Capacitor

Lecture Practice Problems

Capacitor - 5
LEVEL – I
1. Find the charge flowing across the circuit on closing the key K. K
C C
1 2
V
+

2. A capacitor of capacitance 10F is charged to a potential 50V with a battery. The battery is now
disconnected and an additional charge 200C is given to the capacitor. Calculate the potential
difference across the capacitor (in volts).

3. Two capacitors of capacitances 3F and 6 F are charged to a potential of 12 V each. They are
now connected to each other, with the positive plate of each joined to the negative plate of the
other. Calculate the potential difference across each.
C 2C
4. Consider the capacitor circuit shown in the figure. Initially only K1 is closed 1
and after sufficiently long time, K2 is also closed. Calculate the charge
flowing through section 1  2 after closing the K2. K2

K1 2
E

5. A 6 F capacitor is charged from 10 volts to 20 volts. Increase in energy will be

(A) 18  10 4 J (B) 9  104 J (C) 4.5  104 J (D) 9  106 J

6. A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of
the capacitor are moved further apart by means of insulating handles, then
(A) The charge on the capacitor increases
(B) The voltage across the plates decreases
(C) The capacitance increases
(D) The electrostatic energy stored in the capacitor increases

7. Two spherical conductors each of capacity C are charged to potential V and –V. These are then
connected by means of a fine wire. The loss of energy will be
1
(A) Zero (B) CV 2 (C) CV2 (D) 2CV2
2
8. A 2 F capacitor is charged to 100 volt and then its plates are connected by a conducting wire. The
heat produced is
(A) 1 J (B) 0.1 J (C) 0.01 J (D) 0.001 J

9. A parallel plate condenser with oil between the plates (dielectric constant of oil K = 2) has a
capacitance C. If the oil is removed, then capacitance of the capacitor becomes
C C
(A) 2C (B) 2C (C) (D)
2 2
10. A variable condenser is permanently connected to a 100 V battery. If the capacity is changed from 2
F to 10 F, then change in energy is equal to
(A) 2  102 J (B) 2.5  102 J (C) 3.5  10 2 J (D) 4  102 J

11. Two insulated metallic spheres of 3F and 5F capacitances are charged to 300 V and 500 V
respectively. The energy loss, when they are connected by a wire is
(A) 0.012 J (B) 0.0218 J (C) 0.0375 J (D) 3.75 J

12. Work done by an external agent in separating the parallel plate capacitor is
1
Capacitor

1 1
(A) CV (B) C2 V (C) CV 2 (D) None of these
2 2
13. The work done in placing a charge of 8  1018 coulomb on a condenser of capacity 100 micro-farad
is
(A) 32  10 32 J (B) 16  10 32 J (C) 3.1 10 26 J (D) 4  10 10 J

LEVEL – II

1. A 4.00 F capacitor and a 6.00 F capacitor are connected in parallel across a 660 V supply
line.
(a) Find the charge on each capacitor and the voltage across each
(c) The charged capacitors are disconnected from the line and from each other, and then
reconnected to each other with terminals of unlike sign together. Find the final charge on each
and the voltage across each.

2. Two parallel plate capacitors A and B having capacitance 1 F and 5F are charged separately
to the same potential of 100 volt.
Now the positive plate of A is connected to the negative plate of B and negative plate of A to the
positive plate of B.
Find the final charge on each capacitors and total loss of electrical energy in the given system.

4. A battery of 10 V is connected to a capacitor of capacity 0.1F. The battery is now removed and
this capacitor is connected to a second uncharged capacitor. If the charge is equally distributed
on these two capacitors, find the total energy stored in the two capacitors. Find the ratio of final
energy to the initial energy.

8. Find the work W done by an agent against the electric forces to increase the distance between the
plates of a parallel-plate capacitor of area S from x1 to x2 if:
(a) Charge q of the capacitor is constant
(b) Its voltage V is maintained constant.

15. The capacities of two conductors are C1 and C2 and their respective potentials are V1 and V2. If they
are connected by a thin wire, then the loss of energy will be given by
C C (V  V2 ) C C (V  V2 )
(A) 1 2 1 (B) 1 2 1
2(C1  C2 ) 2(C1  C2 )
C1C2 (V1  V2 )2 (C1  C2 )(V1  V2 )
(C) (D)
2(C1  C2 ) C1C2
16. Two condensers of capacities 1 F and 2 F are connected in series and the system is charged to
120 volts. Then the P.D. on 1 F capacitor (in volts) will be
(A) 40 (B) 60 (C) 80 (D) 120

17. Three identical capacitors are combined differently. For the same voltage to each combination, the
one that stores the greatest energy is
(A) Two in parallel and the third in series with it
(B) Three in series
(C) Three in parallel
(D) Two in series and third in parallel with it

18. A capacitor of 20 F is charged to 500 volts and connected in parallel with another capacitor of 10
F and charged to 200 volts. The common potential is
(A) 200 volts (B) 300 volts (C) 400 volts (D) 500 volts

19. Two capacitors each of 1 F capacitance are connected in parallel and are then charged by 200
volts d.c. supply. The total energy of their charges (in joules) is
(A) 0.01 (B) 0.02 (C) 0.04 (D) 0.06
2
Capacitor

ANSWER KEY

LEVEL – I
1. When the key K is kept open, the charge drawn from the source is Q = CV
where C is the equivalent capacitance given by C = C/2
Therefore, Q = (C/2)V
When the key K is closed, the capacitor 2 gets short circuited and thus the charge in that capacitor
comes back to the source.
 Charge flowing is Q = (C/2) V
2. 70 V
3. 4V
2 2 2 2
4. Initially charge configuration (before K2 is closed) is CE from 2 to 1  CE
3 C  3 CE  3 CE  CE
3
1

K2

K1 2
E
Final charge configuration (after K2 is closed) is
The net change of charge in the encircled region can happen only +CE CE 1 0 0

because of charge flowing in section 1  2.

Initial charge = 0 K2
Final charge = CE
 Charge flown = CE from 2 to 1. 2
K1
E
1 2 2
5. (B) E  EFinal  EInitial  C(VFinal  VInitial )
2
1
  6  (202  10 2 )  10 6
2
 3  (400  100)  106  3  300  106  9  104 J
6. (D) When the battery is disconnected, the charge will remains same in any case.
 A
Capacitance of a parallel plate capacitor is given by C  0
d
When d is increased, capacitance will decreases and because the charge remains the same, so
according to q = CV, the voltage will increase. Hence the electrostatics energy stored in the
capacitor will increase.
1 CC
7. (C) V  | V  (  V) |2  CV 2 .
2 (C  C)
1
8. (C) Heat produced = Energy of charged capacitor  CV 2
2
1
  (2  10 6 )  (100)2  0.01J .
2
C C
9. (D) Cair  medium  .
K 2
V2
10. (D) U  U2  U1  (C2  C1 )
2
(100)2
 (10  2)  106  4  102 J
2
1 C1C2 (V2  V1 )2 (3  5)  1012  (500  300)2
11. (C) U  
2 (C1  C2 ) (3  5)  10 6
15  10 12  4  10 4
  0.0375 J
8  10 6

3
Capacitor

12. (C)

Q2 (8  10 18 )2
13. (A) W   6
 32  10 32 J
2C 2  100  10

LEVEL - II
1. (a) 600 V (across each capacitor), (b) 132V

2. Before connection:
Charge on capacitor A = 10-6 x 100 = 10-4 C
Charge on capacitor B = 5 x 10-6 x 100 = 5 x 10-4 C
After connection:
Charge of the system = 5 x 10-4 – 10-4 = 4 x 10-4 C
Capacitance of the system = (1+ 5)F = 6F = 6 x 10-6 F
4 x 10 4 200
Common voltage = 6
 V
6 x 10 3
200
So charge on A = 10-6 x  0.66 x 10 4 C
3
200
Charge on B = 5 x 10-6 x  3.33 x 10 4 C
3
1 2 1 2
Energy,
2 2
 
Initial Total energy = x10 6 100   5x106 100   0.03 J
2
1  200  0.04
Final total energy =
2

6x10 6 
 3 
 
3
J

0.04 5
Loss of energy = 0.03 –  x 10 2 J.
3 3

1 2
4. Energy stored in the system initially =  0.1 10   5J
2
When the 0.1F is capacitor is connected to another 0.1 F capacitor then charge distribution is
10
identical. Hence final voltage =  5 V
2
1 2 1 2
Energy stored in the system finally =  0.1  5    0.1  5 
2 2
= 2.5 J
1
Ratio =
2
q2 1 1 1 
8. (a)  x 2  x1  0 Sv 2   
(b)
2 0 S 2  x x1 
1 1 1
15. (C) Initial energy Ui  C1V12  C2 V22 , Final energy Uf  (C1C2 )V 2
2 2 2
C1V1  C2 V2
(where V  )
C1C2
C1C2
Hence energy loss U  Ui  Uf  (V1  V2 )2 .
2(C1  C2 )
V1
16. (C) Charges developed are same so C1V1  C2 V2  2
V2
V1  V2  120  V1  80 volts.

4
Capacitor

1
17. (C) U  CV 2
2
Now if V is constant, then U is greatest when 'Ceq' is maximum. This is when all the three are in
parallel.

V1C1  V2C2 500  20  200  10

18. (C) V    400 V
C1  C2 20  10
1 1
19. (C) U  CV 2   2  (200)2  106  0.04 J .
2 2