CAPACITANCE
CAPACITANCE
1.
CCCCCCCCCCC CC C CCCCCCCCC

A capcitor is a divice which stores electric energy. It is also named as condenser.

When charge is given to a conductor, its potential increases in the ratio of given charge.
The charge given to the conductor is directly proportional to its potential.
Q  v or Q = cv
where C is a constant called electrical capacitance of the conductor.
If V=1 volt, then Q = C; i.e., the electrical capacitance of the conductor is equal to that amount of charge
which increases the potential of the conductor by a unit amount.

The electrical capacitance of the conductor depends upon the following:
(a) The size of the conductor:
The capacitance is directly proportional to the area of the surface of the conductor, i.e., C  A (area)
(b) The medium around the conductor: If a conductor is placed in a medium other than air or vacuum, its
capacitance will increase.
C medium  KCair or vacuum
where K is a constant called dielectric constant of the medium. This constant is always greater than 1.

Theoretically the charge can be given to any isolated conductor upto an infinite amount. When a very large
amount of charge is given to the conductor, its potential increases to such an extent that dielectric breaks
down and the charge starts leaking from the object and electrical discharge takes place between the conductor
and nearby objects or earth. This decreases the potential of the conductor.

Unit of Capacitance:
(a) In M.K.S. system unit of capacitance is farad.
farad 
coulomb
volt
Dimensions of the capacitance are M 1L2 T 4 A 2
(b) If a charge of one coulomb given to a conductor increases the potential by one volt, then its capacitance
is one farad.
(c) In practice, submultiples of farad are used as units of capacitance.
1 micro farad = 1F  10 6 F
1 nano farad = 1nF  10 9 F
1 micro-micro farad or pico farad = 1F  1pF  10 12 F
(d)
Electrostatic unit of electrical capacitance is stat farad.
1 farad = 9 × 1011statfarad
CAPACITANCE
2.
CCCCCCCCCCC CC CCCCCCCCC CCCCCCCCC
When a charge Q is given to a spherical conductor of radius R, then

Q
The potential of the surface of the spherical conductor will be V  4  R
0

Electrical capacitance of spherical conductor in M.K.S. system will be
C
Q
Q

 4 0 R
V Q / 4 0 R
+
+
+
+
+
++
+ + +
+
+
+
R
+Q
+
+
+
++
++ ++ +
In C.G.S. system, C  R because K = 1 (in C.G.S unit)

The electrical capacitance of a spherical conductor is directly proportional to its radius i.e., C  R

The electrical capacitance of a spherical conductor does not depend on the charge given to a conductor.
3.
PPPPPPPPP PPPPPP PP P PPPPPPP PPPPPPPPP PP PPPPPP PPPPPP

The work done in charging a conductor is equal to its potential energy or the stored energy.

This energy resides in the form of energy of the electric field created.

The potential energy of a charged conductor:
Q

U  Vdq 
0
Q
1
Q2
1
1
qdq 
U  QV  CV 2
C
2C or
2
2
0


Unit of U is joule

Energy stored in the conductor depends upon the given charge, its potential and its capacitance.

Stored energy depends upon the capacitance of the capacitor and the given charge or the potential difference.
It does not depend upon the shape of the capacitor.

If the area of the plates are A and the thickness of dielectric constant is d, then energy stored per unit volume
i.e., energy density of the medium will be
u
1
1
 E 2  0r E 2
2
2
where E is the electric field.

The energy of the charged capacitor resides in the electric field between its plates.
Note : In charging a capacitor by battery half the energy supplied is stored in the capacitor and remaining
1

half energy  QV  is lost in the form of heat.
2


4.
RR RR RR RR RR RR RR R R RR RR RR R R RR RR RR R RR R RR RR RR R RR RR RR RR RR R RR R RR RR R RR RR

When two charged conductors are joined together by a conducting wire of negligible capacity, the charge flows
from higher potential to lower potential.

When the potentials of both conductors become equal, the flow of charge stops.

Law of conservation of charge holds good in the process i.e., total charge on the two conductors will be same
after redistribution.

Let the amounts of charge on the conductors be Q1 and Q 2 and their electrical capacitances be C1 and
C 2 respectively. If their potentials are V1 and V2 , then
CAPACITANCE
Q1  C1V1 and Q 2  C 2 V2
Total charge Q  Q1  Q 2
 C1V1  C 2 V2
Total capacitance C  C1  C 2

Q2
Q1
+
+
+
+
+
+ +
++
+
+
C1
V1
+
++
+
+ ++
+
+
+
+
+
+
+
+
+
+
+
+
+ +
++
+
C2
V2
+
++
+ ++
+
+
+
+
+
+
+
On joining the conductors, the common potential becomes V, then
C1V1  C 2 V2
C1  C 2
V
The charge on the conductors after joining them will be
 Q1'
C C V  C 2 V2 
 C2 V  2 1 1
C1  C 2 
C1C 2
Charge transferred Q  C  C V1  V2 
1
2

+
+
+
+
+
+ +
++
+
+
V
C1
+
++
+
+ ++
+
+
+
+
+ +
++
+
V
+
+
+
+
C2
+
++
+ ++
+
+
+
+
+
+
+
The ratio of the amounts of charge after redistribution
Q1'
Q '2

Q '2
Q1'
((
(
(
((
Q '2
C C V  C 2 V2 
 C1V  1 1 1
C1  C 2 

C1
C2
In the process of redistribution of charge some work is to be done. Total amount of charge remains constant
in this process but their total potential energy decreases i.e., there is loss of energy. This loss of energy is
due to transformation of electrical energy into heat in the connecting wire and in electrical discharge.

Energy loss due to redistribution of charge
U  U initial  U final

1  C1C 2

2  C1  C 2

V1  V2 2 ,

which is always positive.
(a) If V1  V2 , then U  0 that is, there is no loss of energy. Hence no loss of energy occurs in joining the
conductors of equal potentials.
(b) If V1  V2 or V2  V1 , then V1  V2 2 will remain positive. Thus energy will be lost, i.e., there is always
loss of energy in joining the conductors at unequal potentials.
CAPACITANCE
E EE E E EE EE E E E E E E E EE E E E E EE E E EE E EE E EE E EE E EE E EE E E E EE E E EE E EE E E EE E
Ex1. A 20 capacitor is charged to potential of 500V and then connected in parallel to another capacitor of capacity
10F . If the potential of 10F capacitor is 200 volt then the common potential of two will be
[1] 100 V
[2] 200 V
[3] 300 V
[4] 400 V
C1V1  C 2 V2
20  10 6  500  10  10 6
 400 V
Sol. V = C  C
=
1
2
20  10  6  10  10  6
Hence the correct answer will be (4)
5.
CC CCC CCC CCC CC CCC CCC CCC C C C C CCCC CCC C C CCC CC
(i) Parallel plate capacitor : It consists of two parallel plates separated by a small distance d with a dielectric
material in between.
If A is area of each plate, d is the distance between the plates and r is the dielectric constant of the material
placed between the plates, then the capacitance of parallel plate capacitor will be
C
0 A
d
r
(for air or vacuum as medium)
A
where 0  8.85  10 12 F / m
and C 
0 r A
d
d
(for medium of dielectric constant r )
Note : (i) If two plates are placed side by side then three capacitors are formed. One between distant earthed
bodies and the first face of the first plate, the second between the two plates and the third between the second
face of the second plate and distant earthed objects. However the capacitances of the first and third capacitors
are negligibly small in comparison to that between the plates which is the main capacitance.

The capacitance of a capacitor depends upon the following:
(a) Area of the plates : The capacitance is directly proportional to the area of either plate, i.e., C  A .
(b) Distance between the plates : The capacitance of a capacitor decreases as the distance between the
plates increases i.e., C  1
d
(c) Dielectric medium between the plates : If any medium other than air or vacuum is between the plates,
the capacitance of the capacitor increases by r times where r is called relative permittivity or dielectric
constant of the medium.

On keeping the dielectric medium between the plates, the charge on the plates remains unchanged but the
potential difference between the plates decreases.
Note : (ii) It is a very common misconception that a capacitor stores charge but actually a capacitor stores
electric energy in the electrostatic field between the plates. The energy density of the field u =
1
 E2
2
CAPACITANCE
(iii) Two plates of unequal area can also form a capacitor because effective overlapping area is considered.
A
d
(iv) The distance between the plates is kept small to avoid fringing or edge effect (non-uniformity of the field)
at the boundaries of the plates.
Dielectric Medium and Dielectric Constant:
Note : (v) Dielectrics are insulating (non-conducting) materials which transmits electric effect without conducting
we know that in every atom, there is a positively charged nucleus and a negatively charged electron cloud
surrounding it. The two oppositely charged regions have their own centres of charge. The centre of positive
charge is the centre of mass of positively charged protons in the nucleus. The centre of negative charge is
the centre of mass of negatively charged electrons in the atoms/molecules.
(1) Type of Dielectrics : Dielectrics are of two types (i) Polar dielectrics : Like water, Alcohol, CO2, NH3,
HCl etc. are made of polar atoms/molecules
In polar molecules when no electric field is applied
centre of positive charges does not coincide with the
centre of negative charges.
O– –
(ii) Non polar dielectric : Like N2, O2, Benzene,
Met hane et c. ar e made of non- po la r
atoms/moelcules. In non-polar molecules, when no
electric field is applied the centre of positive charge
coincides with the centre of negative charge in the
molecule. Each molecule has zero dipole moment in
its normal state.
105º
–
+
H
H
+ +
–P=0
+
P
A polar molecule has permanent electric dipole
moment (p) in the absence of electric field also. But a
polar dielectric has net dipole moment is zero in the
absence of electric field because polar molecules are
randomly oriented as shown in figure.
+
+
-
E
–
+
-
+
-
+
+
In the presence of electric field polar molecules tends
to line up in the direction of electric field, and the
substance has finite dipole moment.
+
+
+
+
+
+
–
+
P
When electric field is applied, positive charge
experiences a force in the direction of electric field
and negative charge experiences a force in the
direction opposite to the field i.e., molecules
becomes induced electric dipole.
+
–
Note : In general, any non-conducting, material can
be called as a dielectric but broadly non conducting
material having non polar molecules referred to as
dielectric because induced dipole moment is created
in the non polar molecule.
CAPACITANCE
(2) Polarization of a dielectric slab : It is the process of inducing equal and opposite charges on the two
faces of the dielectric on the application of electric field.
Suppose a dielectric slab is inserted between the plates of a capacitor, as shown in figure.
E
++ –
–+ –
Ei
–+ –
–+ –
+
+
+
+
–+ –+
Induced electric field inside the dielectric is E1, hence this induced electric field decreases the main field E
to E – Ei i.e., New electric field between the plates will be E’ = E – Ei.
(3) Dielectric constant : After placing a dielectric slab in an electric field. The net field is decreased in that
region hence.
If E = Original electric field and E’ = Reduced electric field. Then
E
= K where K is called dielectric constant
E'
K is also known as relative permittivity (er) of the material or SIC (specific inductive capacitance)
The value of K is always greater than one. For vacuum there is no polarization and hence E = E’ and K = 1
(4) Dielectric breakdown and dielectric strength : If a very high electric field is created in a dielectric the outer
electrons may get detached from their parent atoms. The dielectric then behaves like a conductor. This phenomenon
is known as dielectric breakdown.
The maximum value of electric field (or potential gradient) that a dielectric material can tolerate without it’s electric
breakdown is called it’s dielectric strength.
S.I. unit of dielectric strength of a material is

On the basis of electrical behavior there are three types of medium:
(a) conductor

V
kV
but practical unit is
.
m
mm
(b) semiconductor
(c) insulator
Free charge carriers do not exist in insulators and their conductivity is of the order of 10 16 mho/metre or
resistivity of the order of 1016 ohm-m.

In insulators there is microscopic local displacement of charges under the influence of electric field. Such
materials are called dielectrics.

Electrical behaviour of a dielectric medium is represented by a dimensionless constant called dielectric constant.
Permittivi ty of medium
Dielectric constant  Permittivi ty of vacuum or free space
r 

0
Dielectric constant is also called relative permittivity or specific inductive capacity of that medium.

Dielectric constants are different for different media.

Following effects are observed when a dielectric medium is placed between the plates of the capacitor.
(a) The capacitance of the capacitor increases while the potential difference between the plates decreases.
CAPACITANCE
(b) The capacitor can be charged upto a higher potential than without the dielectric.
(c) Electrostatic potential energy of the capacitor decreases.
(d) The plates of the capacitor can be placed very close to each other without touching each other.

(a)
The capacitance C, electric field E, potential difference V and the charge q are effected due to introduction
of dielectric medium as follows
(a) C  C 0 r
E0
(b) E   for a given charge
r
V0
(c) V   for a given charge
r
(d) q ' r q for a given potential difference.
If capacitor is partially filled with dielectric between the plates, then
C
0 A

1
d  t  
r 

where t is the thickness of the dielectric medium.
(b)
If several slabs of dielectrics of different thicknesses are placed between
the plates, then
t1
t
A 
r
0 A
t2
t3
r1 r2 r3
C =  t

t
t
1
 2  3  ..... 

 r1 r2 r3

(b)
(a)
(c)
If a slab of metal r    of thickness t is placed between the plates, then C 
0 A
d  t 
Example based on Capacitance of Capacitors of Different Shapes
Ex.2 Two dielectrics of equal size and of constant 2 and 3 respectively fill up space between two plates of a condenser.
The ratio of capacities in two possible arrangements will be
[1]
24
25
[2]
25
24
2K 1K 2
Sol. C1 = [K  K ]
1
2
C2 =
[K 1  K 2 ]
2
4 K 1K 2
C1
 C =
or
(K 1  K 2 ) 2
2
C1
423
24
=
2
C2 =
25
5
Hence the correct answer will be (1)
[3]
4
5
[4]
5
4
CAPACITANCE
Ex.3 The capacity of a parallel plate capacitor in air is 50F and on immersing it into oil it becames 110F . The
dielectric constant of oil is
[1] 0.45
[2] 0.55
[3] 1.10
[4] 2.20
C
Sol. K = C
0
K =
110
11
=
= 2.20
50
5
Hence the correct answer will be (4)
Ex.4 A parallel plate capacitor with air between the plates has a capacitance of 8 pF. What will be the capacitance
if the distance between the plates is reduced by half, and the space between them is filled with a substance
of dielectric constant 6?
[1] 86 pF
[2] 96 pF
[3] 90 pF
[4] 80 pF
Sol. For parallel plate capacitor
C =
K 0 A
, with K = 1 for air
d
If distance between plates is reduced to half and K = 6, the new capacitance is
C'  6 
0 A
 A
 12 . 0
d
d
 
2
= 12 × 8 pF = 96 pF
Ex.5 In a parallel plate capacitor with air between the plates, each plate has an area of 6  10 3 m 2 and the distance
between the plates is 3 mm, in the capacitor a 3 mm thick mica sheet of dielectric constant = 6, is inserted
between the plates, while the voltage supply remains connected. What is the charge on each plate now
8
[1] 1.08  10 Coulomb
8
[2] 2.08  10 Coulomb
8
[3] 3.08  10 Coulomb
8
[4] 4.08  10 Coulomb
Sol. If dielectric is inserted, the capactiance becomes
C’ = KC
 6 × 18 pF
The charge on plates is Q = C’ V
= 108 × 10–12 × 100
= 1.08 × 10–8 coulomb
Ex.6 A parallel plate condenser is charged to a certain potential and then disconnected. The separation of the plates
is now increased by 2.4 mm and a plate of thickness 3 mm is inserted into it keeping its potential constant.
The dielectric constant of the medium will be
[1] 5
[2] 4
[3] 3
[4] 2
Sol. As charge and potential of the condenser both are constant in two cases, hence its capacity must also remain
constant
CAPACITANCE
C0 = C
or
0 A
=
d
0 A
 1
d' t 1  
 K
 1
or d = d’ – t 1  
 K
 1
or (d’ – d) = t 1  
 K
 1
1
or 2.4 × 10 –3 = 3 × 10–3  K 
or 1
1
1
 0.2
= 0.8 or
K
K
K = 5

Hence the correct answer will be (1)
(ii) Spherical Capacitor: It consists of two concentric spheres. The space between the spherical surfaces
is filled by a dielectric.
(a) When outer sphere is earthed and inner is given the charge, then
R 1R 2
C  4 0 r
R 2  R1
R2
R2
O
R1
R1
r
(b) When inner sphere is earthed and outer one is given the charge, then
C  4 0 r
R 1R 2
 4 0 R 2
R 2  R1
(A)
(B)
In order to increase the capacitance of a spherical capacitor:
(1) both spheres should be very close to each other i.e., R 2  R 1  should be small.
(2) medium between the spheres should be such that its dielectric constant is high.
(3) the radii of both spheres should be large.
(iii) Cylindrical capacitor : It consists of two coaxial cylinders with the space between them filled by a dielectric.
If R 1 and R 2 are the radii of inner and outer cylinders respectively and  is the length of the cylinder, then
C

2 0 r l
R
log e  2
 R1



2 0r l
R
2.303 log10  2
 R1



CAPACITANCE
(iv) Capacitance of two parallel transmission lines: If the radius of each wire is r, d is the distance
between them (d > > r) and  is the length of the wires, then
C

 0 r l
d
log e  
r
wire
r

 0 r l
d 
2.3 log10  
r
d
As medium between the wires is usually air so
C
 0 l
d
log e  
r
(v) Multiplate capacitor and capacitance of a variable capacitor : If n is the
number of plates, A is the area of each plate, r is the dielectric constant of medium
and d is the distance between two successive plates, then
C  n  1
0 r A
d
In order to obtain signals of a desired frequency in electronic instruments such as radio, T.V. etc., a parallel
resonance circuit is used. In this circuit a special type of parallel plate capacitor called gang condenser is
connected in parallel with an inductance coil. It consists of two sets of semi circular plates. The plates of
one set are stationary and the plates of other set are rotated by means of a knob. By changing the common
area of the plates, the resultant capacitance can be changed.
6.
CCCCCCCCCCCC CC CCCCCCCCCC
Capacitors can be combined in two ways:
(a)
Series combination
(b)
Parallel combination
(a)
Series combination:

Combination of capacitors on series is shown in following diagram
C1
C2
C3

Amount of charge is same on each capacitor, i.e.,
V1
V2
V3
Q  C1V1  C 2 V2  C 3 V3  .....

. .
V
Potential difference across each capacitor will be different and is inversely proportional to its capacitance.
V1 
Q
Q
Q
, V2 
, V3 
C1
C2
C3
In this combination the potential difference between the plates of the capacitor of least capacitance is maximum.

The potential difference applied in this combination is the sum of potential differences on individual capacitors.
V  V1  V2  V3  .....

If C is the equivalent capacitance of this combination, then
CAPACITANCE
1
1
1
1



 ....
C C1 C 2 C3
That is, if several capacitors are connected in series,then the reciprocal of the capacitors equivalent capacitance
is equal to the sum of the reciprocals of capacitances of the individual capacitors.

On combining the capacitors in series, the total capacitance of the circuit decreases and the equivalent capacitance
is less than the lowest capacitance connected in series.

If n identical capacitors each of capacitance C’ are connected in series, then the equivalent capacitance will
be : C 

C'
n
This combination is used when:
(a) a capacitance less than the lowest value of the given capacitance is needed,
(b) a high voltage is to be divided on several capaacitors.
(b) Parallel combination :

Combination of capacitors in parallel is shown in the diagram given:
v
C1
v
C2
v
C3
. .


V
The potential differance across each capacitors is equal to the applied potential difference in the circuit.
The amount of charge is different on each capacitor and the charge is directly proportional to the capacitance
of the capacitor.
Q1  C1V, Q 2  C 2 V, Q 3  C 3 V....

Total charge in the circuit :
Q  Q1  Q 2  Q 3  ...
or Q  C1V  C 2 V  C 3 V  ...

If C is equivalent capacitance of this combination, then:
C  C1  C 2  C 3  ...
That is, the equivalent capacitance of the capacitors connected in parallel is equal to the sum of their capacitances.

On combining the capacitances in parallel the equivalent capacitance of the circuit increases.

If n identical capacitors each of capacitance C are connected in parallel, then equivalent capacitance will
be :
C’ = nC

This combination is used when :
(a) The capacitance in the circuit is to be increased.
(b) Higher capacitance is required at low potential.
CAPACITANCE
EEEEEEE EEEEE EE CCCCCCCCCCCC CC CCCCCCCCCC
Ex.7 Three capacitors of same capacitance are connected in parallel. When they are connected to a cell of 2 volt,
total charge of 1.8C is accumulated on them. Now they are connected in series and then charged by the
same cell. The total charge stored in them will be
[1] 1.8C
[2] 0.9C
[3] 0.6C
[4] 0.2C
Sol. If the capacitance of each capacitors is C, then equivalent capacitance on joining them en parallel will be 3C.
 Stored charge Q = 3CV
 C =
Q
= 32
3V
Let C’ be the equivalent capacitance on joining them in series, then
1
1
1
1
3




C' 0.3 0.3 0.3 0.3
C' 
0. 3
= 0.1 F
3
Now Q’ = C’V = 0.1 × 2 = 0.2 C
Answer will be (4)
Ex.8 In the adjoining figure, if the capacitance of each capacitor is 1F , then the equivalent capacitance between
A and B will be
C
[1] 2F
C
[2] 2.5F
C
C
C
C
C
[3] 4.5F
C
C
[4] 5F
A
Sol. If equivalent capacitances of the rows of capactors are C1, C2, C3....., then
C 1 = 1 F
C3 =
C2 =
1
F
4
C4 =
1
F
2
1
F etc.
8
Equivalent capactance between A and B
C = C1 + C 2 + C 3 + C 4 + .....
= 1 +
1 1 1
 
2 4 8
This is a geometric series whose first term is a = 1 and common ratio r =
a
1

 C = 1  r 1  1 = 2F
2
Answer will be (1)
1
2
B
CAPACITANCE
Ex.9 The effective capacitance between points A and B is 1F . What is the capacitance of C?
C
[1] 1.39F
1F
A
6 F
4 F
8 F
[2] 2.46F
12F
[3] 2.23F
2 F
[4] None of the above
2 F
B
Sol. The circuit is equivalent to fig, Thus, given data implies that
(32C  9)
= 1
(C  32 / 9)
C = 32/23 = 1.39 F
The correct answer is (1)
Ex.10 Four indentical capacitors are connected in series with a battery of emf 10 V. The point X is earthed. Then
the potential of point A is
+ – 10 V
[1] 10 V
[2] 7.5 V
C
[3]  7.5 V
C
C
X
C
A
[4] 0 V
B
Sol. The equivalent capacitance of the circuit is C/4 and the charge on each capacitor is
Q = C’V =
C
 10  2.5C
4
Thus potential difference acros each capacitor is V =
Q
 2.5 volt.
C
Starting, from grounded point X, we go to point A, with change in p.d of V across each capacitor.
Thus VA – VX = 7/5 V
since Vx = 0, thus VA = 7.5 volt.
The correct anser is (2)
Ex.11 The equivalent capacity between the points X and Y in the following circuit will be
[1] 6F
6F
[2] 1F
6 F
6F
6 F
Y
X
[3] 24F
6 F
[4] 3F
Sol. Because the bridge is balanced, hence the central capacitance between Z and T is ineffective.
C1 and C2 are connected in series, hence their resultant C’ =
C
= 3 F
2
C3 and C4 are connected in series, hence their resultant C” =
C
 3F
2
Now the two branches are connected in parallel
 Ceq = 3 +3 = 6 F
Hence the correct answer will be (1)
CAPACITANCE
Ex.12 Five similar condenser plates, each of area A, are placed at equal distance d apart and are connected to
a source of e.m.f. E as shown in the following diagram. The charge on the plates 1 and 4 will be
[1]
[3]
0 A 2 0 A
,
d
d
[2]
0 AV 3 0 AV
,
d
d
[4]
0 AV 2 0 AV
,
d
d
1 2 4 5
0 AV 4 0 AV
,
d
d
–
E
+
Sol. Equivalent circuit diagram charge on first plate
Q = CV
1
2
3
2
0 AV
Q =
d
3
4
charge on fourth plate
5
4
Q' 
V
+ -
 0 AV
d
As okate 4 us reoeated twice, hence charge on 4 will be Q’’ = 2Q’
Hence Q'  
2 0 AV
d
Hence the correct answer will be (2)
Ex.13 Four condensers of equal capacity are connected as shown in the figure. If the capacity of the system between
P and Q is 1F then the capacity of each condenser will be
[1] 1F
[2] 2F
[3] 3F
[4] 4F
P
Q
Sol. All the four condensers are connected in series between P and Q and capacity of each is say C. Hence
resultant capacity of the combination
C’ =
C
4
But C’ = 1 f
C = f
Hence the correct answer will be (2)
Ex.14 Two condensers are joined as shown in the figure. Their central rigid part is movable. The capacity of the
combination will be
d
[1]
0 A
ab
[2]
2 0 A
ab
[3]
0 A
ab
a  b
[4]  A
0
b
a
CAPACITANCE
Sol. For the first condenser
C1 =
0 A
d
For second condenser
C2 =
0 A
a  ( b  d)
C1 and C2 are connected in series
0 A
C1C 2
0 A
a  (b  d)
 C = C  C =
=
d
0 A
0 A
1
2

d
a  (b  d)

0 A
(a  b)
Hence the correct answer will be (3)
7.
8.
UUU UU UUUU UUUU UU
(a)
In storing charge .
(b)
In storing energy .
(c)
In electronic devices for tuning.
(d)
In electrical appliances as circuit elements, for smoothing rectified current, in timing circuits etc.
(e)
In scientific studies.
C CC CC C CC C C C C CC CC C CC CC C C C C C C CC CC C CC C C CC CC C C CC CC C CC CC C
(i) Charging process:
C
R
(a) When a capacitor of capacitance C connected in series with a resistance R is
2
charged from a source of voltage V0 , then the voltage across the capacitor increases
from zero to the maximum value.The voltage at any time t is given by
vt = v0 (1 – e–t/CR)
V0
+ –
1
K
(b) In charging, initially the current is maximum. Afterwards the current decreases exponentially with time as :

I  I 0 e t / CR

where I 0 = maximum current
=
V0
R
(c) The charge on the capacitor at time t is

q  Q 1  e  t / CR


In charging a capaciator the voltage and the charger rise exponentially whereas the current decreases exponentially.

When t  , the current becomes zero and the potential difference across the capacitor becomes equal to
the voltage of the applied voltage source.
CAPACITANCE

RC is called time constant of the circuit. Time constant is the time in which the charge on the capacitor and
the voltage across the capacitor become equal to 0.632 or 63.2% times their final values,i.e., at t=RC
V  0.632V0 and q  0.632Q

Time constant is also the time in which the current in the circuit reaches a value 0.37 or 37% of its initial
maximum value i.e., at t =RC
I  0.37I 0

In charging, the variation of voltage and current with time are shown in the following figures:
I0
V0
I
V
t
t
(ii) Discharging process:
(a) In discharging the potential difference decreases exponentially with time as : V  V0 e  t / RC
(b) The charge also decreases exponentially with time as : q  Qe  t / RC
(c) In discharging the current decreases exponentially from maximum with time, as I  I 0 e t / RC
V0

In discharging the variation of the voltage and
the current with time are shown in the following
figure:
V
t
I
I0
EXAMPLE BASED ON CHARGING AND DISCHARGING OF A CAPACITOR THROUGH A RESISTANCE
Ex.15 A capacitor of capacity 10F is charged through a resistance of 100 k by a dc source of 2 volts. The potential
difference across the capacitor after 2 sec will be
[1] 2.0 V
[2] 1.73 V
Sol. Time constant of the RC circuit
 = RC = 100 × 103 × 10 × 10–6
= 1 sec
During the process of charging
V = V0 (1 – e–t/RC), (e = 2.718)
= 2 (1 – e–2/1) = 2 (1– 0.135)
= 2 × 0.865 = 1.73 volts
Answer will be (2)
[3] 1.52 V
[4] 1.0 V
CAPACITANCE
Ex.16 A 2500F capacitor is charged through a 1 K resistor by a 12V d.c. source. What is the voltage across
the capacitor after 5 sec?
[1] 10.38 volt
[2] 11.38 volt
[3] 12.38 volt
[4] 13.38 volt
Sol. The time constant of the circuit is
 = RC
= 103 × 2500 × 10–6
= 2.5 sec
For charging
V = V0 (1 – e–t/RC)
put t = 5 sec, and r = 2.5 sec, then
V = 12 (1 – e–2)
= 12 (1 – 0.135)
= 12 × 0.865
= 10.38 volt
9.
SSSS SSSSSSSSS SSSSSS SSSSSSSSS SSSSSSSSSS

An arrangement of conductors in which capacitance can be increased without changing the size of the conductor
is called condenser or capacitor. It has two conductors, one is charged and other is earthed. A suitable dielectric
material may be placed between them.

There are several types of capacitors based on the dielectric such as paper capacitor, electrolytic capacitor,
ceramic capacitor etc. Similarly there are several types of capacitors based on the shape of the conductors
such as parallel plate capacitor, spherical capacitor, cylindrical capacitor etc.

Alternating current flows easily through a capacitor while direct current does not flow through the capacitor.

Both plates of the charged capacitor have equal and opposite charges. Force of attraction acts between these
plates and it is equal to:
F

If C is the capacitance of the capaacitor,V is the potential difference and d is the distance
between the plates, then the attracative force between the plates:
F

1
 AE 2
2
1 CV 2
2 d
The relation between the voltage and the current is :
–
–
–
–
–
–
–
–
–
+
+
+
+
+
+
+
+
+
 dQ  d
 dV 
i
   VC  C 

 dt  dt
 dt 
Following conclusions can be drawn from the above relations.
(a) Since Q  CV , therefore voltage is directly proportional to the charge but not current.
(b) Capacitor has the ability to accumulate charge, therefore it has the capability to store information.
(c) There can be potential difference across the capacitor even when current is not flowing through the capacitor.
CAPACITANCE
(d) Capacitor acts as an open circuit for direct current. Instantaneous current flows during charging or discharging
of the capacitor.
(e) In the above relation
dV 1
 , that is, the rate of change of voltage is inversely proportional to the capacitance
dt C
of the capacitor. If rate of change of voltage is less, then C will be more. Thus the voltage in the capacitor
does not change suddenly.

Comparison of the energy of charged capacitor and the potential energy of the spring:

1 q2 
1 2

(a) Comparing the relation for energy  U  2 C  of the capacitor and the energy of the spring  U  kx  ,
2




the charge q is equivalent to displacement x.
(b) The reciprocal of the capacitance of the capacitor
1
is equivalent to the force constant k of the spring.
C
(c) The potential difference between the plates of the capacitor V 
q
is equivalent to the restoring force F  kx
C
of the spring.

Metallic plate can not be used as a dielectric in the capacitor because it will short circuit the plates.

The dielectric constant of a metal is infinite.
10.
PPPPP PP PP PPPPPPPP

Capacity or Capacitance of a Conductor :
C
Q
,
V
Unit is farad =
coulomb
volt
1F = 10–6 F, 1 pF = 10–12 F

Capacitance of a spherical conductor : C = 40R,
0is permittivity of vacuum.
Cm = 4R = r (40R)
is permittivity of medium and r is dielectric constant of the medium.

Work done in charging or energy stored :
U

1 Q2 1
1
 CV 2  QV
2 C
2
2
Common potential when two charged conductors are connected :
C = C1 + C 2 , Q = Q 1 + Q2 = C 1V1 + C2V2
Common potential
V
C1V  C 2 V2
C1  C 2
Charge transferred
Q 
C1C 2
( V1  V2 )
C1  C 2
CAPACITANCE
Energy loss
U 

C1C 2
1

( V1  V2 ) 2
2 C1  C 2
Capacitor or condenser : An arrangement of conductors for increasing the capacitance. It has two conductors
placed nearby, one is charged and the other is earthed.
Capacity of a condenser

C =
Q
V
Parallel plate capacitor :
C
0 A
 A 
, C m r  0  r C
d
 d 
If a dielectric of thickness t is placed in between, then
C
0 A

t
 d  t 
r




If slabs of thicknesses t1, t2, t3 ... tn of dielectric constants r1 ,r2 ,.... rn are placed in between
C
and

0 A
 t1
t
t

 2  ...  n
 r r
rn
2
 1




d  t 1  t 2  ....  t n
Spherical condenser :
When outer spherical shell is earthed
C  4 0 r
R1R 2
R 2  R1
When inner sphere is earthed
C  4  0 r

Cylindrical capacitor :
C


R1R 2
 4  0 R 2
R 2  R1
2 0 r 
loge (R 2 / R1)
2 0r 
2.303 log10 (R 2 / R1 )
Multiplate capacitor :
C = (n – 1)
0r A
d
CAPACITANCE

Energy stored in a capacitor :
U
1
1
1 Q2
CV 2  QV 
2
2
2 C
This energy resides in electric field. Energy density of electric field
U

1
 E2
2
Combination of capacitors :
Series combination :
1
1
1
1


 .... 
, Q1  Q 2  ...  Q
C C1 C 2
Cn
Parallel combination : C  C1  C2  ...  Cn , V1  V2  ...  V

Charging and discharging of a capacitor through a resistance :
Chargingq = Q (q – e–t/RC)
V = V0 (1 – e–t/RC)
I  I0 e  t / RC , I 0 
V0
R
Discharging q  Qe t / RC
V = V0 e –t/RC
I = –I0e–t/RC
Time constant :
= RC, time in which q = 0.63 Q, V = 0.63 V0 and I = 0.37 I0, while charging, q = 0.37 Q, V = 0.37 V0,
I = 0.37I0 while discharging.

Force of attraction between the plates of a capacitor :
F

1
2
 E2 A 
A
2
2
Q2
1 CV 2

2 A 2 d
CAPACITANCE
SSSSSS
SSSSSSS
Ex.1 8 similar charged drops combine to form a bigger drop. The ratio of the capacity of bigger drop to that of
smaller drop will be
Sol. Cbigger drop = (Csmall drop )n1/3
C bigger drop
C small drop

.............(1)
2
1
.............(2)
Ex.2 Two charged metal spheres of radii R and 2R are temporarily placed in contact and then separated. At the
surface of each, which sphere has the greater value for the following (a) charge, (b) charge density, (c) potential
and (d) electric field?
q1 C1 R1
R
1
Sol. (a) In sharing, charge divides in proportion to capacity so q  C  R  2R  2
2
2
2
i.e. smaller sphere has greater charge density
2
2
(b) Charge density,  = q/(4R )
so
2
1 q1  R 2 
1  2R 

      2
 2 q 2  R1 
2 R 
i.e, smaller sphere has greater charge density
V1 q1 R 2 1
2R
1
(c) In case of spherical conductor, VS = q/(4R) so V  q  R  2 ×
R
2
2
1
i.e., both the sphere are at same potential
2
2
E1 q1  R 2 
1  2R 

     2
(d) In case of spherical conductor, ES = q/(4R)
E 2 q 2  R1 
2 R 
i.e., filed at the surface of smaller sphere has greater value
Ex.3 Two uniformly charged spherical drops at potential V coalesce to form a large drop. If capacity of each smaller
drop is C then find capacity and potential of larger drop
Sol. When drops coalesce to form a larger drop then total charge and volume remains conserved. If r is radius
and q us charge of smaller drop then C = 4 and
Equating volume we get
4
4
R 3  2  r 3 = 21/3 C
3
3
Capacitance of larger drop C’ = 40R = 21/3C
Charge on larger drop Q = 2q = 2CV
Potential of larger drop V ' 
Q
2CV
 1/ 3  22 / 3 V
C' 2 C
q = CV
CAPACITANCE
Ex.4 A capacitor is made of a flat plate of area A and a second plate having
A/3
a stair-like structure as shown in figure. If the area of each stair is
A/3
A
and the height is d, find the capacitance of the arrangement
3
Sol. The given arrangement of flat plate and the starilike structure is equivalent
d
d
A/3
d
to the parallel combination of three capacitors C1, C2 and C3. The plate
AREA = A
area of each of the three capacitors is A/3, while the separation between
the plates of the three capacitors is d, 2d and 3d respectively.
 C1 
0 A / 3
0 A
0 A / 3
0 A
=
; C2 =
=
d
3d
2d
6d
Equivalent capacitance, C = C1 + C2 + C3 =
abd C 3
0 A / 3 0 A

3d
9d
0 A 0 A 0 A 
1 1  11 0 A


1    
3d
6d
3d 
2 3
18d
Ex.5 If capacitors C, 3C, 5C ......  in one network and 2C, 4C, 6C .....  in another network are connected in series.
C1C 2
If the effective capacitance of first network be C1 and that of network second be C 2 then C  C equal to
2
1
1
1
1
1
Sol. For first network, C  C  3C  5C  ...
1
1
1
1
1
For second network, C  2C  4C  6C  ...
2
1
1
1
1
1
1
1

 1




 ......  – 


 ...... 
=

C1 C 2
 C 3C 5C

 2C 4C 6C

1
1
1
1


1  1 1 1
1


 ...... =
loge (1  1)
=  
1     ......  =
C  2 3 4
C
 C 2C 3C 4C


[log e (1 +x) = x –
X2 x 3 x 4


 ..... ]
2
3
4
C 2  C1 loge 2
C1C 2
C
 C C  C
 C  C  log 2
1 2
2
1
e
C
C
C C
A
C
Sol. First n capacitor are connected in sereis so, their resulted capacity C1 = C/n
second n capacitor are connected in parallel so, their resultant capacitance C2 = nC
1
1
1
Now C1 and C2 are in series, so resultant capacity = C'  C  C
1
2

1 n
1
n2  1
 

C' C nC
nC
(B) Q1 
C1
Q = C1V
C1  C 2

and
C' 
n
C
n 1
C2
Q2 = C  C Q = CV
1
2
2
(C) The total energy stored in parallel combination of two capacitors is
U = U1 + U 2 =
1
1
1 2
2
C V2 + C 2 V = V (C1  C 2 )
2 1
2
2
C
up to n
up to n
Ex.6 In the given figure the resultant capacity will be
C
C
B
CAPACITANCE
Ex.7 In the following circuit, the resultant capacitance between A and B is 1F . Then value of C is
Sol. 12 F and 6 F are in series and again are in parallel with 4F
C
1F
A
Therefore resultant of these three will be
Equivalent of 8F, 2F and 2F =
12  6
= 4 = 8F
12  6
4  8 32 8

 F .....(2)
4  8 12 3
8F
B
2F
6F
4F
2F 12F
(1) and (2) are in parallel and are in series with C
32
C
8 8
32
9

C

1


=
and eq
9 3
9
32
C
9


32
32
C 
C
9
9
32
32
32 9
32
CC 


F
 C =
9
9
9 23 23
Ex.8 The capacitance of a parallel plate capacitor is 400 pico farad and its plates are separated by 2 mm in air
(i) What will be the energy when it is charged to 1500 volts, (ii) What wil be the potential difference with the
same charge if plate separation is doubled?, (iii) How much energy is needed to doubled the distance between
its plates?
Sol. Here C = 400 pico-farad = 400 × 10–12 farad and d = 2mm = 0.002 meters
(i) The capacitor is charged to a potential V = 1500 volts. The energy W of the capacitor is given by
W
1
1
CV 2 
× (400 × 10–12) (1500)2 = 4.5 × 10–4 joules
2
2
(ii) We know that the capacity of parallel plate condenser C = 0 A/d. When d is doubled, the new capacity.
C’ becomes halved i.e. C' 
1
1
C   400  10 12  200  10 12 farad
2
2
Charge on the capacitor q = CV = 400 × 10–12 × 1500 = 6 × 10–7 coul.
Let the new potential difference be V’ then for the same charge q, we have
q = C’ V’ = 6 × 10–7 or 200 × 10—12 × V’ = 6 × 10–7 
V’ = 3000 volts
(iii) The energy required to double the distance between the plate = final energy – initial energy
=
1
1
C' V' 2  CV 2
2
2
=
1
(200 × 10–12) (3000)2 – (4.5 × 10–4) = 9 × 10–4 – 4.5 × 10–4 = 4.5 × 10–4 joules
2
CAPACITANCE
EXERCISE # 1
Q.1
The capacitance of a spherical conductor is proportional to
[1] C  R2
Q.2
Q.4
[3] C  R
[4] C  R–1
Stored energy in a charged conductor is
[1]
Q.3
[2] C  R–2
1
CV 2
2
[2]
1 2 2
Q V
2
[3]
1 Q2
2 C2
[4]
1 Q
2 C2
If the two plates of the charged capacitor are connected by a wire, then
[1] potential will become infinite
[2] charge will become infinite
[3] capacitor will get discharged
[4] charge will become double that of earlier one
The capacity of a parallel plate capacitor is 0.5mF. When a mica sheet is placed between the plates the
1
potential difference reduces to   of the previous value. The dielectric constant of mica is
8
[1] 1.6
Q.5
[3] 8
[4] 40
[2] volt
[3] henry
[4] farad
Unit of capacitance is
[1] coulomb
Q.6
[2] 5
The capacitance of a capacitor is
[1] directly proportional to the dielectric constant of the medium between the plates
[2] inversely proportional to the dielectric constant of the medium between the plates
[3] proportional to the square of the dielectric constant of the medium between the plates
[4] independent of the dielectric constant of the medium between the plates
Q.7
If the energy of a capacitor of capacitance 2F is 0.16 joule, then its potential difference will be
[1] 800 V
Q.8
[2] 400 V
[3] 16  10 4 V
[4] 16  104 V
The capacitance of a parallel plate capacitor is 12F . If the distance between its plates is reduced to half
and the area of plates is doubled, then the capacitance of the capacitor will become
[1] 24F
Q.9
[2] 12F
[3] 16F
[4] 48F
A capacitor of 6F is charged to such an extent that the potential difference between the plates becomes
50 V. The work done in this process will be
[1] 7.5  10 2 J
[2] 7.5  10 3 J
[3] 3  10 6 J
[4] 3  10 3 J
Q.10 The radius of the circular plates of a parallel plate capacitor is R. Air is dielectric medium between the plates.
If the capacitance of the capacitor is equal to the capacitance of a sphere of radius R, then the distance
between the plates is
[1]
R
4
[2]
R
2
[3] R
[4] 2R
CAPACITANCE
Q.11 A capacitor of capacitance 500F is charged at the rate of 100C / s . The time in which the potential difference
will become 20 V, is
[1] 100 s
[2] 50 s
[3] 20 s
[4] 10 s
Q.12 The capacitances of spherical iron, copper and alluminium conductors of same radii are C1 , C 2 and C 3 , then
[1] C 2  C1  C 3
[2] C 3  C 2  C1
[3] C1  C 2  C 3
[4] C1  C 2  C 3
Q.13 Eight drops of mercury of same radius and having same charge coalesce to form a big drop. Capacitance
of big drop relative to that of small drop will be
[1] 16 times
[2] 8 times
[3] 4 times
[4] 2 times
Q.14 64 drops each charged to a potential of 100 volt, coalesce to form a big drop. The potential of the big drop
will be
[1] 6400 volt
[2] 3200 volt
[3] 1600 volt
[4] 800 volt
Q.15 Two spheres of capacitances 3F and 5F are charged to 300 V and 500 V respectively and are connected
together. The common potential will be
[1] 400 V
[2] 425 V
[3] 350 V
[4] 375 V
Q.16 The equivalent capacitance between A and B of the combination, shown in the figure, will be
[1] 1.5F
[3]
[2] 3.0F
6
F
11
A
1F 2F
3F
B
[4] 6F
Q.17 The potential of earth is zero because it is
[1] unchanged
[2] an object of zero capacitance
[3] an object of infinite capacitance
[4] having infinite charge
Q.18 In the combination shown in the figure, the voltmeter reading will be
[1] 4.5 V
 18 
[2]  V
 11 
[3] 3 V
[4] 2 V
V
2F
6F
3F
6V
Q.19 If a thin metal foil of same area is placed between the two plates of a parallel plate capacitor of capacitance
C, then new capacitance will be
[1] C
[2] 2C
[3] 3C
[4] 4C
Q.20 On charging a capacitor of 20F upto 500 V and a capacitor of 10F upto 200 V, they are connected in
parallel, their common potential will be
[1] 500 V
[2] 400 V
[3] 350 V
[4] 250 V
CAPACITANCE
Q.21 Two capacitors of capacitances 1F and 2F are connected in series and this combination is charged upto
a potential difference of 120 V. The potential difference on the capacitor of 1F will be
[1] 40 V
[2] 60 V
[3] 80 V
[4] 120 V
Q.22 To reduce the capacitance of a parallel plate capacitor, the space between the plates is
[1] filled with dielectric material
[2] reduced and area of the plates is increased
[3] increased and area of the plates is decreased
[4] increased and area is increased relatively
Q.23 The effective capacitance between A and B is
[1] 1F
2F
2F
[2] 2F
1F
[3] 1.5F
A
B
[4] 2.5F
Q.24 In the figure given, the effective capacitance between A and B will be
[1] C
[2]
C
2
C
C
C
C
C
A
[3] 2C
C
C
C
C
B
[4] 3C
Q.25 In the given circuit, the equivalent capacitance between A and B is
[1] C
[2] 2C
[3] 3C
[4] 4C
Q.26 In the circuit shown in the figure the equivalent capacitance between A and B will be
3C
3C
[1] C
A
[2] 9 C
3C
3C
2C
2C
2C
3C
3C
3C
3C
3C
[3] 0.86 C
[4] 5.0 C
B
Q.27 In the given figure, the equivalent capacitance between a and b will be
[1] 2F
6F
a
6F
l
[2] 4F
4F
[3] 6F
4F
[4] 16F
bl
6F
6F
6F
CAPACITANCE
Q.28 In the given figure the capacitance between A and B will be
[1] C
[2] 2C
[3] 3C
[4] 4C
Q.29 The vertical plates of a parallel plate capacitor are just in front of each other and the capacitance is C. If the plates
are shifted relatively, then capacitance
[1] will remain C
[2] will be more than C
[3] will be less than C
[4] nothing can be said
Q.30 Three capacitors of same capacitance are connected according to the following figures. For which combination
the equivalent capacitance will be maximum?
[1] in C and A
[2] in A and D
[3] in B
[4] in D
Q.31 In the above question, in which combination the energy stored will be maximum?
[1] in B and C
[2] in D
[3] in C and D
[4] in A and B
Q.32 Three capacitors of capacitances 12F each are available. The minimum and maximum capacitances, which
may be obtained from these are
[1] 12F,36F
[2] 4F,12F
[3] 4F,36F
[4] 0F, F
Q.33 n Capacitors each having capacitance C and breakdown voltage V are joined in series. The capacitance and
the breakdown voltage of the combination is
[1] C and V
V
n
[2] nC and nV
[3] nC and
[2] farad/metre
[3] henry/metre
[4]
C
and nV
n
Q.34 The SI unit of 0 is
[1] farad
[4] none of the above
Q.35 Capacitors are used for
[1] smoothing rectified current from power supplies
[2] elimination of sparking in switches
[3] storing large quantities of charge for use in research such as nuclear fusion
[4] all of the above
Q.36 Five capacitors of 10F capacity each are connected to a D.C. potential of 100 volts as shown in figure. The
equivalent capacitance between the points A and B will be equal to
[1] 40F
[2] 20F
[3] 30F
[4] 10F
CAPACITANCE
Q.37 A parallel plate air capacitor is connected to a battery. the quantities charge, voltage, electric field and energy
associated with this capacitor are given by Q 0 , V0 , E 0 and U 0 respecitvely. A dielectric slab is now introduced to
fill the space between the plates with battery still in connection. The corresponding quantities now given by Q, V,
E and U are related to the previous ones as
[1] Q  Q 0 , U  U 0
[2] V  V0 , Q  Q 0
[3] E  E 0 , U  U 0
[4] V  V0 , E  E 0
Q.38 For the section AB of a circuit shown in figure, C1  1F , C 2  2F , E  10V and the potential difference
VA  VB  10 volt. Then, the charge on capacitor C1 is A
+ –
C1
 20 
[2]  C
 3 
[1] 0C
E
B
C2
 40 
[3]  C
 3 
[4] none of the above
Q.39 Two spherical conductors A and B of radii a and b b  a  are placed concentrically in air. A is given a charge +Q
while B is earthed. Then the equivalent capacitance of the system is
[1] 4 0
ab
ba
[2] 4 0 b  a 
[3] 4 0 b
[4]
4 0 b 2
ba
Q.40 Force acting upon a charged particle kept between the plates of a charged condenser is F. If one of the plates of
the condenser is removed, force acting on the same particle becomes
[1] 0
[2]
F
2
[3] F
[4] 2F
Q.41 For the circuit shown in figure, the equivalent capacitance of the combination is
[1] 13F
 30 
[2]  F
 13 
 54 
[3]  F
 10 
[4] 1F
Q.42 For the circuit given above, the charge on 4F capacitor is
[1] 30C
[2] 40C
[3] 24C
[4] 54C
Q.43 Two capacitors, 3F and 4F , are individually charged across a 6V battery. After being disconnected from
the battery, they are connected together with the negative plate of one attached to the positive plate of the
other. What is the common potential?
[1] 6V
6
[2]   V
7
[3] 2V
3
[4]  V
2
CAPACITANCE
Q.44 In the above question what is the final charge on the 3F capacitor?
[1] 2.57C
[2] 3.43C
[3] 6C
[4] none of the above
Q.45 On charging a capacitor, the energy stored in it will be
[1] kinetic
[2] magnetic
[3] potential
[4] thermal
Q.46 The capacitance between terminals A and B is
9
[2]  F
 4
[1] 36F
[3]
27
F
4
[4] 12F
Q.47 The capacitance between terminals x and y is
[1]
C
2
[2] 5C
[3]
25
C
6
[4] zero
Q.48 A 4F capacitor is charged to 400 volts and then its plates are joined through a resistor of resistance 1K .
The heat produced in the resistor is
[1] 0.16 J
[2] 0.32 J
[3] 0.64 J
[4] 1.28 J
Q.49 Seven capacitors each of capacitance 2F are to be connected to obtain a capacitance of
10
F . Which
11
of the following combination is possible
[1] 2 in parallel 5 in series
[2] 3 in parallel 4 in series
[3] 4 in parallel 3 in series
[4] 5 in parallel 2 in series
Q.50 We wish to obtain a capacitance of 5F , by using some capacitors, each of 2F . Then the minimum number
of capacitors required is
[1] 3
[2] 4
[3] 5
[4] not possible
Q.51 Capacitance of a conducting sphere will be 1F if its radius is approximately
[1] 1.11 km
[2] 111 km
[3] 0.111 mm
[4] 9 km
Q.52 A 10 V battery is connected to two capacitors C1 and C 2 and a resistor R. When the capacitors are fully
charged, then the potential difference across the resistor will be
[1] 10 V
[2] 5 V
[3] 3.33 V
[4] zero
Q.53 In the above question, when the capacitors are fully charged, then the potential drop across the capacitor C1 will be
10
volt
3
Q.54 In the above question the potential of point A is
20
volt
3
[1] zero
[2]
[3]
[1] 3V
[2] 6V
[3] 9V
[4] 10 volt
[4] Zero V
CAPACITANCE
Q.55 The capacitance of the capacitors of plate areas A1 and A2(A1 < A2) at a distance d is
[1]
[3]
0 A1
d
[2]
0 ( A1  A 2 )
2d
[4]
0 A 2
d
A1
A2
 0 A 1A 2
d
d
Q.56 A parallel plate capacitor has separation t and capacitance 100 pF. If a metallic foil of thickness t/3 is introduced
between the plates, the capacitance would become (in pF unit)
[1] 3 × 100
[2] (3/2) × 100
[3] 100
[4] (2/3) × 100
Q.57 A 4 F condenser is charged to 400 volts and then its plates are joined through a resistance of 1 k. The heat
produced in the resistance is
[1] 0.16 J
[2] 1.28 J
[3] 0.64 J
[4] 0.32 J
Q.58 Two capacitors 2 F and 4 F are connected in parallel. A third capacitor of 6 F capacity is connected in series.
The combination is then connected across a 12 V battery. The voltage across 2 F capacity is
[1] 2 V
[2] 6 V
[3] 8 V
[4] 1 V
Q.59 As shown in the figure, a very thin sheet of aluminium in placed in between the plates of the condenser. Then the
capacity -
[1] will increase
[2] will decrease
[3] Remains unchanged
[4] May increase or decrease
Q.60 Work done by an external agent in separating the parallel plate capacitor is [1] CV
[2]
1 2
C V
2
[3]
1
CV 2
2
[4] None of these
Q.61 If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance [1] Decreases two times [2] increase two times
[3] increases four times
[4] Remain the same
Q.62 A capacitors is used to store 24watt hour of energy at 1200 volt. What should be the capacitance of the capacitor
[1] 120 mF
[2] 120 mF
[3] 12 mF
[4] 24 mF
Q.63 A solid conducting sphere of radius R1 is surrounded by another concentric hollow conducting sphere of radius R2.
The capacitance of this assembly is proportional to R 2  R1
[1] R R
1 2
R 2  R1
[2] R R
1 2
R1 R2
[3] R  R
2
1
R1 R2
[4] R  R
2
1
ANSWER KEY
Qus.
Ans.
Qus.
Ans.
Qus.
Ans.
Qus.
Ans.
Qus.
Ans.
1
3
16
4
31
2
46
4
61
3
EXERCISE # 1
2
1
17
3
32
3
47
1
62
1
3
3
18
4
33
4
48
2
63
4
4
1
19
1
34
2
49
4
5
4
20
2
35
4
50
2
6
1
21
3
36
4
51
4
7
2
22
3
37
1
52
4
8
4
23
2
38
3
53
3
9
2
24
3
39
1
54
1
10
1
25
3
40
2
55
1
11
1
26
1
41
3
56
2
12
4
27
3
42
3
57
4
13
4
28
2
43
2
58
2
14
3
29
3
44
1
59
3
15
2
30
4
45
3
60
3
CAPACITANCE
EXERCISE # 2
Q.1
A capacitor of capacitance 10F is charged to a potential of 100 V. Now connecting it in parallel with an
uncharged capacitor, the resultant potential difference becomes 40 volt. The capacitance of this capacitor is
[1] 2.5F
Q.2
[2] 5F
[3] 10F
[4] 15F
Three capacitors of same capacitance are connected in parallel. When they are connected to a cell of 2 volt,
total charge of 1.8C is accumulated on them. Now they are connected in series and then charged by the
same cell. The total charge stored in them will be
[1] 1.8C
Q.3
[2] 0.9C
[3] 0.6C
[4] 0.2C
A large number of metal plates and mica sheets are given. If a mica sheet is placed between two metal plates,
then it becomes a capacitor of capacitance C. How many such plates will be required to make a capacitor
of capacitance 5C
[1] 5
[2] 6
[3] 9
[4] 10
Q.4
In the arrangements shown in figure, dielectric constant K1  2 and K 2  3 . If the capacitances are C1 and
C1
C 2 respectively, then C will be
2
Q.5
K1
K1
[1] 1 : 1
[2] 2 : 3
[3] 9 : 5
[4] 25 : 24
K2
K2
After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates
[1] the potential difference between the plates and the energy stored will decrease but the charge on plates
will remain same
[2] the charge on the plates will decrease and the potentaial difference between the plates will increase
[3] the potential difference between the plates will increase and energy stored will decrease but the charge
on the plates will remain same
[4] the potential difference, energy stored and the charge will remain unchanged
Q.6
The distance between two neighbouring plates of a gang condenser is 0.5
d
cm and the effective area of overlap of adjacent plates is 6cm 2 . Its capacitance
will be
d
[1] 1.06 pF
d
a
[2] 3.18 pF
Q.7
b
d
[3] 6.36 pF
d
[4] 12.72 pF
d
In the circuit given, charge (in C ) on each capacitor in steady state will be
[1] 12
4
^v^v^v
4
^v^v^v
4
^v^v^v
3F
[2] 10
[3] 8
[4] 6
3F
+
10V
–
3F
3F
^v^v^v
1
CAPACITANCE
Q.8
In the figure given, the potential difference between A and B in steady state will be
3F
B
l
[1] 20 V
3F
1 F
1 F
^v^v^v
[2] 25 V
1F
[3] 75 V
20
^v^v^v^v
Al
[4] 100 V
l
10
C
100 V
Q.9
In the above question, the potential difference between B and C in steady state will be
[1] 20 V
[2] 25 V
[3] 50 V
[4] 75 V
Q.10 In the given figure, the equivalent capacitance between P and Q will be
P
[1] zero
[2] infinite
C
C
C
C
l
C
C
C
C
 
[3] 0.62 C
Ql
[4] 1.62 C
Q.11 Four plates are arranged as shown in the diagram. If area of each plate is A and the distance between two
neighbouring parallel plates is d, then the capacitance of this system will be
[1]
4 0 A
d
[2]
3 0 A
d
[3]
2 0 A
d
[4]
0 A
d
Q.12 If four plates each of area A are arranged according to the given diagram with distance d between neighbouring
plates then the capacitance of the system will be
[1]
4 0 A
d
[2]
3 0 A
d
[3]
2 0 A
d
[4]
0 A
d
Q.13 An air capacitor is charged upto a potential V1 . It is connected in parallel to anidentical uncharged capacitor
filled with a dielectric medium. After redistribution of charge if the potential difference of this combination is
V, then the dielectric constant of the substance will be
[1] V 2 V1  V
[2] VV1  V 
[3]
V
V1  V 
[4]
V1  V 
V
CAPACITANCE
Q.14 As shown in the figure an insulating material of dielectric constant K is inserted into half of the space between
the plates. If initial capacitance of the capacitor was C, then its new capacitance will be
[1]
1
CK  1
2
[2]
1 C
2 1  K 
[3]
1 1  K 
2 C
[4] C1  K 
Q.15 Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a potential
difference V. If the battery is disconnected and the space between the plates of the capacitor of capacitance
C is completely filled with a material of dielectric constant K, then the potential difference across the capacitors
will be come
[1] 3VK  2
[2]
K  2
[3]
3V
3V
K  2
[4]
3K  2
V
Q.16 Three parallel plates are arranged as shown in figure. The area of each plate is A and the distance between
two neighbouring parallel plates is d. If upper and lower plates are connected by a wire, then the capacitance
of this system will be
[1]
4 0 A
d
[2]
3 0 A
d
[3]
2 0 A
d
[4]
0 A
d
Q.17 A parallel plate capacitor of plate area A and separation d is filled with two materials each of thickness
d
2
and dielectric constants 1 and 2 respectively. The equivalent capacitance will be
[1]
0 A
1  2 
d
[2]
0 A 1  2 
d
12
[3]
2 0 A 12
d 1  2 
[4]
2 0 A 1  2 
d
12
Q.18 In the given figure A  1cm 2 , d  2mm, K1  5.0, K 2  5.5 and K 3  2.1 . The capacitance of the system will be
a
d
K2
K1
K3
b
[1] 0.44 pF
[2] 0.88 pF
[3] 1.77 pF
[4] 2.20 pF
CAPACITANCE
Q.19 Four identical metal plates each of area A (on one side) and each separated by a distance d are connected
as shown in figure. The capacitance of the system between points A and B is
[1]
0 A
d
[2]
3 0 A
d
[3]
2 0 A
3d
[4]
3 0 A
2d
Q.20 Four identical metal plates each of area A on one side and each separated by a distance d are connected
as shown in figure. The capacitance of the system between points A and B is
[1]
0 A
d
[2]
3 0 A
d
[3]
2 0 A
3d
[4]
3 0 A
2d
Q.21 Capacitor C1 of capacitance 1 microfarad and capacitor C2 of capacitance 2 microfarad are separately charged fully
by a common battery. The two capacitors are then separately allowed to discharge through equal resistors at time
t = 0 Consider the statements
(a) the current in each of the two discharging circuits is zero at t = 0
(b) the currents in the two discharging circuits at t = 0 are equal but not zero
(c) the currents in the two discharging circuits at t = 0 are unequal
(d) capacitor C1 loses 50% of its initial charge sooner than C2 loses 50% of its initial charge.
Then, the correct statement (is) are
[1] (a) and (d) only
[2] (b) and (d) only
[3] (d) only
[4] none of the above
Q.22 The time constant of charging of the circuit shown in fig. is
[1]
2
RC
3
[2] 2 RC
[3] 3 RC
[4]
3
RC
2
2
second. Capacitor is discharged at time t = 0. The ratio of charge on the
ln( 2)
capacitor at time t = 2s and t = 6s is
Q.23 Time cosntant of a C-R circuit is
[1] 3 : 1
[2] 8 : 1
[3] 4 : 1
[4] 2 : 1
Q.24 In the circuit shown in fig. switch S is closed at time t = 0. Let i1 and i2 be the currents at any finite time t then the
i1
ratio i
2
[1] is constant
[2] increases with time
[3] decreases with time
[4] first increases and then decreases
CAPACITANCE
Q.25 In the circuit shown in fig. charge stored in the capacitor of capacity 5 F is
[1] 60 C
[2] 20 C
[3] 30 C
[4] zero
Q.26 Two indentical capacitors 1 and 2 connected in series to a battery as shown in fig. Capacitor 2 contains a dielectric
slab of dielectric constant k as shown. Q1 and Q2 are the charges stored in the capacitors. Now the dielectric slab
is removed and the corresponding charges are Q’1 and Q’2. Then
1
2
E
Q'1 k  1
[1] Q  k
1
Q' 2 k  1
[2] Q  2
2
Q' 2 k  1
[3] Q  2 k
2
Q'1 k
[4] Q  2
1
Q.27 The plates of a parallel plate capacitor are charged upto 100 volt. A 2 mm thick plate is inserted between the
plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by
1.6m.m. The dielectric constant of the plate is
[1] 5
[2] 1.25
[3] 4
[4] 2.5
Q.28 A capacitor of capacitance 1 F withstands a maximum voltage of 6 kV, while another capacitor of capacitance
2 F, the maximum voltage 4 kV. If they are connected in series, the combination can withstand a maximum of
[1] 6 kV
[2] 4 kV
[3] 10 kV
[4] 9 kV
Q.29 In order to obtain a time constant of 10 seconds in an RC circuit containing a resistance of 103  the capacity of
a condenser should be
[1] 10 F
[2] 100 F
[3] 1000 F
[4] 10,000 F
Q.30 Two identical capacitors, have the same capacitance C. One of them is charged to potential V1 and the other to V2.
The negative ends of the capacitors are connected together. When the positive ends are also connected, the
decrease in energy of the combined system is
[1]
1
C( V12  V22 )
4
[2]
1
C( V12  V22 )
4
[3]
1
C( V1  V2 )2
4
[4]
1
C( V1  V2 ) 2
4
Q.31 A parallel plate capacitor carries a charge q. The distance between the plate is doubled by application of a force.
The work done by the force is [1] zero
[2]
q2
C
[3]
q2
2C
[4]
q2
4C
Q.32 A parallel plate capacitor has an electric field of 105V/m between the plates. If the charge on the capacitor plate is
1mC, the force on each capacitor plate is [1] 0.5N
[2] 0.05N
[3] 0.005N
[4] none of these
Q.33 A conducting sphere of radius 10cm is charged 10mC. Another uncharged sphere of radius 20cm is allowed to
touch it for some time. After that if the spheres are separated, then surface density of charges on the spheres will
be in the ratio of [1] 1 : 4
[2] 1 : 3
[3] 2 : 1
[4] 1 : 1
CAPACITANCE
Q.34 N identical spherical drops are charged to the same potential V. They are combine to form a bigger drop. The
potential of the big drop will be [1] VN1/3
[2] VN2/3
[3] V
[4] VN
Q.35 When we touch the terminals of a high voltage capacitors, even after a high voltage has been cut off, then the
capacitor has a tendency to [1] Restore energy
[2] discharge energy
[3] Affect dangerously
[4] Both 2 and 3
Q.36 1000 small water drops each of radius r and charge q coalesce together to form one spherical drop. The potential
of the big drop is larger than that of the smaller drop by a factor of [1] 1000
[2] 100
[3] 10
[4] 1
Q.37 Two metal spheres of radii 1cm and 2cm are given charges of 10–2C and 5 × 10–2 C respectively. If both spheres are
joined by a metal wire, then the final charge on the smaller spheres will be [1] 2 × 10–2 C
[2] 4 × 102 C
[3] 3 × 10–2 C
[4] 4 × 10–2 C
Q.38 64 drops each having the capacity C and potential V are combined to form a big drop. If the charge on the small
drop is q, then the charge on the big drop will be [1] 2q
[2] 4q
[3] 16q
[4] 64q
Q.39 A capacitor is kept connected to the battery and a dielectric slab is inserted between the plates. During this
process [1] No work is done
[2] Work is done at the cost of the energy already stored in the capacitor before the slab is inserted
[3] Work is done at the cost of the battery
[4] Work is done at the cost of both the capacitor and the battery
Q.40 Force acting upon a charged particle kept between the plates of charged condenser is F. If one plate of the
condenser is removed, then the force acting on the same particle will become [1] 0
[2] F/2
[3] F
[4] 2F
ANSWER KEY
Qus.
Ans.
Qus.
Ans.
Qus.
Ans.
1
4
16
3
31
3
EXERCISE # 2
2
4
17
3
32
2
3
2
18
3
33
3
4
4
19
3
34
2
5
1
20
4
35
4
6
3
21
2
36
2
7
1
22
2
37
1
8
2
23
3
38
4
9
4
24
2
39
4
10
3
25
4
40
4
11
3
26
3
12
2
27
1
13
4
28
4
14
1
29
4
15
3
30
3
CAPACITANCE
EXERCISE # 3
Q1.
Q2.
When a dielectric material is introduced between the plates of a charged condenser, the electric field between
the plates
(EAMCET Engg. 1996)
[1] decreases
[2] increases
[3] does not change
[4] may increase or decrease
Three capacitors 3F,10F and 15F are connected in series to a voltage source of 100 V. The charge on
15F is
[1] 22C
Q3.
Q4.
Q5.
Q6.
(EAMCET Med.1998, CPMT 1998)
[2] 100C
[3] 2800C
[4] 200C
A condenser is charged and then battery is removed. A dielectric plate is put between the plates of condenser,
then correct statement is
(RAJ PET 1998)
[1] Q constant V and U decrease
[2] Q constant V increases U decreases
[3] Q increases V decreases U increases
[4] None of these
When air is replaced by a dielectric medium of constant K, the capacity of the condenser (AIIMS 1998)
[1] increases K time
[2] increases K 2 times
[3] remains unchanged
[4] decreases K times
If we increase the distance between two plates of the capacitor, the capacitance will (AIIMS 1998)
[1] decrease
[2] remain same
[3] increase
[4] first decrease then increase
In the electric circuit given below, capacitance of each capacitor is 1F . The effective capacitance between
the points A and B is (in F )
[1] 6
[2]
1
6
3
2
[4]
2
3
[3]
Q7.
A parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates. The
quantity that remains unchanged is
(MP PMT 1998)
[1] Charge Q
Q8.
Q9.
(Karnataka CET 1998)
[2] Potential V
[3] Capacity C
How do adjust three capacitors to get high energy on same potential
[1] Two parallel and one in seires
[2] three in series
[3] three in parallel
[4] None of these
[4] Energy U
(RAJ PMT 1998)
Five capacitors of 10F capacity each are connected to a d.c. potential difference of 100 volts as shown in
the figure. The equivalent capacitance between the points A and B will be equal to[1] 40 F
(RPET 98, MP PMT 99)
[2] 20 F
[3] 30 F
[4] 10 F
+
-
CAPACITANCE
Q10. A capacitor of 20F is charged up to 500 V is connected in parallel with another capacitor of 10F which
is charged up to 200 V. The common potential is
[1] 500 V
[2] 400 V
(CPMT 1999)
[3] 300 V
[4] 200 V
Q11. When a P.D. of 103 V is applied between A and B, a charge of 0.75 mC is stored in the system of capacitors.
The value of C is F 
(Karnataka CET-1999)
[1] 2.5
[2] 2
[3]
1
2
[4] 3
Q12. Three capacitors of 3F , 2F and 6F are connected in series. When a battery of 10 V is connected to
this combination then charge on 3F capacitor will be
[1] 5C
[2] 10C
(Raj. PMT 1999)
[3] 15C
[4] 20C
Q13. The equivalent capacity between A and B will be
(Raj. PMT 1999)
[1] 0.2F
[2] 0.3F
[3] 0.4F
[4] 0.5F
Q14. A parallel capacitor of capacitance C is charged and disconnected from the battery. The energy stored in it
is E. If a dielectric slab of dielectric constant 6 is ineserted between the plates of the capacitor then energy
and capacitance will become
(CBSE PMT 1999)
[1] 6E, 6C
[2] E, C
[3]
E
,6C
6
[4] E, 6C
Q15. A parallel plate capacitor has the space between its plates filled by two slabs of thickness
d
each and dielectric
2
constant K1 and K 2 . d is the plate separation of the capacitor. The capacitance of the capacitor is
(MP PET 1999)
[1]
2 0 A  K1  K 2

d  K 1K 2



[2]
2 0 A
K 1  K 2 
d
[3]
2 0 A  K 1K 2

d  K1  K 2



[4]
2 0 A  K1  K 2

d  K 1K 2



Q16. A parallel plate condenser with oil between the plates (dielectric constant of oil K=2) has a capacitance C.
If the oil is removed, the capacitance of the capacitor becomes
(EAMCET Engg. 1999)
[1]
2C
[2] 2C
[3]
C
2
[4]
C
2
CAPACITANCE
Q17. A parallel plate capacitor has charge Q coulomb, potential V volt, and energy E joule. A dielectric slab is
now inserted between the two plates, then
(UPSEAT 2000)
[1] V and E both decreases
[2] V and E both increases
[3] V decreases, E increases
[4] V increases, E decreases
Q18. A capacitor is fully charged with a battery. Then the battery is removed and a coil is connect with the capacitor
in parallel. Current varies as
(RAJ PET 2000)
[1] increases monotonically
[2] decreases monotonically
[3] zero
[4] oscillates indefinitely
Q19. Ten capacitors are joined in parallel and charged with a battery up to a potential V. They are then disconnected
from battery and joined again in series then the potential of this combination will be(RAJ PET 2000)
[1] V
[2] 10V
[3] 5V
[4] 2V
Q20. In the given diagram equivalent capacitance between A and B will be
(RAJ PET 2000)
[1] 20F
[2] 15F
[3] 30F
[4] 60F
Q21. Capacity of a parallel plate condenser is 10F when the distance between its plates is 8 cm. If the distance
between the plates is reduced to 4 cm, its capacity will be
[1] 10F
[2] 15F
[3] 20F
(CPMT 1997, AFMC 2000)
[4] 40F
Q22. The capacitance between the points P and Q in the following circuit, is (CPMT 1998, J & K CET 2000)
[1] 1.5F
[2] 3F
[3] 2F
[4] 1F
Q23. Given a number of capacitors labelled as 8F,250V . Find the minimum number of capacitors needed to get
an arrangement equivalent to 16F,1000V
[1] 4
[2] 16
(AIIMS 2000)
[3] 32
[4] 64
Q24. A capacitor of capacity C, is charged by connecting it across a battery of e.m.f. V. The battery is then removed
and the capacitor is connected in parallel with an uncharged capacitor of capacity C 2 . The potential difference
across this combination is
C2
[1] C  C .V0
1
2
(MP PET 2000)
C1
[2] C  C .V0
1
2
[3]
C1  C 2
.V0
C2
[4]
C1  C 2
.V0
C1
Q25. A capacitor is connected with a battery and stores energy U. After removing the battery, it is connected with
another similar capacitor in parallel. The new stored energy in each capacitor will be-(CBSE PMT 2000)
[1]
U
2
[2] U
[3]
U
4
[4]
3U
2
CAPACITANCE
Q26. While a capacitor remains connected to a battery, a dielectric slabis slipped between the plates. Then
[1] the energy stored in the capacitor decreases
(Karnataka CET 2001)
[2] the electric field between the plates increases
[3] charges flow from the battery to the capacitor
[4] the potential difference between the plates is changed
Q27. Time constant of a series R-C circuit is
[1]  RC
[2]  RC
(Raj. PMT 2001)
[3]
R
C
[4]
C
R
Q28. If the charge on a body is increased by 2C , the energy stored in it increases by 21%. The original charge
on the body in micro-coulombs is
[1] 10
[2] 20
(EAMCET (BE) 2001)
[3] 30
[4] 40
Q29. In a parallel plate capacitor of capacitance C, a metal sheet is inserted between the plates, parallel to them.
The thickness of the sheet is half of the seperation between the plates. The capacitance now becomes
(Karnataka CET 2001)
[1]
C
4
[2]
C
2
[3] 2C
[4] 4C
Q30. A capacitor of capacitance C1 is charged to potential V and battery is disconnected. Now the capacitor is
connected to a uncharged capacitor of capacitance C 2 then what will be common potential of the combination
(CBSE PMT 2002)
C1V
[1] C  C
1
2
C2 V
[2] C  C
1
2
[3]
C1  C 2 V
C1  C 2
C2
[4] C  C
1
2
Q31. Two capacitors of capacitances 3F and 6F are charged to a potential of 12V each. They are now connected
to each other, with the positive plate of each joined to the negative plate of the other. The potential difference
across each will be
(Karnataka CET 2002)
[1] 6V
[2] 4V
[3] 3V
[4] Zero
Q32. What fraction of the energy drawn from the charging battery is stored in a capacitor(Karnataka CET 2002)
[1] 100%
[2] 75%
[3] 50%
[4] 25%
Q33. A 0.2F capacitor is charged to 600 V. After removing it from battery it is connected to antoher capacitor
1.0F in parallel. The voltage on the capacitor will become
[1] 300 V
[2] 600 V
[3] 100 V
(MP PET 2002)
[4] 120 V
Q34. Two capacitors of 1F and 2F are connected in series, the resultant capacitance will be
(MP PET 2002)
[1] 4F
[2]
2
F
3
[3]
3
F
2
[4] 3F
CAPACITANCE
Q.35 Charge on 1F capacitor will be if V=3 volt
(Raj. PET 2002)
2 F
[1] 9C
3 F
[2] 6C
1 F
3 F
[3] 3C
+
–
[4] none
Q.36 The equivalent capacity between P and Q is
(RPET 2002)
[1] C
C
1
[2] C
2
[3] 4C
[4]
C
P
C
1
C
4
C
Q
C
Q.37 The capacity of a spherical conductor is 1F. Its radius will be
[1] 9 10 9 cm
[2] 9  109 m
[3] 9  10 9 km
(RPET 2002)
[4] none
Q.38 A parallel plate capacitor is made by stacking n equally spaced plates connected alternatively. If the capacitance
between any two adjacent plates is ‘C’ then the resultant capacitance is (AIEEE 2005)
[1] C
[2] nC
[3] (n – 1)C
[4] (n + 1) C
Q.39 A fully charged capacitor has a capacitance ‘C’. It is discharged through a small coil of resistance wire embedded
in a thermally insulated block of specific heat capacity ‘s’ and mass ‘m’. If the temperature of the block is
raised by DT, the potential difference V across the capacitance is (AIEEE 2005)
[1]
msT
C
2msT
C
[2]
[3]
2mCT
s
[4]
mCT
s
Q.40 A network of four capacitors of capacitor equal to C1 = C, C2 = 2C, C3 = 3C and C4 = 4C are conducted to
a battery as shown in figure. The ratio of the charges on C2 and C4 is (CPMT 2005)
C2
C1
C3
C4
V
[1]
3
22
[2]
7
4
[3]
22
3
[4]
4
7
Q.41 A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging
battery the distance between the plates of the capacitor is increased using an insulating handle. As a result
the potential difference between the plates
(CPMT-2006)
[1] does not change
[2] becomes zero
[3] increases
[4] decreases
Q.42 A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is
charged to a potential V volts. The dielectric slab is slowly removed from between the plates and then
reinserted. The net work done by the system in this process is
(AIEEE 2007)
(1) (K – 1) CV2
(2) zero
(3) ½ (K – 1) CV2
(4) CV2 (K –1)/K
CAPACITANCE
Q.43 A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes
equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done
by the battery will be
(AIEEE 2007)
(1)
1
4
(2)
1
2
(3) 1
(4) 2
Q.44 Two condensers, one of capacity C and the other of capacity C/2, are connected to a V-volt battery, as shown..
The work done in charging fully both the condensers is
(CPMT 2007)
(1)
1
CV 2
2
C/2
C
V
(2) 2CV2
(3)
1
CV 2
4
(4)
ANSWER KEY
Qus.
Ans.
Qus.
Ans.
Qus.
Ans.
1
1
16
4
31
2
3
CV 2
4
EXERCISE # 3
2
4
17
1
32
3
3
1
18
4
33
3
4
4
19
2
34
2
5
1
20
1
35
3
6
4
21
3
36
1
7
1
22
4
37
2
8
3
23
2
38
3
9
4
24
2
39
2
10
2
25
2
40
1
11
2
26
3
41
3
12
2
27
1
42
2
13
4
28
2
43
2
14
3
29
3
44
4
15
3
30
1