CHAPTER 6 CAPACITORS AND CAPACITANCE
EXERCISE 26, Page 65
1. Find the charge on a 10 μF capacitor when the applied voltage is 250V
Charge, Q = C  V = 10 106  250  2.5 103 C = 2.5 mC
2. Determine the voltage across a 1000 pF capacitor to charge it with 2 C.
Q
2 106

Q = CV hence voltage, V =
= 2000 V or 2 kV
C 1000 1012
3. The charge on the plates of a capacitor is 6 mC when the potential between them is 2.4 kV. Determine the
capacitance of the capacitor.
Charge, Q = CV hence capacitance, C =
Q 6 103

 2.5 10 6 F = 2.5 μF
V 2.4 103
4. For how long must a charging current of 2 A be fed to a 5 F capacitor to raise the p.d. between
its plates by 500 V.
Charge Q = I  t
from which,
and
time, t =
Q=CV
hence
It=CV
C  V 5 10 6  500

= 1.25 ms
I
2
5. A direct current of 10 A flows into a previously uncharged 5 μF capacitor for 1 ms. Determine the p.d.
between the plates.
Charge Q = I  t
from which,
and
Q=CV
hence
It=CV
I  t 10  1 103

 2000 V = 2 kV
p.d., V =
C
5 106
© John Bird Published by Taylor and Francis
44
6. A 16 F capacitor is charged at a constant current of 4 A for 2 min. Calculate the final p.d.
across the capacitor and the corresponding charge in coulombs.
It=CV
from which,
voltage, V =
I  t 4 106  2  60

= 30 V
C
16 106
Charge, Q = C  V = 16 106  30 = 480 C
7. A steady current of 10 A flows into a previously uncharged capacitor for 1.5 ms when the p.d.
between the plates is 2 kV. Find the capacitance of the capacitor
It=CV
from which,
capacitance, C =
I  t 10 1.5 10 3

= 7.5 F
V
2 103
© John Bird Published by Taylor and Francis
45
EXERCISE 27, Page 67
1. A capacitor uses a dielectric 0.04 mm thick and operates at 30 V. What is the electric field
strength across the dielectric at this voltage?
Electric field strength, E =
V
30

 750 kV/m
d 0.04 10 3
2. A two-plate capacitor has a charge of 25 C. If the effective area of each plate is 5 cm2 find the electric flux
density of the electric field.
Electric flux density, D =
Q
25C

 50000 C / m 2 = 50 kC / m 2
A 5 104 m 2
3. A charge of 1.5 C is carried on two parallel rectangular plates each measuring 60 mm by
80 mm. Calculate the electric flux density. If the plates are spaced 10 mm apart and the voltage
between them is 0.5 kV determine the electric field strength.
Electric flux density, D =
Q
1.5 10 6

= 312.5 C / m2
6
A 60  80 10
Electric field strength, E =
V 0.5 103

= 50 kV/m
d 10 10 3
4. Two parallel plates are separated by a dielectric and charged with 10 μC. Given that the area of each plate is
50 cm2, calculate the electric flux density in the dielectric separating the plates.
Electric flux density, D =
Q 10 106

 2 103 C / m 2 = 2 mC / m 2
4
A 50 10
5. The electric flux density between two plates separated by polystyrene of relative permittivity 2.5
is 5 C / m2 . Find the voltage gradient between the plates.
© John Bird Published by Taylor and Francis
46
D
D
5 106
  0  r from which, voltage gradient, E =

= 226 kV/m
E
0r 8.85 1012  2.5
6. Two parallel plates having a p.d. of 250 V between them are spaced 1 mm apart. Determine the
electric field strength. Find also the electric flux density when the dielectric between the plates is
(a) air and (b) mica of relative permittivity 5.
Electric field strength, E =
(a) For air,  r = 1, hence
V
250

= 250 kV/m
d 110 3
D
  0 from which,
E
electric flux density, D = E 0  250 103  8.85 1012 = 2.213 C / m2
(b)
D
  0  r from which, D = E 0  r  250 103  8.85 1012  5 = 11.063 C / m2
E
© John Bird Published by Taylor and Francis
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EXERCISE 28, Page 68
1. A capacitor consists of two parallel plates each of area 0.01 m2, spaced 0.1 mm in air. Calculate the
capacitance in picofarads.
Capacitance, C 
0  r A
8.85 1012 1 0.01
(n  1) 
(2  1)  885 1012 F = 885 pF
3
d
0.110
since  r = 1 for air
2. A waxed paper capacitor has two parallel plates, each of effective area 0.2 m 2 . If the capaci-tance is 4000 pF determine the effective thickness of the paper if its relative permittivity is 2.
C
0r A
  A 8.85 1012  2  0.2
 885 106 m
hence, thickness of the paper, d = 0 r 
12
d
C
4000 10
= 0.885 mm
3. Calculate the capacitance of a parallel plate capacitor having 5 plates, each 30 mm by 20 mm and separated
by a dielectric 0.75 mm thick having a relative permittivity of 2.3
0  r A
8.85 1012  2.3  30  20 106
(n  1) 
(5  1)  65.14 1012 F = 65.14 pF
Capacitance, C 
3
d
0.75 10
4. How many plates has a parallel plate capacitor having a capacitance of 5 nF, if each plate is
40 mm by 40 mm and each dielectric is 0.102 mm thick with a relative permittivity of 6.
0r A
Cd
5 109  0.102 103
C
(n  1) from which, n – 1 =

=6
d
0  r A 8.85 1012  6  40  40 106
Hence, the number of plates, n = 6 + 1 = 7
5. A parallel plate capacitor is made from 25 plates, each 70 mm by 120 mm, interleaved with mica
of relative permittivity 5. If the capacitance of the capacitor is 3000 pF determine the thickness of
the mica.
© John Bird Published by Taylor and Francis
48
C
0r A
(n  1) from which,
d
0  r A
8.85 1012  5  70 120 106
(n  1) 
(25  1) = 0.00297 m
dielectric thickness, d =
C
3000 1012
= 2.97 mm
6. A capacitor is constructed with parallel plates and has a value of 50 pF. What would be the capacitance of
the capacitor if the plate area is doubled and the plate spacing is halved?
If the plate area is doubled then the capacitance will double, i.e. C = 50 pF × 2 = 100 pF
If also the plate spacing is halved, then the capacitance will again double, i.e. capacitance, C = 100 pF× 2
= 200 pF
7. The capacitance of a parallel plate capacitor is 1000 pF. It has 19 plates, each 50 mm by 30 mm
separated by a dielectric of thickness 0.40 mm. Determine the relative permittivity of the
dielectric.
C
0r A
(n  1) hence,
d
relative permittivity, r 
Cd
1000 1012  0.40 103

= 1.67
0 A (n  1) 8.85 1012  50  30 106  (19  1)
8. The charge on the square plates of a multiplate capacitor is 80 C when the potential between
them is 5 kV. If the capacitor has twenty five plates separated by a dielectric of thickness
0.102 mm and relative permittivity 4.8, determine the width of a plate.
C
0r A
(n  1) from which,
d
 80 106 
Q
 0.102 103
d

3 
 
5

10
Cd
V

  

plate area, A =
= 0.0016 m 2
12
0  r (n  1) 0  r (n  1) 8.85 10  4.8  (25  1)
Since the plates are square, the width of a plate =
0.0016  0.040 m = 40 mm
© John Bird Published by Taylor and Francis
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9. A capacitor is to be constructed so that its capacitance is 4250 pF and to operate at a p.d. of
100 V across its terminals. The dielectric is to be polythene ( r  2.3 ) which, after allowing a
safety factor, has a dielectric strength of 20 MV/m. Find (a) the thickness of the polythene
needed, and (b) the area of the plate.
(a) Dielectric strength, E =
V
from which,
d
thickness of polythene, d =
V
100

 5  10 6 m = 0.005 mm
6
E 20 10
0r A
C d 4250 1012  5 106

(b) C 
from which, cross-sectional area, A =
d
0  r
8.85 1012  2.3
= 1.044 103 m 2  1.044 103 104 cm 2
= 10.44 cm 2
© John Bird Published by Taylor and Francis
50
EXERCISE 29, Page 72
1. Capacitors of 2 μF and 6 μF are connected (a) in parallel and (b) in series. Determine the equivalent
capacitance in each case.
(a) In parallel, equivalent capacitance, CT = 2 + 6 = 8 μF
(b) In series,
1
1
1 1 1 4


  
CT C1 C2 2 6 6
from which, CT =
6
= 1.5 μF
4
2. Find the capacitance to be connected in series with a 10 F capacitor for the equivalent
capacitance to be 6 F
For series connection,
1
1
1


C1 C 2 CT
i.e.
1
1 1


10 C 2 6
and
C2 =
from which,
1 1 1
   0.06666666
C2 6 10
1
= 15 F
0.06666666
3. What value of capacitance would be obtained if capacitors of 0.15 μF and 0.10 μF are connected (a) in series
and (b) in parallel
(a) In series,
3
1
1
1
1
1
50





from which, CT =
= 0.06 μF
50
CT C1 C2 0.15 0.10 3
(b) In parallel, equivalent capacitance, CT = 0.15 + 0.10 = 0.25 μF
4. Two 6 F capacitors are connected in series with one having a capacitance of 12 F. Find the
total equivalent circuit capacitance. What capacitance must be added in series to obtain a
capacitance of 1.2 F?
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Two 6 F capacitors in series has a total capacitance of
66
= 3 F. (Two equal value capacitors in series
66
will have a total capacitance of half the value of one of the capacitors).
3 F in series with 12 F has a total capacitance of
3  12
= 2.4 F = total circuit capacitance.
3  12
Let new capacitance be C X then if new total capacitance is to be 1.2 F then
1
1
1


1.2 2.4 C X
1
1
1


 0.41666
CX 1.2 2.4
from which
Hence, capacitance to be added, C X =
1
= 2.4 F
0.41666
5. Determine the equivalent capacitance when the following capacitors are connected (a) in parallel and (b) in
series: (i) 2 μF, 4 μF and 8 μF
(iii) 50 pF and 450 pF
(ii) 0.02 μF, 0.05 μF and 0.10 μF
(iv) 0.01 μF and 200 pF
(a)(i) CT = 2 + 4 + 8 = 14 μF
(ii) CT = 0.02 + 0.05 + 0.10 = 0.17 μF
(iii) CT = 50 + 450 = 500 pF
(iv) CT = 0.0110 6  200 10 12 = 10.2 nF or 0.0102 μF
(b)(i)
1
1
1
1 1 1 1 7



   
CT C1 C2 C3 2 4 8 8
from which, CT =
8
= 1.143 μF
7
(ii)
1
1
1
1
1
1
1
1






 80 from which, CT =
= 0.0125 μF
80
CT C1 C2 C3 0.02 0.05 0.10
(iii)
1
1
1
1
1
1





from which, CT = 45 pF
CT C1 C 2 50 450 45
(iv)
1
1
1
1
1
1




 5.1109 from which, CT =
= 196.1 pF
6
12
5.1 10 9
CT C1 C2 0.0110
200 10
© John Bird Published by Taylor and Francis
52
6. For the arrangement shown below find (a) the equivalent circuit capacitance and (b) the voltage
across a 4.5 F capacitor.
(a) Three 4.5 F capacitors in series gives 1.5 F and two 1 F capacitors in series gives 0.5 F
1.5 F and 0.5 F capacitors in parallel gives 1.5 + 0.5 = 2 F
2 F in series with 3 F gives:
23 6
 = 1.2 F = equivalent circuit capacitance
23 5
 3 
(b) The equivalent circuit is shown below where V1  
  500  = 300 V = voltage across three
 23
4.5 F capacitors in series. Hence, voltage across each 4.5 F capacitor = 300/3 = 100 V.
(Alternatively, to find V1 :
Since CT = 1.2 F then QT  CT  V  1.2 106  500  600 C . This is the charge on each
capacitor of the circuit shown below. Hence, V1 
QT 600 106

= 300 V)
C1
2 106
7. Three 12 F capacitors are connected in series across a 750 V supply. Calculate (a) the
equivalent capacitance, (b) the charge on each capacitor and (c) the p.d. across each capacitor.
(a)
1
1
1 1 1
    0.25 from which, equivalent capacitance, CT 
= 4 F
0.25
CT 12 12 12
(Alternatively, the total capacitance of three capacitors each having the same value, will be one
© John Bird Published by Taylor and Francis
53
third the value of one of the capacitors, i.e. CT 
12
= 4 F)
3
(b) Total charge, QT  CT  V  4 106  750  3000 10 6 C = 3 mC = the charge on each
capacitor since they are connected in series.
(c) P.d. across each capacitor =
750
= 250 V since each capacitor has the same value.
3
8. If two capacitors having capacitances of 3 F and 5F respectively are connected in series across a 240 V
supply, determine (a) the p.d. across each capacitor and (b) the charge on each capacitor.
The circuit is shown below.
 5 
(a) Similar to question 5(b), V1  
  240  = 150 V
 35
and
 3 
V2  
  240  = 90 V
 35
(b) Charge, Q1  C1  V1  3 106 150 = 450 C or 0.45 mC
and
Q 2  C2  V2  5 106  90 = 450 C or 0.45 mC
(Note that in a series circuit the charge is the same on each capacitor).
9. In the circuit below, capacitors P, Q and R are identical and the total equivalent capacitance of
the circuit is 3 F. Determine the values of P, Q and R.
© John Bird Published by Taylor and Francis
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3.5 F and 4.5 F in parallel gives an equivalent capacitance of 3.5 + 4.5 = 8 F
2 F in series with 8 F gives
2  8 16

= 1.6 F
2  8 10
Let the equivalent capacitance of P, Q and R in series be C X
Then
1.6 + C X = 3 from which, C X = 3 – 1.6 = 1.4 F
Thus,
1
1
1
1
3
(since CP  CQ  CR )




1.4 CP CQ CR CP
i.e.
C P  3 1.4 = 4.2 F = CQ  CR
10. Capacitance’s of 4 μF, 8 μF and 16 μF are connected in parallel across a 200 V supply. Determine (a) the
equivalent capacitance, (b) the total charge and (c) the charge on each capacitor.
(a) Equivalent capacitance, CT = 4 + 8 + 16 = 28 μF
(b) Total charge, Q = CV = 28 106  200  5.6 103 C = 5.6 mC
(c) Q1  C1  V  4 106  200  800 10 6 C = 0.8 mC
Q 2  C2  V  8 106  200  1600  106 C = 1.6 mC
Q3  C3  V  16 106  200  3.2 103 C = 3.2 mC
11. A circuit consists of two capacitors P and Q in parallel, connected in series with another
capacitor R. The capacitances of P, Q and R are 4 F, 12 F and 8 F respectively. When the
circuit is connected across a 300 V d.c. supply find (a) the total capacitance of the circuit,
(b) the p.d. across each capacitor and (c) the charge on each capacitor
© John Bird Published by Taylor and Francis
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The circuit is shown below in diagram (a).
(a) 4 F in parallel with 12 F gives: 4 + 12 = 16 F as shown in the equivalent circuit below in
diagram (b).
The total capacitance, CT 
16  8
= 5.33 F
16  8
(b) Total charge on circuit, QT  CT  V  5.33333 106  300  1.6 mC = the charge on each
capacitor in diagram (b).
Q1 1.6 103
Q2 1.6 103

 100 V and V2 

 200 V
Hence, voltage V1 
C1 16 106
C2
8 106
Hence, the voltage across P = 100 V, the voltage across Q = 100 V and the voltage across
R = 200 V
(c) Charge on P, QP  CP  VP  4 106 100 = 400 C or 0.4 mC
Charge on Q, QQ  CQ  VQ  12 106 100 = 1200 C or 1.2 mC
Charge on R, Q R  CR  VR  8 106  200 = 1600 C or 1.6 mC
© John Bird Published by Taylor and Francis
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12. For the circuit shown below, determine (a) the total circuit capacitance, (b) the total energy in
the circuit, and (c) the charges in the capacitors shown as C1 and C2
(a) 2 F in series with 2 F gives 1 F
Hence, 1 F, 2 F, 2 F and 1 F in parallel gives: 1 + 2 + 2 + 1 = 6 F
The circuit is now as shown below in (i).
(i)
Total capacitance, CT , is obtained from:
1
1 1 1
    1.16666...
CT 2 6 2
and
CT =
(b) Total energy stored, W =
1
= 0.857 F
1.16666..
1
1
2
CV 2   0.857 106   50  = 1.071 mJ
2
2
(c) Total charge, QT  CT  V  0.857 106  50 = 42.85 C
This is the charge on each of the capacitors in circuit (i) above.
Hence, charges on capacitors C1 and C2 = 42.85 C on each
© John Bird Published by Taylor and Francis
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EXERCISE 30, Page 74
1. When a capacitor is connected across a 200 V supply the charge is 4 C. Find (a) the
capacitance and (b) the energy stored.
(a) Q = CV from which, capacitance, C =
(b) Energy stored, W =
Q 4 10 6

= 20 nF or 0.02 F
V
200
1
1
CV 2   0.02 106  2002 = 400 J or 0.4 mJ
2
2
2. Find the energy stored in a 10 μF capacitor when charged to 2 kV
Energy stored, W =
1
1
CV 2  10 106  20002 = 20 J
2
2
3. A 3300 pF capacitor is required to store 0.5 mJ of energy. Find the p.d. to which the capacitor
must be charged.
1
Energy, W = CV 2 from which, p.d., V =
2
 2  0.5 103 
2W
= 550 V
 
12 
C
 3300 10 
4. A capacitor is charged with 8 mC. If the energy stored is 0.4 J find (a) the voltage and (b) the
capacitance.
(a) Energy, W =
1
1Q
1
CV 2    V 2  QV
2
2V
2
Hence, voltage, V =
2W 2  0.4

= 100 V
Q 8 10 3
Q 8 10 3

(b) Capacitance, C =
= 80 F
V
100
© John Bird Published by Taylor and Francis
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5. A capacitor, consisting of two metal plates each of area 50 cm 2 and spaced 0.2 mm apart in air,
is connected across a 120 V supply. Calculate (a) the energy stored, (b) the electric flux density
and (c) the potential gradient.
(a) Energy stored, W =
1
CV 2
2
0 r A 8.85 1012 1 50 104

 221.25 1012 F
and capacitance, C =
3
d
0.2 10
Hence, energy stored, W =
1
 221.24 10 12 120 2 = 1.593 J
2
Q C  V 221.25 1012 120


(b) Electric flux density, D =
= 5.31 C / m2
4
A
A
50 10
(c) Potential gradient or electric field strength, E =
V
120

= 600 kV/m
d 0.2 10 3
6. A bakelite capacitor is to be constructed to have a capacitance of 0.04 F and to have a steady
working potential of 1 kV maximum. Allowing a safe value of field stress of 25 MV/m find (a)
the thickness of bakelite required, (b) the area of plate required if the relative permittivity of
bakelite is 5, (c) the maximum energy stored by the capacitor and (d) the average power
developed if this energy is dissipated in a time of 20 s.
V
from which,
d
V
1000

 40 106 m  40  103 mm = 0.04 mm
thickness of dielectric, d =
E 25 106
(a) Field stress, E =
0 r A
from which,
d
C d 0.04 106  0.04 103

 0.03616 m 2 = 361.6 cm 2
cross-sectional area, A =
12
0  r
8.85 10  5
(b) Capacitance, C =
(c) Maximum energy, Wmax =
1
1
CV 2   0.04 106 1000 2 = 0.02 J
2
2
© John Bird Published by Taylor and Francis
59
(d) Energy = power  time, hence, power, P =
energy
0.02 J

= 1000 W or 1 kW
time
20  106 s
© John Bird Published by Taylor and Francis
60