 + C2 = 5 pF + 10 pF = 15 pF
(b) The same potential difference V = 1000 V is across each capacitor (Fig. 26-5). The charge on the 5-pF
capacitor is

Q1 = C1V = (5 × 10−12 F)(103 V) = 5 × 10−9 C

and that on the 10-pF capacitor is

Q2 = C2V = (10 × 10−12 F)(103 V) = 10−8 C

ENERGY OF A CHARGED CAPACITOR

To produce the electric field in a charged capacitor, work must be done to separate the positive and negative
charges. This work is stored as electric potential energy in the capacitor. The potential energy W of a capacitor
of capacitance C whose charge is Q and whose potential difference is V is given by
1 Q2
W = 21 QV = 21 CV 2 =
2 C

SOLVED PROBLEM 12
How much energy is stored in a 50-pF capacitor when it is charged to a potential difference of 200 V?

W = 1 CV 2 = ( 1 )(50 × 10−12 F)(200 V)2 = 10−6 J

2 2
SOLVED PROBLEM 13
A 100-F capacitor is to have an energy content of 50 J to operate a flashlamp. (a) What voltage is
required to charge the capacitor? (b) How much charge passes through the flashlamp?
(a) Since W = 12CV 2,
, ,
2W (2)(50 J)
V= = = 1000 V
C 10−4 F
(b) Q = CV = (10−4 F)(103 V) = 0.1 C

CHARGING A CAPACITOR
When a capacitor is being charged in a circuit such as that of Fig. 6, at any moment the voltage Q/C across it
is in the opposite direction to the battery voltage V and thus tends to oppose the flow of additional charge. For
this reason a capacitor does not acquire its final charge the instant it is connected to a battery or other source of
emf. The net potential difference when the charge on the capacitor is Q is V − (Q/C), and the current is then
∆Q V − (Q/C)
I = =
∆t R
As Q increases, its rate of increase I = ∆Q/∆t decreases. Figure 26-7 shows how Q, measured in percent of
final change, varies with time when a capacitor is being charged; the switch of Fig. 26-6 is closed at t = 0.
The product RC of the resistance R in the circuit and the capacitance C governs the rate at which the
capacitor reaches its ultimate charge of Q0 = CV. The product RC is called the time constant T of the circuit.
After a time equal to T, the charge on the capacitor is 63 percent of its final value.

Fig. 6

Fig. -7
 The formula that governs the growth of charge in the circuit of Fig. 6 is

Q = Q0(1 − e−t/T )

where Q0 is the final charge CV and T is the time constant RC. Figure 7 is a graph of this formula. The
quantity e has the value
e = 2.718 · · ·

and is often found in equations in engineering and science. A quantity that consists of e raised to a power is
called an exponential. To find the value of ex or e−x , an electronic calculator or a suitable table can be used.
Exponentials are sometimes written exp x or exp(−x) instead of ex or e−x . The meaning is exactly the same.
It is easy to see why Q reaches 63 percent of Q0 in time T. When t = T, t/T = 1 and
1
Q = Q 0 (1 − e−1) = Q 0 1 − = Q 0(1 − 0.37) = 0.63Q0
e

DISCHARGING A CAPACITOR
When a charged capacitor is discharged through a resistance, as in Fig. 8, the decrease in charge is governed by
the formula
Q = Q0e−t/T

where again T = RC is the time constant. The charge will fall to 37 percent of its original value after time T
(Fig. -9). The smaller the time constant T, the more rapidly a capacitor can be charged or discharged.

Fig. -8

Fig. -9