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 @Lets_Learn_Channel
Pre-Medical : Physics
ALLEN
CAPACITOR
CAPACITOR & CAPACITANCE
A capacitor consists of two conductors carrying charges of equal magnitude and opposite sign. The capacitance
C of any capacitor is the ratio magnitude of the charge Q on either conductor to the potential difference V between
Q
them C =
V

The capacitance depends only on the geometry of the conductors and not charge given to conductor or potential
difference.
CAPACITANCE OF AN ISOLATED S PHERICAL CONDUCTOR
C = 4p Î0Îr R in a medium C = 4 p Î0 R in air

S PHERICAL CAPACITOR :

B
A
a
C1
b
C2

PARALLEL PLATE CAPACITOR :

(i) UNIFORM DI-ELECTRIC M EDIUM :If two parallel plates each of area A & separated by a distance d are
charged with equal & opposite charge Q, then the system is called a parallel plate capacitor & its capacitance
Î0 Îr A Î A
is given by, C = in a medium; C = 0 with air as medium
d d

This result is only valid when the electric field between plates of capacitor is constant.

Î0 A
(ii) M EDIUM P ARTLY AIR : C =
æ tö
d - çt - ÷
è Îr ø
Z:\NODE02\B0AI-B0\SPARK KOTA\PHYSICS\ENG\CLASS_XII\15-CAPACITOR.P65

d
t

Îr Î0

When a di-electric slab of thickness t & relative permittivity Îr is introduced between the plates of an air
æ t ö
capacitor, then the distance between the plates is effectively reduced by ç t - Î ÷ irrespective of the position
è r ø
of the di-electric slab .

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Pre-Medical : Physics
ALLEN
(iii) COMPOSITE M EDIUM :
d
Îr1 Îr2 Îr3

Î0 A
C=
t1 t2 t
+ + 3
Îr1 Îr2 Îr3
t1 t2 t3
COMBINATION O F CAPACITORS :
(i) CAPACITORS IN S ERIES :In this arrangement all the capacitors when uncharged get the same charge Q but
the potential difference across each will differ (if the capacitance are unequal).
Then series conbination the potential difference capacitor will be inversely proportional to its capacitance.
Q Q Q

C1 C2 C3
1 1 1 1 1
= + + + ..... + V1 V2 V3
C eq. C1 C 2 C 3 Cn

(ii) CAPACITORS IN P ARALLEL : When one plate of each capacitor is connected to the positive terminal of the
battery & the other plate of each capacitor is connected to the negative terminals of the battery, then the
capacitors are said to be in parallel connection. The capacitors have the same potential difference, V
but the charge on each one is different (if the capacitors are unequal). C eq. = C1 + C2 + C3 + ...... + Cn.
In parallel combination charge on capacitor will be distributed in the ratio of their capacitance.

Q1 C1,V

Q2 C2,V

Q3 C3,V Q = Q 1 + Q2 + Q3
Q V

E NERGY S TORED IN A CHARGED CAPACITOR :

Capacitance C, charge Q & potential difference V; then energy stored is

1 1 1 Q2
U= CV2 = QV =
2 2 2 C

This energy is stored in the electrostatic field set up in the di-electric medium between the conducting
plates of the capacitor.

1
Z:\NODE02\B0AI-B0\SPARK KOTA\PHYSICS\ENG\CLASS_XII\15-CAPACITOR.P65

Energy density u = Î0 E 2 , E is electric field between plate of capacitor

2

HEAT PRODUCED IN SWITCHING IN CAPACITIVE CIRCUIT :

Due to charge flow always some amount of heat is produced when a switch is closed in a circuit which can be
obtained by energy conservation as –
Heat = Work done by battery – Change in energy of capacitor.

Q2
w Work done by battary to charge a capacitor W = CV 2 = QV =
C

S HARING OF CHARGES :
When two charged conductors of capacitance C1 & C2 at potential V1 & V2 respectively are connected by
a conducting wire, the charge flows from higher potential conductor to lower potential conductor, until the
potential of the two condensers becomes equal. The common potential (V) after sharing of charges;
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Pre-Medical : Physics
ALLEN
(i) When the positive plate of one connected to positive plate of other then common potential.

net ch arg e q + q 2 C1V1 + C 2 V2

V= = 1 =
net capaci tan ce C1 + C 2 C1 + C2 .

Charges after sharing q1 = C1V & q2 = C2V. In this process energy is lost in the connecting wire as heat.

C1 C 2
This loss of energy is U initial - U final = (V - V2)2 .
2 ( C1 + C 2 ) 1

C1V1 - C 2 V2
(ii) When positive of one with negative of other then common potential. VC =
C1 + C 2

q1 C1
After connection ratio of charge would be in ratio of capacitance q = C
2 2

æ s ö Q2
• Attractive force between capacitor plate F = ç ÷ ( sA ) =
è 2 Î0 ø 2 Î0 A

• Charging of a capacitor : q = q 0 (1 - e - t / RC ) where q 0 = CV0

q R
q = charge at any time
C

t = RC is time constant of circuit

V0

• Discharging of a capacitor : q = q 0e –t/RC

q
q = charge at any time
C

q0 = initial charge
t = RC = time constant of circuit
R
KEY POINTS
• The energy of a charged conductor resides outside the conductor in its electric field, where as in a
condenser it is stored within the condenser in its electric field.
Z:\NODE02\B0AI-B0\SPARK KOTA\PHYSICS\ENG\CLASS_XII\15-CAPACITOR.P65

• The energy of an uncharged condenser = 0.

• The capacitance of a capacitor depends only on its size & geometry & the di-electric between the conducting
surface. (i.e. independent of the nature of material of conductor, like, whether it is copper, silver, gold etc)
• The two adjacent conductors carrying same charge can be at different potential because the conductors may
have different sizes and hence different capacitance.
• When a capacitor is charged by a battery, both the plates received charge equal in magnitude, no matter sizes
of plates are identical or not because the charge distribution on the plates of a capacitor is in accordance with
charge conservation principle.
• On filling the space between the plates of a parallel plate air capacitor with a dielectric, capacity of the
capacitor is increased because the same amount of charge can be stored at a reduced potential.
• The potential of a grounded object is taken to be zero because capacitance of the earth is very large.
If an isolated conductor is brought near charged conductor then it increases the capacity of system. This is
principle of parallel plate capacitor.

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Pre-Medical : Physics
ALLEN
Illustration The plates of a capacitor are charged to a potential difference of 100 V and then connected
across a resistor. The potential difference across the capacitor decays exponentially with respect
to time. After one second. the potential difference between the plates of the capacitor is 80
V. What is the fraction of the stored energy which has been dissipated ?
1 1
Solution Energy losses DU = CV02 – CV2
2 2

1 1
CV02 - CV2 V2 - V2
DU 2 2
Fractional energy loss = = 0 2
U0 1 V0
CV02
2

(100)2 - (80)2 20 ´ 180 9

= = =
(100)2 (100)2 25
Illustration Two uniformly charged spherical drops each at a potential V coalesce to form a larger drop.
If the capacity of each smaller drop is C then find the capacity and potential of larger drop.
Solution When drops coalesce to form a larger drop then total charge and volume remains conserved.
If r is the radius and q is the charge on smaller drop then C = 4pe0r and q = CV
4 3 4 3 1/3
Equating volume we get pR = 2 × pr Þ R = 2 r
3 3
1/3
Capacitance of larger drop C' = 4pe0R = 2 C
Charge on larger drop Q = 2q = 2CV
Q 2CV 2/3
Potential of larger drop V' = = = 2 V..
C' 21 / 3
C
Illustration The stratosphere acts as a conducting layer for the earth. If the stratosphere extends beyond
50 km from the surface of earth, then calculate the capacitance of the spherical capacitor formed
between the stratosphere and earth's surface. Take radius of earth as 6400 km.

Solution The capacitance of a spherical capacitor is C = 4pe0 æç ab ö÷

è b - aø
6
b = radius of the stratosphere layer = 6400 km + 50 km = 6450 km = 6.45 × 10 m
6
a = radius of earth = 6400 km = 6.4 × 10 m

1 6.45 ´ 106 ´ 6.4 ´ 106

\ C= ´ = 0.092 F
9 ´ 10 (6.45 ´ 106 - 6.4 ´ 106 )
9

Illustration If the distance between the plates of a capacitor of capacitance C1 is halved and the area of
plates is doubled then what will be the capacitance ?
e0 A C1 A1 d2 A æ 1 öæ d ö
Z:\NODE02\B0AI-B0\SPARK KOTA\PHYSICS\ENG\CLASS_XII\15-CAPACITOR.P65

Solution C= r = = 1 ´ ç ÷ ç 1 ÷ = 1 rC = 4C
d C2 A2 d1 2A1 è d1 ø è 2 ø 4 2 1

Illustration A parallel-plate capacitor is formed of two plates, each of area 100 cm , separated by a distance
2

of 1mm. A dielectric of dielectric constant 5.0 and dielectric strength 1.9 × 107 V/m is introduced
between the plates. Find the maximum charge that can be stored in the capacitor without causing
any dielectric breakdown.
Solution If the charge on the capacitor = Q

Q Q
the surface charge density s = and the electric field = .
A KAe 0
This electric field should not exceed the dielectric strength 1.9 × 107 V/m.

Q
\ if the maximum charge which can be given is Q then = 1.9 ´ 107 V/m
KAe0
Q A = 100 cm2 = 10–2 m2 Þ Q = (5.0) × (10–2) × (8.85 × 10–12) × (1.9 × 107) = 8.4 × 10–6 C.

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Pre-Medical : Physics
ALLEN
Illustration The distance between the plates of a parallel-plate capacitor is 0.05 m. A field of 3 × 104 V/
m is established between the plates. It is disconnected from the battery and an uncharged metal
plate of thickness 0.01 m is inserted into the gap between the plates. Find the potential difference
of the capacitor (i) before the introduction of the metal plate and (ii) after its introduction. What
would be the potential difference if a plate of dielectric constant K = 2 is introduced in place
of metal plate ?
Solution
(i) In case of a capacitor as E = (V/d), the potential difference between the plates before the
introduction of metal plate
V = E × d = 3 × 104 × 0.05 = 1.5 kV
(ii) Now as battery is removed after charging, capacitor is isolated so q = constant. If C' and

C
V' are the capacity and potential after the introduction of plate then, q = CV = C'V' i.e., V ' = V
C'

e0 A e0 A (d - t) + (t / K)
And as C = and C' = , V' = V
d (d - t) + (t / K) d

éd - tù é 0.05 - 0.01 ù
So in case of metal plate as K = ¥, VM = ê ú V=ê ú 1.5 = 1.2 kV
ë d û ë 0.05 û

And if instead of metal plate a dielectric with K = 2 is

é (0.05 - 0.01) + (0.01/ 2) ù

introduced then VD = ê ú ´ 1.5 = 1.35 kV .
ë 0.05 û

Illustration Twenty seven charged water droplets, each of radius 10–3 m and having a charge of 10–12 C,
coalesce to form a single drop. Calculate the potential of the bigger drop.

4 4
Solution Volume of bigger drop=N ×volume of smaller drop Þ p R3 =N × p r3 Þ R3 = Nr3 Þ R= N1 3r
3 3

R = (27)1/3 r = 3r

9 –12
kQ kNq 9 ´ 10 ´ 27 ´ 10
Potential of bigger drop V = = = –3 = 81 volts
R 3r 3 ´ 10
Z:\NODE02\B0AI-B0\SPARK KOTA\PHYSICS\ENG\CLASS_XII\15-CAPACITOR.P65

Illustration What is the effective capacitance between the points X and Y ?

6mF
6mF 6mF 6mF
B
X D Y
6mF

Solution This is a balanced Wheatstone bridge, so BD can be removed. equivalent circuit

B
6 mF and 6 mF are in series so equivalent capacitance = 3 mF 6mF 6mF

then 3 mF and 3 mF are in parallel X 6mF F

so equivalent capacitance between X and Y is 6mF.
6mF 6mF
D
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Pre-Medical : Physics
ALLEN
Illustration What is the the equivalent capacitance of the combination ?

C
C
C

Solution The capacitor shown as a vertical element is shorted, so it can be removed.

C
Þ Ceq = C + C = 2C

C
R T
Illustration Twelve identical capacitors each of capacitance C are connected as shown in figure.
Q S
Find the effective capacitance between P and T.
U
W
Solution P V
T

2Q
2Q 2Q

P 2Q Q

6Q E

2Q Q 2Q 5Q 6Q 6C
E= + + = , Ceff = = .
C C C C E 5

Illustration Find the equivalent capacitance of the infinite ladder shown in figure between the points A & B.

2 mF 2 mF 2 mF 2 mF 2 mF
A

1 mF 1 mF 1 mF 1 mF 1 mF

Solution Let equivalent capacitance be x.

2µF
Z:\NODE02\B0AI-B0\SPARK KOTA\PHYSICS\ENG\CLASS_XII\15-CAPACITOR.P65

A
2x
x = +1
2+x x 1µF x

2x + 2 + x 2
x = Þ x(2 + x) = 3x + 2 Þ 2x + x = 3x + 2
2+ x B

2
Þx – x – 2 = 0

–b ± b2 - 4ac 1± 1+ 8 1± 3
Using x = = = = 2 and –1
2a 2 2

x = 2, Ceq = 2 µF (Q C ¹ –ve)

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ALLEN
Illustration

There are six plates of equal area A and separation between the adjacent plates is d or
2d (d<<A). They are arranged as shown in figure. Find the equivalent capacitance between
points 2 and 5.

Solution Redrawing the circuit

1, 4

2 5

3, 6

It is a wheatstone bridge with points (3, 6) and (1, 4) being equipotential. So, the capacitance
C/2 can be removed.

\ Ceq = C =
Î0 A
d
Illustration Calculate the potential of point O in terms of C1, C2, C3, VA, VB, & VC in the following curcuit.

Z:\NODE02\B0AI-B0\SPARK KOTA\PHYSICS\ENG\CLASS_XII\15-CAPACITOR.P65

Solution Let the potential of the junction O be V0. Now apply Kirchhoff's current law at a junction.

C1VA + C2 VB + C3 VC
C1 (VA–V 0) + C 2 (VB –V0) + C 3 (VC –V0 ) = 0 Þ V0 = .
C1 + C2 + C3

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 @Lets_Learn_Channel
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ALLEN
Illustration A parallel plate capacitor with no dielectric has a capacitance of 0.5 mF. Half of the space
between the plates is filled with a medium of dielectric constant 2 and remaining half is filled
with a medium of dielectric constant of 3 as shown in figure. Find its net capacity.

K=2

K=3

Î0 A
Solution Given that original capacitance C = = 0.5 mF
d

2 Î0 A / 2 Î0 A
Capacitance of part with dielectric constant 2 is C1 = = =0.5 mF
d d

3 Î0 A / 2 3 Î0 A
Capacitance of part with dielectric constant 3 is C2 = = =0.75 mF
d 2d

As both the capacitors are connected in parallel

so Ceq = C1 + C2 = 0.5 + 0.75 = 1.25 mF
Illustration In the given circuit calculate the potential difference across the
2 mF capacitor in the steady state condition, if internal resistance
of battery is 1 ohm.

4 ,
Solution Current in the steady state condition = = 1A
2+1+1

Potential difference between x and y = 3 V.

Divide this 3V in inverse ratio of capacity

1
Voltage on 2 µF capacitor = × 3 = 1V..
2+1 ,

Illustration In the given circuit find the :-

(a) initial current supplied by the battery.
Z:\NODE02\B0AI-B0\SPARK KOTA\PHYSICS\ENG\CLASS_XII\15-CAPACITOR.P65

(b) final current supplied by the battery.

Solution
(a) Initially capacitor behaves (b) Finally it behaves as open circuit resistance wire
just like a zero resistance wire
Total resistance R = 5W Total resistance R = 7.5 W

5 5 2
I = = 1A I = = A
5 75
. 3

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CAPACITOR EXERCISE
1. The earth has Volume 'V' and Surface area 'A' ; then 5. Two condensers, one of capacity 2C and the other
its capacitance would be : C
of capacity , are connected to a V-volt battery,
2
A V as shown.
(1) 4pÎ0 (2) 4pÎ0
V A
2C
V A
(3) 12 pÎ0 (4) 12 pÎ0
A V
The work done by battery in charging fully both the
condensers is :-
2. There are two metallic plates of a parallel plate
capacitor. One plate is given a charge +q while the 1
(1) CV2 (2) 2 CV2
other is earthed as shown. Points P, P1 and P2 are 2
taken as shown in adjoining figure. Then the electric 5 3
intensity is not zero at : (3) CV2 (4) CV2
2 2
6. A series combination of n 1 capacitors, each of value
C1, is charged by a source of potential difference
8V. When another parallel combination of n 2
capacitors, each of value C2, is charged by a source
P2 P P1
of potential difference 2V, it has the same (total)
energy stored in it, as the first combination has.
The value of C2, in terms of C1, is then :-

16C1 2C1
(1) n n (2) n n
(1) P only 1 2 1 2

(2) P1 only n2 n2
(3) 16 n C1 (4) 2 n C1
(3) P2 only 1 1

7. Two parallel metal plates having charges + Q

(4) P, P1 and P2 and – Q face each other with a certain separation
between them. If the plates are now dipped in
3. Two spherical conductors A and B of radius a and b kerosene oil tank, the electric field between the
(b > a) are placed in air concentrically. B is given a plates will :-
charge + Q coulombs and A is grounded. The (1) increase (2) decrease
equivalent capacitance of these is :
(3) remain same (4) become zero
ab 8. Find energy stored between the plates of parallel
(1) 4pÎ0
(b - a)
Z:\NODE02\B0AI-B0\SPARK KOTA\PHYSICS\ENG\CLASS_XII\15-CAPACITOR.P65

plate capacitor of area A and separated by a

(2) 4pÎ0 (a + b) distance d, if electric field between its plates is E,
(3) 4pÎ0b also find energy required to charge it :-
b2 1 1 1
(4) 4pÎ0 (1) e0 E2 Ad, e0 E2 Ad (2) e0 E 2 Ad, e0 E2 Ad
(b - a) 2 2 2
4. A parallel plate air capacitor is charged to a 1 d e E2A 1 Ad Ad
potential difference of V volts. After disconnecting (3) e0 2 , 0 (4) 2 e E 2 , e E2
2 E A d 0 0

the charging battery the distance between the plates 9. A 300 pF capacitor is charged with a 100 V battery.
of the capacitor is decreased using an insulating After disconnecting the battery this capacitor is
handle. As a result the potential difference between connected with another uncharged 300 pF
the plates :- capacitor. Then the energy loss is :-
(1) decreases (2) does not change (1) 25 × 10–8 J (2) 50 × 10–8 J
(3) becomes zero (4) increases (3) 75 × 10–8 J (4) 100 × 10–8 J
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10. Two thin dielectric slabs of dielectric constants K1 14. A 3 mF capacitor is charged to a potential of 100 V and
and K2 (K1 > K2) are inserted between the plates 2 mF capacitor is charged to 100 V. The capacitors
of a parallel plate capacitor, as shown in the figure. are then connected in parallel with plates of opposite
The variation of electric field 'E' between the plates polarities joined together. What amount of charge will
with distance 'd' as measured from plate P is flow, when the plates are so connected?
correctly shown by :- (1) 1300 mC (2) 800 mC
P + – Q
+ – (3) 600 mC (4) 240 mC
+ –
+ – 15. An air capacitor of capacity C = 10 mF is connected
+ –
+ – to a constant voltage battery of 12 V. Now the space
+ –
+ – between the plates is filled with a liquid of dielectric
+ –
K1 K2 constant 3. The additional charge that flows now,
from the battery to the capacitor is :
­ ­
E E (1) 120 mC (2) 600 mC
(1) (2) (3) 480 mC (4) 240 mC
0 0
d® d® 16. In a capacitor of capacitance 20 mF the distance
­ ­ between the plates is 2 mm. If a dielectric slab of
E E width 2 mm and dielectric constant 2 is inserted
(3) (4) between the plates, then the new capacitance will
0 be :
d® 0
d®
(1) 22 mF (2) 26.6 mF (3) 52.2 mF (4) 40 mF
11. A parallel plate air capacitor of capacitance C is
connected to a cell of emf V and then disconnected
from it. A dielectric slab of dielectric constant K, 17. Two capacitors C1 = 4 mF and C2 = 4 mF in series,
which can just fill the air gap of the capacitor, is now are connected in parallel to a third capacitor
inserted in it. Which of the following is incorrect ? C3 = 4 mF. This arrangement is then connected to a
battery of e.m.f. = 4 V, as shown in the figure. How
(1) The energy stored in the capacitor decreases
much energy is given by the battery in charging the
K times.
capacitors ?
1 2æ 1 ö
(2) The change in energy stored is CV ç – 1 ÷ . C1 C2
2 èK ø
(3) The charge on the capacitor is not conserved. C3
(4) The potential difference between the plates
decreases K times.
12. A parallel plate air capacitor has capacity 'C' and 4V
separation between the plates is 'd'. A potential (1) 96 ´ 10–6 J (2) 48 ´ 10–6 J
difference 'V' is applied between the plates. Force
FG 32IJ FG 16 IJ ´ 10
Z:\NODE02\B0AI-B0\SPARK KOTA\PHYSICS\ENG\CLASS_XII\15-CAPACITOR.P65

of attraction between the plates of the parallel plate

air capacitor is :-
(3) H3K ´ 10–6 J (4) H3K –6
J

d
C2 V2 C2 V2 CV 2 CV 2 18. A slab of copper of thickness is introduced
(1) (2) (3) (4) 3
2d 2 2d 2d d
between the plates of a parallel plate capacitor
13. A capacitance of value 8 mF charged to 50 V is where d is the separation between its two
connected with another capacitance of value 4 mF plates. If the capacitance of the capacitor
charged to 100 V, in such a way that plates of similar without the copper slab is C and with copper
charges are connected together. The total energy C'
in multiples of 10–2 J before joining and after joining slab is C' then is :-
C
will be :
(1) 1.5 and 1.33 (2) 1.33 and 1.5 3 1
(1) 2 (2) 2 (3) (4)
2 2
(3) 3.0 and 2.67 (4) 2.67 and 3.0
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19. Three parallel plates of the same metal and same 24. Distance between the plates of a parallel plate
area are placed between the plates of a parallel capacitor is 'd' and area of each plate is A. When
plate capacitor of capacity C, If the thickness of each a slab of dielectric constant K and thickness t is placed
between the plates, its capacity becomes:
1
plate is equal to th of the distance between the
5 e0 A e0 A
(1) (2)
plates of the original capacitor then the capacity of é ì 1 üù é ì 1 üù
ê d + t í1 - K ý ú ê d + t í1 + K ý ú
the new capacitor is :- ë î þû ë î þû

5 5 3 10
(1) C (2) C (3) C (4) C e0 A e0 A
3 2 10 3 (3) (4)
é ì 1 üù é ì 1 üù
20. When a capacitor of value 200 mF charged to 200V ê d - t í1 + K ý ú ê d - t í1 - K ý ú
ë î þû ë î þû
is discharged separately through resistance of 2
ohms and 8 ohms, then heat produced in joule will
25. A parallel plate capacitor is connected with a battery
respectively be:
whose potential difference remains constant . If the
(1) 4 and 16 (2) 16 and 4 plates of the capacitor are shifted apart then the
(3) 4 and 8 (4) 4 and 4 intensity of electric field :
(1) decreases and charge on plates also decreases.
21. The charge q on a capacitor varies with voltage as
shown in figure. The area of the triangle AOB (2) remains constant but charge on plates decreases.
represents : (3) remains constant but charge on the plates
increases.
A (4) increases but charge on the plates decreases.
V 26. A parallel plate air capacitor
has a capacitance C. When it
is half filled with a dielectric of
dielectric constant 5, the
B percentage increase in the
O q d
capacitance will be :-
(1) electric field between the plates (1) 400% (2) 66.6% (3) 33.3% (4) 200%
(2) electric flux between the plates
27. Two conducting spheres of radii R1 and R2 are
(3) energy density charged with charges Q1 and Q2 respectively. On
(4) energy stored by the capacitor. bringing them in contact there is :
22. A charged parallel plate capacitor of distance (d) has
(1) no change in the energy of the system
U0 energy. A slab of dielectric constant (K) and
thickness (d) is then introduced between the plates (2) an increase in the energy of the system if
Z:\NODE02\B0AI-B0\SPARK KOTA\PHYSICS\ENG\CLASS_XII\15-CAPACITOR.P65

of the capacitor. The new energy of the system is Q1R2 ¹ Q2R1

given by :
(3) always a decrease in the energy of the system
U U0
(1) KU0 (2) 0K2U (3) 0 (4) 2 (4) a decrease in the energy of the system if
K K
23. The energy and capacity of a charged parallel plate Q1R2 ¹ Q2R1
capacitor are U and C respectively. Now a dielectric
28. A capacitor is connected to a 10 V battery. The
slab of Îr=6 is inserted in it then energy and
charge on plates is 40 mC when medium between
capacity becomes : (Assuming charge on plates
plates is air. The charge on the plates become
remains constant)
100 mC when the space between the plates is filled
(1) 6U, 6C (2) U, C with oil. The dielectric constant of oil is :

U (1) 2.5 (2) 4 (3) 6.25 (4) 10

(3) , 6C (4) U, 6C
6

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29. A 40 µF capacitor in a defibrillator is charged to 33. The electrostatic force between the metal plates of
3000 V. The energy stored in the capacitor is sent an isolated parallel plate capacitor C having a
through the patient during a pulse of duration
charge Q and area A, is :-
2 ms. The power delivered to the patient is :
(1) independent of the distance between the plates.
(1) 45 kW (2) 90 kW (2) linearly proportional to the distance between the
plates
(3) 180 kW (4) 360 kW
(3) proportional to the square root of the distance
30. A capacitor of 2µF is charged as shown in the between the plates.
(4) inversely proportional to the distance between
diagram. When the switch S is turned to position 2,
the plates.
the percentage of its stored energy dissipated is: 34. Three capacitors each of capacitance C and of
1 2 breakdown voltage V are joined in series. The
capacitance and breakdown voltage of the
S combination will be :-
V C V
2µF 8µF (1) 3C, 3V (2) ,
3 3
C
(1) 0% (2) 20% (3) 75% (4) 80% (3) 3C, V (4) , 3V
3
31. A parallel-plate capacitor of area A, plate C1
35. C3
separation d and capacitance C is filled with four
dielectric materials having dielectric constants k1, k2, C2
k3 and k4 as shown in the figure below. If a single V
dielectric material is to be used to have the same Potential difference across C1 is:

capacitance C in this capacitor, then its dielectric (C1 – C2 )V C3 V

(1) C + C + C (2) C + C + C
constant k is given by :- 1 2 3 1 2 3

A/3 A/3 A/3

(C1 + 2C2 )V (C1 – 2C2 )V
(3) (4) C + C + C
C1 + C 2 + C 3 1 2 3
d/2

d 36. Seven capacitors each of capacitance 2 mF are to

be connected so as to have a total capacity of
8
A mF . Which will be the combination shown ?
13
2 3 1
(1) k = k + k + k + k
Z:\NODE02\B0AI-B0\SPARK KOTA\PHYSICS\ENG\CLASS_XII\15-CAPACITOR.P65

1 2 3 4

1 1 1 1 3 (1)
(2) k = k + k + k + 2k
1 2 3 4

(3) k = k1 + k2 + k3 + 3k4
2
(4) k = (k + k2 + k3) + 2k4
3 1 (2)
32. A capacitor is charged by a battery. The battery is
removed and another identical uncharged capacitor
(3)
is connected in parallel. The total electrostatic
energy of resulting system :-
(1) Decreases by 50% (2) Remains the same (4)
(3) Increases by 50% (4) Increases by 200 %
E 33
 @Lets_Learn_Channel
Pre-Medical : Physics
ALLEN
37. An infinite number of identical capacitors each of (1) Q1 = Q2 = Q3 and V1 = V2 = V3 = V
capacitance 8 mF are connected as in the adjoining
(2) Q1 = Q2 + Q3 and V = V1+ V2 + V3
figure. Then the equivalent capacitance between A
and B is : (3) Q1 = Q2 + Q3 and V = V1 + V2
(4) Q2 = Q3 and V2 = V3
(Symbols have their usual meanings)
42. The effective capacitance between the points P and
8 capacitors Q of the arrangement shown in the figure is :
16 capacitors
A B 2 mF 2 mF 1 mF
¥ P Q
1 2 mF 5 mF
(1) 16 mF (2) 8 mF (3) mF (4) ¥
2
38. In the given circuit if point C is connected to the 2 mF 1 mF
earth and a potential of + 2000 V is given to point (1) (1/2) mF (2)1 mF
A, then potential at B is :
(3) 2 mF (4) 1.33 mF
5mF 20mF 20mF
C 43. Two spheres of radii R1 and R2 having equal charges
A are joined together with a copper wire. If V is the
B potential of each sphere after they are separated
10mF from each other, then the initial charge on both
(1) 1500 V (2) 1000 V (3) 500 V (4) 400 V spheres was :
39. Minimum number of 8 mF and breakdown voltage
V V
250 V capacitors required to make a combination (1) ( R1 + R2 ) (2) ( R1 + R2 )
k 2k
of 8 mF and 1000 V are :
(1) 32 (2) 16 (3) 8 (4) 4
V V ( R1R 2 )
40. A 10 volts battery is connected to three capacitors; (3) ( R1 + R 2 ) (4) k R + R
( 1 2)
3k
C1 = 2 µF , C2 = 4 µF and C3 = 6 µF, as shown.
The charges on the capacitors C1, C2 and C3 are
44. Half of the space between a parallel plate capacitor
respectively :
is filled with a medium of dielectric constant K
C2 parallel to the plates. If initially the capacity was C,
C1
+ then the new capacity will be :
10V–
C3 (1) 2KC/(1+K) (2) C (K+1)/2
(3) CK/(1+K) (4) KC
(1) 2 µC, 3 µC, 5 µC
Z:\NODE02\B0AI-B0\SPARK KOTA\PHYSICS\ENG\CLASS_XII\15-CAPACITOR.P65

(2) 5 µC, 10 µC, 15 µC 45. Two capacitances C1 and C2 in a circuit are joined
as shown in figure The potential of point A is V1
(3) 10 µC, 15 µC, 25 µC
and that of B is V2. The potential of point D will
(4) 10 µC, 20 µC, 30 µC
be :
41. In an adjoining figure three capacitors C1, C2 and
C3 are joined to a battery. The correct condition will A D B
be : V1 C1 C2 V2
V2C2Q2
V1C1Q1 1 C2 V1 + C1 V2
(1) (V1+V2) (2)
2 C1 + C 2

V3C3Q3
C1 V1 + C2 V2 C2 V1 - C1 V2
(3) (4)
V C1 + C 2 C1 + C2

34 E
 @Lets_Learn_Channel
Pre-Medical : Physics
ALLEN
46. Two capacitors A and B are connected in series 49. The voltage drop across a capacitor connected with
a resistance and a battery of 30 V in series, after
with a battery as shown in the figure. When the
a long time is :-
switch S is closed and the two capacitors get charged (1) 0 V (2) 60 V (3) 30 V (4) 38 V
fully, then : 50. Current drawn by the battery long time after switch
'S' is closed is :-
2mF 3mF R C

A B
V R R

10V S S
(1) the potential difference across the plates of A is 2V 3V 2V V
(1) (2) (3) (4)
4 V and across the plates of B is 6 V 3R 2R R 2R
(2) the p.d. across the plates of A is 6 V and across 51. In the given figure the steady state current I is :
the plates of B is 4 V 3W

(3) the ratio of electrical energies stored in A and B 3W

is 2 : 3
C=0.2 mF
(4) the ratio of charges on A and B is 3 : 2
4W
I
47. The charge flowing through the cell on closing the
key k is equal to :
6V 1.5 W
k

C
(1) zero (2) 0.6 A (3) 0.9 A (4) 2 A
C C 52. A capacitor of capacitance 5 mF is connected as
C
shown in the figure. The internal resistance of the
V
cell is 0.5 W. The amount of charge on the capacitor
CV 4 3
(1) (2) 4CV (3) CV (4) CV plate is :
4 3 4

48. Figure shows two capacitors connected in series 1W 1W

and joined to a battery. The graph shows the 5mF 2W
variation in potential as one moves from left to right 2.5V
along the branch containing the capacitors. Then:
Z:\NODE02\B0AI-B0\SPARK KOTA\PHYSICS\ENG\CLASS_XII\15-CAPACITOR.P65

(1) 0 (2) 5 mC 3) 10 mC (4) 25 mC

V
53. In the circuit shown in figure, the battery is an ideal
one with emf V. The capacitor is initially uncharged.
Switch S is closed at time t = 0.
C1 C2 X R/2 A 5R/2
S
(1) C1 > C2 (2) C1 = C2 V R C
(3) C1 < C2 (4) The information is R/2
B
not sufficient to decide the relation between C1 and
The final charge Q on the capacitor is :
C2 CV CV CV
(1) (2) (3) CV (4)
2 3 6
E 35
 @Lets_Learn_Channel
Pre-Medical : Physics
ALLEN
54. In previous question, what is the current in the steady 57. When the key is pressed at time t = 0 then which of
state is ?
the following statements about the current i in the
V V 2V V
(1) (2) (3) (4) resistor AB of the given circuit is true ?
2R R R 3R
55. A bulb, a capacitor and a battery are connected K A B
together as shown here, with switch S initially open. 2V 1000 W
When the switch S is closed, which one of the
i 1 mF C 1000 W
following is true ?

(1) at t = 0, i = 2 mA and it reduces to 1 mA at t = ¥

S (2) i oscillates between 1 mA and 2 mA
(1) The bulb will light up for an instant when the (3) i = 2 mA at all time
capacitor starts charging. (4) i = 1 mA at all time
(2) The bulb will light up when the capacitor is fully 58. In an isolated parallel plate capacitor of capacitance
charged
C, the four surfaces have charges Q1, Q2, Q3, and
(3) The bulb will not light up at all
Q4 as shown. The potential difference between the
(4) The bulb will light up and go off at regular
intervals. plates is :
56. Two charged spheres having radii a and b are joined Q1 + Q2 + Q3 + Q 4
(1) Q1 Q3
with a wire then the ratio of electric field Ea/Eb on 2C
their respective surfaces is : Q2 + Q3 Q2 Q4
(1) a/b (2) b/a (3) a2/b2 (4) b2/a2 (2)
2C
Q2 – Q3
(3)
2C
Q1 + Q4
(4)
2C

Z:\NODE02\B0AI-B0\SPARK KOTA\PHYSICS\ENG\CLASS_XII\15-CAPACITOR.P65

ANSWER KEY
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 3 1 4 1 3 1 2 1 3 2 3 3 3 4 4
Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Ans. 4 1 3 2 4 4 3 3 4 1 4 4 1 2 4
Que. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
Ans. 1 1 1 4 2 2 1 4 2 4 3 2 2 1 3
Que. 46 47 48 49 50 51 52 53 54 55 56 57 58
Ans. 2 1 3 3 4 4 3 1 1 1 2 1 3

36 E
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