Capacitance JEE Notes | EduRev

JEE : Capacitance JEE Notes | EduRev

 Page 1


J E E - P h y s i c s
E
1
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Capacitance\English\theory.p65
CAPACITANCE
CONCEPT OF CAPACITANCE
Capacitance of a conductor is a measure of ability of the conductor to store charge on it. When a conductor
is charged then its potential rises. The increase in potential is directly proportional to the charge given to
the conductor. Q ? V ? Q = CV
The constant C is known as the capacity of the conductor.
Capacitance is a scalar quantity with dimension 
2 2 2
1 2 2
Q Q A T
C
V W
M L T
?
? ? ?
 = M
–1
 L
–2
 T
4
 A
2
Unit :- farad, coulomb/volt
The capacity of a conductor is independent of the charge given or its potential raised. It is also  independent
of nature of material and thickness of the conductor. Theoretically infinite amount of charge can be given to
a conductor. But practically the electric field becomes so large that it causes ionisation of medium surrounding
it. The charge on conductor leaks reducing its potential.
THE CAPACITANCE OF A SPHERICAL CONDUCTOR
When a charge Q is given to a isolated spherical conductor then its potential rises.
1 Q
V
4 R
0
?
? ?
 ? 
Q
C 4 R
0
V
? ? ? ?
R
O
+
+
+
+
+
+ +
+
+
+
+
If conductor is placed in a medium thenC
medium
 = 4 ??R  =  4 ??
?
?
r
R
Capacitance  depends upon :
• Size and Shape of Conductor    • Surrounding medium   • Presence of other conductors nearby
CONDENSER/CAPACITOR
The pair of conductor of opposite charges on which sufficient quantity of charge may be accommodated is
defined as condenser.
• Principle of a Condenser
It is based on the fact that capacitance can be increased by reducing potential keeping the charge
constant.
Consider a conducting plate M which is given a charge Q such that its potential rises to V then
Q
C
V
?
+
+
+
+
+
+
+
+
+
+
+
+
+
+
M
Q
Let us place another identical conducting plate N parallel to it such that charge is induced
on plate N (as shown in figure). If V
–
 is the potential at M due to induced negative charge
on N and V
+
 is the potential at M due to induced positive charge on N, then
Q Q
C '
V ' V V V
? ?
? ?
? ?
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ +
+ +
+ +
+ +
+ +
+ +
+ +
M N
Since V' < V (as the induced negative charge lies closer to the plate M in comparison to
induced positive charge). ? ? C' > C Further, if N is earthed from the outer side
(see figure) then V" = V
+
 – V
–
 ( ? the entire positive charge flows to the earth)
Q Q
C "
V " V V
?
? ?
?
 ? C" >> C                 
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+
+
+
+
+
+
+
M N
If an identical earthed conductor is placed in the vicinity of a charged conductor then the capacitance of the
charged conductor increases appreciable.  This is the principle of a parallel plate capacitor.
JEEMAIN.GURU
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J E E - P h y s i c s
E
1
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Capacitance\English\theory.p65
CAPACITANCE
CONCEPT OF CAPACITANCE
Capacitance of a conductor is a measure of ability of the conductor to store charge on it. When a conductor
is charged then its potential rises. The increase in potential is directly proportional to the charge given to
the conductor. Q ? V ? Q = CV
The constant C is known as the capacity of the conductor.
Capacitance is a scalar quantity with dimension 
2 2 2
1 2 2
Q Q A T
C
V W
M L T
?
? ? ?
 = M
–1
 L
–2
 T
4
 A
2
Unit :- farad, coulomb/volt
The capacity of a conductor is independent of the charge given or its potential raised. It is also  independent
of nature of material and thickness of the conductor. Theoretically infinite amount of charge can be given to
a conductor. But practically the electric field becomes so large that it causes ionisation of medium surrounding
it. The charge on conductor leaks reducing its potential.
THE CAPACITANCE OF A SPHERICAL CONDUCTOR
When a charge Q is given to a isolated spherical conductor then its potential rises.
1 Q
V
4 R
0
?
? ?
 ? 
Q
C 4 R
0
V
? ? ? ?
R
O
+
+
+
+
+
+ +
+
+
+
+
If conductor is placed in a medium thenC
medium
 = 4 ??R  =  4 ??
?
?
r
R
Capacitance  depends upon :
• Size and Shape of Conductor    • Surrounding medium   • Presence of other conductors nearby
CONDENSER/CAPACITOR
The pair of conductor of opposite charges on which sufficient quantity of charge may be accommodated is
defined as condenser.
• Principle of a Condenser
It is based on the fact that capacitance can be increased by reducing potential keeping the charge
constant.
Consider a conducting plate M which is given a charge Q such that its potential rises to V then
Q
C
V
?
+
+
+
+
+
+
+
+
+
+
+
+
+
+
M
Q
Let us place another identical conducting plate N parallel to it such that charge is induced
on plate N (as shown in figure). If V
–
 is the potential at M due to induced negative charge
on N and V
+
 is the potential at M due to induced positive charge on N, then
Q Q
C '
V ' V V V
? ?
? ?
? ?
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ +
+ +
+ +
+ +
+ +
+ +
+ +
M N
Since V' < V (as the induced negative charge lies closer to the plate M in comparison to
induced positive charge). ? ? C' > C Further, if N is earthed from the outer side
(see figure) then V" = V
+
 – V
–
 ( ? the entire positive charge flows to the earth)
Q Q
C "
V " V V
?
? ?
?
 ? C" >> C                 
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+
+
+
+
+
+
+
M N
If an identical earthed conductor is placed in the vicinity of a charged conductor then the capacitance of the
charged conductor increases appreciable.  This is the principle of a parallel plate capacitor.
JEEMAIN.GURU
J E E - P h y s i c s
2
E
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Capacitance\English\theory.p65
ENERGY STORED IN A CHARGED CONDUCTOR/CAPACITOR
Let C is capacitance of a conductor. On being connected to a battery. It charges to a potential V from zero
potential. If q is charge on the conductor at that time then q = CV. Let battery supplies small amount of charge
dq to the conductor at constant potential V.  Then small amount of work done by the battery against the force
exerted by exsiting charge is
q
dW Vdq dq
C
? ?
??
Q
Q
2
0 0
q 1 q
W dq
C C 2
? ?
? ?
? ?
? ?
?
?
2
Q
W
2C
?
where Q is the final charge acquired by the conductor. This work done is stored as potential energy, so
2
Q
U
2C
? = 
2
2
1 (CV ) 1
CV
2 C 2
? = 
2
1 Q
V
2 V
? ?
? ?
? ?
 = 
1
QV
2
?? ? ?
2
2
Q 1 1
U CV QV
2C 2 2
? ? ?
GOLDEN KEY POINTS
? • As the potential of the Earth is assumed to be zero, capacity of earth or a conductor
C= ?
connceted to earth will be infinite
q q
C
V 0
? ? ? ?
? • Actual capacity of the Earth  C = 4 ? ?
??
R = 
5
9
1
64 10 711 F
9 10
? ? ? ?
?
? • Work done by battery W
b
 = (charge given by battery) ×  (emf) = QV but Energy stored in conductor 
1
QV
2
?
so 50% energy supplied by the battery is lost in form of heat.
REDISTRIBUTION OF CHARGES AND LOSS OF ENERGY
When two charged conductors are connected by a conducting wire then charge flows from a conductor at higher
potential to that at lower potential. This flow of charge stops when the potential of two conductors became equal.
Let the amounts of charges after the conductors are connected are Q
1
' and Q
2
' respectively and potential is V
then
+
+
+
+
+
+
+
+
C
1
Q
1
V
1
+
+
+
+
+
+
+
+
+
+
+
+
C
2
Q
2
V
2
+
+
+
+
+
+
+
+
C
1
Q
1
'
V
+
+
+
+
+
+
+
+
+
+
+
C
2
Q'
2
V
          (Before connection) (After connection)
• Common potential
According to law of Conservation of charge Q
before connection
 = Q 
after connection
 ?? ? ?C
1
V
1
 + C
2
V
2
 =  C
1
V + C
2
V
Common potential after connection
1 1 2 2
1 2
C V C V
V
C C
?
?
?
JEEMAIN.GURU
Page 3


J E E - P h y s i c s
E
1
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Capacitance\English\theory.p65
CAPACITANCE
CONCEPT OF CAPACITANCE
Capacitance of a conductor is a measure of ability of the conductor to store charge on it. When a conductor
is charged then its potential rises. The increase in potential is directly proportional to the charge given to
the conductor. Q ? V ? Q = CV
The constant C is known as the capacity of the conductor.
Capacitance is a scalar quantity with dimension 
2 2 2
1 2 2
Q Q A T
C
V W
M L T
?
? ? ?
 = M
–1
 L
–2
 T
4
 A
2
Unit :- farad, coulomb/volt
The capacity of a conductor is independent of the charge given or its potential raised. It is also  independent
of nature of material and thickness of the conductor. Theoretically infinite amount of charge can be given to
a conductor. But practically the electric field becomes so large that it causes ionisation of medium surrounding
it. The charge on conductor leaks reducing its potential.
THE CAPACITANCE OF A SPHERICAL CONDUCTOR
When a charge Q is given to a isolated spherical conductor then its potential rises.
1 Q
V
4 R
0
?
? ?
 ? 
Q
C 4 R
0
V
? ? ? ?
R
O
+
+
+
+
+
+ +
+
+
+
+
If conductor is placed in a medium thenC
medium
 = 4 ??R  =  4 ??
?
?
r
R
Capacitance  depends upon :
• Size and Shape of Conductor    • Surrounding medium   • Presence of other conductors nearby
CONDENSER/CAPACITOR
The pair of conductor of opposite charges on which sufficient quantity of charge may be accommodated is
defined as condenser.
• Principle of a Condenser
It is based on the fact that capacitance can be increased by reducing potential keeping the charge
constant.
Consider a conducting plate M which is given a charge Q such that its potential rises to V then
Q
C
V
?
+
+
+
+
+
+
+
+
+
+
+
+
+
+
M
Q
Let us place another identical conducting plate N parallel to it such that charge is induced
on plate N (as shown in figure). If V
–
 is the potential at M due to induced negative charge
on N and V
+
 is the potential at M due to induced positive charge on N, then
Q Q
C '
V ' V V V
? ?
? ?
? ?
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ +
+ +
+ +
+ +
+ +
+ +
+ +
M N
Since V' < V (as the induced negative charge lies closer to the plate M in comparison to
induced positive charge). ? ? C' > C Further, if N is earthed from the outer side
(see figure) then V" = V
+
 – V
–
 ( ? the entire positive charge flows to the earth)
Q Q
C "
V " V V
?
? ?
?
 ? C" >> C                 
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+
+
+
+
+
+
+
M N
If an identical earthed conductor is placed in the vicinity of a charged conductor then the capacitance of the
charged conductor increases appreciable.  This is the principle of a parallel plate capacitor.
JEEMAIN.GURU
J E E - P h y s i c s
2
E
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Capacitance\English\theory.p65
ENERGY STORED IN A CHARGED CONDUCTOR/CAPACITOR
Let C is capacitance of a conductor. On being connected to a battery. It charges to a potential V from zero
potential. If q is charge on the conductor at that time then q = CV. Let battery supplies small amount of charge
dq to the conductor at constant potential V.  Then small amount of work done by the battery against the force
exerted by exsiting charge is
q
dW Vdq dq
C
? ?
??
Q
Q
2
0 0
q 1 q
W dq
C C 2
? ?
? ?
? ?
? ?
?
?
2
Q
W
2C
?
where Q is the final charge acquired by the conductor. This work done is stored as potential energy, so
2
Q
U
2C
? = 
2
2
1 (CV ) 1
CV
2 C 2
? = 
2
1 Q
V
2 V
? ?
? ?
? ?
 = 
1
QV
2
?? ? ?
2
2
Q 1 1
U CV QV
2C 2 2
? ? ?
GOLDEN KEY POINTS
? • As the potential of the Earth is assumed to be zero, capacity of earth or a conductor
C= ?
connceted to earth will be infinite
q q
C
V 0
? ? ? ?
? • Actual capacity of the Earth  C = 4 ? ?
??
R = 
5
9
1
64 10 711 F
9 10
? ? ? ?
?
? • Work done by battery W
b
 = (charge given by battery) ×  (emf) = QV but Energy stored in conductor 
1
QV
2
?
so 50% energy supplied by the battery is lost in form of heat.
REDISTRIBUTION OF CHARGES AND LOSS OF ENERGY
When two charged conductors are connected by a conducting wire then charge flows from a conductor at higher
potential to that at lower potential. This flow of charge stops when the potential of two conductors became equal.
Let the amounts of charges after the conductors are connected are Q
1
' and Q
2
' respectively and potential is V
then
+
+
+
+
+
+
+
+
C
1
Q
1
V
1
+
+
+
+
+
+
+
+
+
+
+
+
C
2
Q
2
V
2
+
+
+
+
+
+
+
+
C
1
Q
1
'
V
+
+
+
+
+
+
+
+
+
+
+
C
2
Q'
2
V
          (Before connection) (After connection)
• Common potential
According to law of Conservation of charge Q
before connection
 = Q 
after connection
 ?? ? ?C
1
V
1
 + C
2
V
2
 =  C
1
V + C
2
V
Common potential after connection
1 1 2 2
1 2
C V C V
V
C C
?
?
?
JEEMAIN.GURU
J E E - P h y s i c s
E
3
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Capacitance\English\theory.p65
• Charges after connection
Q
1
' = C
1
V = C
1 
1 2 1
1 2 1 2
Q Q C
Q
C C C C
? ? ? ? ?
?
? ? ? ?
? ?
? ? ? ?
  (Q : Total charge on system)
Q
2
' = C
2
V = C
2 
1 2 2
1 2 1 2
Q Q C
Q
C C C C
? ? ? ? ?
?
? ? ? ?
? ?
? ? ? ?
Ratio of the charges after redistribution  
1 1 1
2 2 2
Q C V R
Q C V R
'
(in case of spherical conductors)
'
? ?
• Loss of energy in redistribution
When charge flows through the conducting wire then energy is lost mainly on account of Joule effect,
electrical energy is converted into heat energy, so change in energy of this system,
?U = U
f
 – U
i
? ?
2 2 2 2
1 2 1 1 2 2
1 1 1 1
C V C V C V C V
2 2 2 2
? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? 
2
2 1 2
1 2
1
1 C C
U (V V )
2 C C
? ? ? ?
?
? ?
? ?
? ?
Here negative sign indicates that energy of the system decreases in the process.SOLVED EXAMPLES
Example
A conductor gets a charge of 50 ?C when it is connected to a battery of e.m.f. 5 V. Calculate capacity of the
conductor.
Solution
Capacity of the conductor 
6
Q 50 10
C 10 F
V 5
?
?
? ? ? ?
Example
The capacity of a spherical capacitor in air is 50 ?F and on immersing it into oil it becomes 110 ?F. Calculate the
dielectric constant of oil.
Solution
Dielectric constant of oil r
medium
air
C 110
2.2
C 50
? ? ? ?
Example
A radio active source in the form of a metal sphere of diameter 10
–3
m emits ? particles  at a constant rate of
6.25 ×  10
10
 particles per second. If the source is electrically insulated, how long will it take for its potential to
rise by 1.0 volt, assuming that 80% of emitted ? particles escape from the surface.
Solution
Capacitance of sphere C  = 
3
0
9
0.5 10
4 R
9 × 10
?
?
? ? ? =
12
1
10 F
18
?
?
Rate to escape of charge from surface
10 19
80
6.25 10 1.6 10
100
?
? ? ? ? ?
 = 8 ×  10
–9
  C/s
therefore q = (8 ×  10
–9
) t  and  q  = CV  ??8 ×  10
–9 
× t=
12
1
10 1
18
?
? ? ? t = 
12
9
10
8 10 18
?
?
? ?
=
3
10
144
?
= 6.95 ?s
JEEMAIN.GURU
Page 4


J E E - P h y s i c s
E
1
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Capacitance\English\theory.p65
CAPACITANCE
CONCEPT OF CAPACITANCE
Capacitance of a conductor is a measure of ability of the conductor to store charge on it. When a conductor
is charged then its potential rises. The increase in potential is directly proportional to the charge given to
the conductor. Q ? V ? Q = CV
The constant C is known as the capacity of the conductor.
Capacitance is a scalar quantity with dimension 
2 2 2
1 2 2
Q Q A T
C
V W
M L T
?
? ? ?
 = M
–1
 L
–2
 T
4
 A
2
Unit :- farad, coulomb/volt
The capacity of a conductor is independent of the charge given or its potential raised. It is also  independent
of nature of material and thickness of the conductor. Theoretically infinite amount of charge can be given to
a conductor. But practically the electric field becomes so large that it causes ionisation of medium surrounding
it. The charge on conductor leaks reducing its potential.
THE CAPACITANCE OF A SPHERICAL CONDUCTOR
When a charge Q is given to a isolated spherical conductor then its potential rises.
1 Q
V
4 R
0
?
? ?
 ? 
Q
C 4 R
0
V
? ? ? ?
R
O
+
+
+
+
+
+ +
+
+
+
+
If conductor is placed in a medium thenC
medium
 = 4 ??R  =  4 ??
?
?
r
R
Capacitance  depends upon :
• Size and Shape of Conductor    • Surrounding medium   • Presence of other conductors nearby
CONDENSER/CAPACITOR
The pair of conductor of opposite charges on which sufficient quantity of charge may be accommodated is
defined as condenser.
• Principle of a Condenser
It is based on the fact that capacitance can be increased by reducing potential keeping the charge
constant.
Consider a conducting plate M which is given a charge Q such that its potential rises to V then
Q
C
V
?
+
+
+
+
+
+
+
+
+
+
+
+
+
+
M
Q
Let us place another identical conducting plate N parallel to it such that charge is induced
on plate N (as shown in figure). If V
–
 is the potential at M due to induced negative charge
on N and V
+
 is the potential at M due to induced positive charge on N, then
Q Q
C '
V ' V V V
? ?
? ?
? ?
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ +
+ +
+ +
+ +
+ +
+ +
+ +
M N
Since V' < V (as the induced negative charge lies closer to the plate M in comparison to
induced positive charge). ? ? C' > C Further, if N is earthed from the outer side
(see figure) then V" = V
+
 – V
–
 ( ? the entire positive charge flows to the earth)
Q Q
C "
V " V V
?
? ?
?
 ? C" >> C                 
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+
+
+
+
+
+
+
M N
If an identical earthed conductor is placed in the vicinity of a charged conductor then the capacitance of the
charged conductor increases appreciable.  This is the principle of a parallel plate capacitor.
JEEMAIN.GURU
J E E - P h y s i c s
2
E
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Capacitance\English\theory.p65
ENERGY STORED IN A CHARGED CONDUCTOR/CAPACITOR
Let C is capacitance of a conductor. On being connected to a battery. It charges to a potential V from zero
potential. If q is charge on the conductor at that time then q = CV. Let battery supplies small amount of charge
dq to the conductor at constant potential V.  Then small amount of work done by the battery against the force
exerted by exsiting charge is
q
dW Vdq dq
C
? ?
??
Q
Q
2
0 0
q 1 q
W dq
C C 2
? ?
? ?
? ?
? ?
?
?
2
Q
W
2C
?
where Q is the final charge acquired by the conductor. This work done is stored as potential energy, so
2
Q
U
2C
? = 
2
2
1 (CV ) 1
CV
2 C 2
? = 
2
1 Q
V
2 V
? ?
? ?
? ?
 = 
1
QV
2
?? ? ?
2
2
Q 1 1
U CV QV
2C 2 2
? ? ?
GOLDEN KEY POINTS
? • As the potential of the Earth is assumed to be zero, capacity of earth or a conductor
C= ?
connceted to earth will be infinite
q q
C
V 0
? ? ? ?
? • Actual capacity of the Earth  C = 4 ? ?
??
R = 
5
9
1
64 10 711 F
9 10
? ? ? ?
?
? • Work done by battery W
b
 = (charge given by battery) ×  (emf) = QV but Energy stored in conductor 
1
QV
2
?
so 50% energy supplied by the battery is lost in form of heat.
REDISTRIBUTION OF CHARGES AND LOSS OF ENERGY
When two charged conductors are connected by a conducting wire then charge flows from a conductor at higher
potential to that at lower potential. This flow of charge stops when the potential of two conductors became equal.
Let the amounts of charges after the conductors are connected are Q
1
' and Q
2
' respectively and potential is V
then
+
+
+
+
+
+
+
+
C
1
Q
1
V
1
+
+
+
+
+
+
+
+
+
+
+
+
C
2
Q
2
V
2
+
+
+
+
+
+
+
+
C
1
Q
1
'
V
+
+
+
+
+
+
+
+
+
+
+
C
2
Q'
2
V
          (Before connection) (After connection)
• Common potential
According to law of Conservation of charge Q
before connection
 = Q 
after connection
 ?? ? ?C
1
V
1
 + C
2
V
2
 =  C
1
V + C
2
V
Common potential after connection
1 1 2 2
1 2
C V C V
V
C C
?
?
?
JEEMAIN.GURU
J E E - P h y s i c s
E
3
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Capacitance\English\theory.p65
• Charges after connection
Q
1
' = C
1
V = C
1 
1 2 1
1 2 1 2
Q Q C
Q
C C C C
? ? ? ? ?
?
? ? ? ?
? ?
? ? ? ?
  (Q : Total charge on system)
Q
2
' = C
2
V = C
2 
1 2 2
1 2 1 2
Q Q C
Q
C C C C
? ? ? ? ?
?
? ? ? ?
? ?
? ? ? ?
Ratio of the charges after redistribution  
1 1 1
2 2 2
Q C V R
Q C V R
'
(in case of spherical conductors)
'
? ?
• Loss of energy in redistribution
When charge flows through the conducting wire then energy is lost mainly on account of Joule effect,
electrical energy is converted into heat energy, so change in energy of this system,
?U = U
f
 – U
i
? ?
2 2 2 2
1 2 1 1 2 2
1 1 1 1
C V C V C V C V
2 2 2 2
? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? 
2
2 1 2
1 2
1
1 C C
U (V V )
2 C C
? ? ? ?
?
? ?
? ?
? ?
Here negative sign indicates that energy of the system decreases in the process.SOLVED EXAMPLES
Example
A conductor gets a charge of 50 ?C when it is connected to a battery of e.m.f. 5 V. Calculate capacity of the
conductor.
Solution
Capacity of the conductor 
6
Q 50 10
C 10 F
V 5
?
?
? ? ? ?
Example
The capacity of a spherical capacitor in air is 50 ?F and on immersing it into oil it becomes 110 ?F. Calculate the
dielectric constant of oil.
Solution
Dielectric constant of oil r
medium
air
C 110
2.2
C 50
? ? ? ?
Example
A radio active source in the form of a metal sphere of diameter 10
–3
m emits ? particles  at a constant rate of
6.25 ×  10
10
 particles per second. If the source is electrically insulated, how long will it take for its potential to
rise by 1.0 volt, assuming that 80% of emitted ? particles escape from the surface.
Solution
Capacitance of sphere C  = 
3
0
9
0.5 10
4 R
9 × 10
?
?
? ? ? =
12
1
10 F
18
?
?
Rate to escape of charge from surface
10 19
80
6.25 10 1.6 10
100
?
? ? ? ? ?
 = 8 ×  10
–9
  C/s
therefore q = (8 ×  10
–9
) t  and  q  = CV  ??8 ×  10
–9 
× t=
12
1
10 1
18
?
? ? ? t = 
12
9
10
8 10 18
?
?
? ?
=
3
10
144
?
= 6.95 ?s
JEEMAIN.GURU
J E E - P h y s i c s
4
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NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Capacitance\English\theory.p65
Example
The plates of a capacitor are charged to a potential difference of 100 V and then connected across a resister.
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 ?
Solution
Energy losses ?U = 
1
2
CV
0
2
 –
1
2
 CV
2
Fractional energy loss  
2 2
0
2
0
0
1 1
CV CV
U
2 2
1
U
CV
2
?
?
? = 
2 2
0
2
0
V V
V
?
 = 
2 2
2
(100) (80)
(100)
?
 = 
2
20 180
(100)
?
=
9
25
Example
Two uniformly charged spherical drops at potential V coalesce to form a larger drop. If capacity of each smaller
drop is C then find 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 radius and
q is charge on smaller drop then    C = 4 ? ? ?
0
 r and q = CV
Equating volume we get
4
3
  ? R
3
 = 2 × 
4
3
 ? r
3
??R = 2
1/3
 r
Capacitance of larger drop C' = 4 ? ? ?
0
R = 2
1/3
 C
Charge on larger drop Q = 2q = 2CV
Potential of larger drop V' = 
1 / 3
Q 2CV
C '
2 C
?
 = 2
2/3
 V
PAR ALLEL PLATE CAPACITOR
(i) Capacitance
It consists of two metallic plates M and N each of area A at
M
N
–
–
–
–
–
–
–
–
–
–
+
+
+
+
+
+
+
+
+
+
M N
area=A
P
E+
E–
d
Edge effect
+ ? – ?
separation d. Plate M is positively charged and plate N is
earthed. If ?
r
 is the dielectric constant of the material medium
and E is the field at a point P that exists between the two
plates, then
I step : Finding electric field E = E
+ 
+ E
– 
 = 
2
?
?
?+ ?
2
?
?
= 
?
?
=
0 r
?
? ?
[ ? = ?
0
 ?
r 
]
II step : Finding potential difference V = Ed = 
0 r
?
? ?
d= 
0 r
qd
A ? ?
V q
( E and )
d A
? ? ? ?
?III step : Finding capacitance
r 0
A q
C
V d
? ?
? ?
If medium between the plates is air or vacuum, then ?
r
= 1 ? ?
0
0
A
C
d
?
?
so
r 0 0
C C KC ? ? ? (where ?
r 
= K = dielectric constant)
JEEMAIN.GURU
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J E E - P h y s i c s
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NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Capacitance\English\theory.p65
CAPACITANCE
CONCEPT OF CAPACITANCE
Capacitance of a conductor is a measure of ability of the conductor to store charge on it. When a conductor
is charged then its potential rises. The increase in potential is directly proportional to the charge given to
the conductor. Q ? V ? Q = CV
The constant C is known as the capacity of the conductor.
Capacitance is a scalar quantity with dimension 
2 2 2
1 2 2
Q Q A T
C
V W
M L T
?
? ? ?
 = M
–1
 L
–2
 T
4
 A
2
Unit :- farad, coulomb/volt
The capacity of a conductor is independent of the charge given or its potential raised. It is also  independent
of nature of material and thickness of the conductor. Theoretically infinite amount of charge can be given to
a conductor. But practically the electric field becomes so large that it causes ionisation of medium surrounding
it. The charge on conductor leaks reducing its potential.
THE CAPACITANCE OF A SPHERICAL CONDUCTOR
When a charge Q is given to a isolated spherical conductor then its potential rises.
1 Q
V
4 R
0
?
? ?
 ? 
Q
C 4 R
0
V
? ? ? ?
R
O
+
+
+
+
+
+ +
+
+
+
+
If conductor is placed in a medium thenC
medium
 = 4 ??R  =  4 ??
?
?
r
R
Capacitance  depends upon :
• Size and Shape of Conductor    • Surrounding medium   • Presence of other conductors nearby
CONDENSER/CAPACITOR
The pair of conductor of opposite charges on which sufficient quantity of charge may be accommodated is
defined as condenser.
• Principle of a Condenser
It is based on the fact that capacitance can be increased by reducing potential keeping the charge
constant.
Consider a conducting plate M which is given a charge Q such that its potential rises to V then
Q
C
V
?
+
+
+
+
+
+
+
+
+
+
+
+
+
+
M
Q
Let us place another identical conducting plate N parallel to it such that charge is induced
on plate N (as shown in figure). If V
–
 is the potential at M due to induced negative charge
on N and V
+
 is the potential at M due to induced positive charge on N, then
Q Q
C '
V ' V V V
? ?
? ?
? ?
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ +
+ +
+ +
+ +
+ +
+ +
+ +
M N
Since V' < V (as the induced negative charge lies closer to the plate M in comparison to
induced positive charge). ? ? C' > C Further, if N is earthed from the outer side
(see figure) then V" = V
+
 – V
–
 ( ? the entire positive charge flows to the earth)
Q Q
C "
V " V V
?
? ?
?
 ? C" >> C                 
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+
+
+
+
+
+
+
M N
If an identical earthed conductor is placed in the vicinity of a charged conductor then the capacitance of the
charged conductor increases appreciable.  This is the principle of a parallel plate capacitor.
JEEMAIN.GURU
J E E - P h y s i c s
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NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Capacitance\English\theory.p65
ENERGY STORED IN A CHARGED CONDUCTOR/CAPACITOR
Let C is capacitance of a conductor. On being connected to a battery. It charges to a potential V from zero
potential. If q is charge on the conductor at that time then q = CV. Let battery supplies small amount of charge
dq to the conductor at constant potential V.  Then small amount of work done by the battery against the force
exerted by exsiting charge is
q
dW Vdq dq
C
? ?
??
Q
Q
2
0 0
q 1 q
W dq
C C 2
? ?
? ?
? ?
? ?
?
?
2
Q
W
2C
?
where Q is the final charge acquired by the conductor. This work done is stored as potential energy, so
2
Q
U
2C
? = 
2
2
1 (CV ) 1
CV
2 C 2
? = 
2
1 Q
V
2 V
? ?
? ?
? ?
 = 
1
QV
2
?? ? ?
2
2
Q 1 1
U CV QV
2C 2 2
? ? ?
GOLDEN KEY POINTS
? • As the potential of the Earth is assumed to be zero, capacity of earth or a conductor
C= ?
connceted to earth will be infinite
q q
C
V 0
? ? ? ?
? • Actual capacity of the Earth  C = 4 ? ?
??
R = 
5
9
1
64 10 711 F
9 10
? ? ? ?
?
? • Work done by battery W
b
 = (charge given by battery) ×  (emf) = QV but Energy stored in conductor 
1
QV
2
?
so 50% energy supplied by the battery is lost in form of heat.
REDISTRIBUTION OF CHARGES AND LOSS OF ENERGY
When two charged conductors are connected by a conducting wire then charge flows from a conductor at higher
potential to that at lower potential. This flow of charge stops when the potential of two conductors became equal.
Let the amounts of charges after the conductors are connected are Q
1
' and Q
2
' respectively and potential is V
then
+
+
+
+
+
+
+
+
C
1
Q
1
V
1
+
+
+
+
+
+
+
+
+
+
+
+
C
2
Q
2
V
2
+
+
+
+
+
+
+
+
C
1
Q
1
'
V
+
+
+
+
+
+
+
+
+
+
+
C
2
Q'
2
V
          (Before connection) (After connection)
• Common potential
According to law of Conservation of charge Q
before connection
 = Q 
after connection
 ?? ? ?C
1
V
1
 + C
2
V
2
 =  C
1
V + C
2
V
Common potential after connection
1 1 2 2
1 2
C V C V
V
C C
?
?
?
JEEMAIN.GURU
J E E - P h y s i c s
E
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NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Capacitance\English\theory.p65
• Charges after connection
Q
1
' = C
1
V = C
1 
1 2 1
1 2 1 2
Q Q C
Q
C C C C
? ? ? ? ?
?
? ? ? ?
? ?
? ? ? ?
  (Q : Total charge on system)
Q
2
' = C
2
V = C
2 
1 2 2
1 2 1 2
Q Q C
Q
C C C C
? ? ? ? ?
?
? ? ? ?
? ?
? ? ? ?
Ratio of the charges after redistribution  
1 1 1
2 2 2
Q C V R
Q C V R
'
(in case of spherical conductors)
'
? ?
• Loss of energy in redistribution
When charge flows through the conducting wire then energy is lost mainly on account of Joule effect,
electrical energy is converted into heat energy, so change in energy of this system,
?U = U
f
 – U
i
? ?
2 2 2 2
1 2 1 1 2 2
1 1 1 1
C V C V C V C V
2 2 2 2
? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? 
2
2 1 2
1 2
1
1 C C
U (V V )
2 C C
? ? ? ?
?
? ?
? ?
? ?
Here negative sign indicates that energy of the system decreases in the process.SOLVED EXAMPLES
Example
A conductor gets a charge of 50 ?C when it is connected to a battery of e.m.f. 5 V. Calculate capacity of the
conductor.
Solution
Capacity of the conductor 
6
Q 50 10
C 10 F
V 5
?
?
? ? ? ?
Example
The capacity of a spherical capacitor in air is 50 ?F and on immersing it into oil it becomes 110 ?F. Calculate the
dielectric constant of oil.
Solution
Dielectric constant of oil r
medium
air
C 110
2.2
C 50
? ? ? ?
Example
A radio active source in the form of a metal sphere of diameter 10
–3
m emits ? particles  at a constant rate of
6.25 ×  10
10
 particles per second. If the source is electrically insulated, how long will it take for its potential to
rise by 1.0 volt, assuming that 80% of emitted ? particles escape from the surface.
Solution
Capacitance of sphere C  = 
3
0
9
0.5 10
4 R
9 × 10
?
?
? ? ? =
12
1
10 F
18
?
?
Rate to escape of charge from surface
10 19
80
6.25 10 1.6 10
100
?
? ? ? ? ?
 = 8 ×  10
–9
  C/s
therefore q = (8 ×  10
–9
) t  and  q  = CV  ??8 ×  10
–9 
× t=
12
1
10 1
18
?
? ? ? t = 
12
9
10
8 10 18
?
?
? ?
=
3
10
144
?
= 6.95 ?s
JEEMAIN.GURU
J E E - P h y s i c s
4
E
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Capacitance\English\theory.p65
Example
The plates of a capacitor are charged to a potential difference of 100 V and then connected across a resister.
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 ?
Solution
Energy losses ?U = 
1
2
CV
0
2
 –
1
2
 CV
2
Fractional energy loss  
2 2
0
2
0
0
1 1
CV CV
U
2 2
1
U
CV
2
?
?
? = 
2 2
0
2
0
V V
V
?
 = 
2 2
2
(100) (80)
(100)
?
 = 
2
20 180
(100)
?
=
9
25
Example
Two uniformly charged spherical drops at potential V coalesce to form a larger drop. If capacity of each smaller
drop is C then find 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 radius and
q is charge on smaller drop then    C = 4 ? ? ?
0
 r and q = CV
Equating volume we get
4
3
  ? R
3
 = 2 × 
4
3
 ? r
3
??R = 2
1/3
 r
Capacitance of larger drop C' = 4 ? ? ?
0
R = 2
1/3
 C
Charge on larger drop Q = 2q = 2CV
Potential of larger drop V' = 
1 / 3
Q 2CV
C '
2 C
?
 = 2
2/3
 V
PAR ALLEL PLATE CAPACITOR
(i) Capacitance
It consists of two metallic plates M and N each of area A at
M
N
–
–
–
–
–
–
–
–
–
–
+
+
+
+
+
+
+
+
+
+
M N
area=A
P
E+
E–
d
Edge effect
+ ? – ?
separation d. Plate M is positively charged and plate N is
earthed. If ?
r
 is the dielectric constant of the material medium
and E is the field at a point P that exists between the two
plates, then
I step : Finding electric field E = E
+ 
+ E
– 
 = 
2
?
?
?+ ?
2
?
?
= 
?
?
=
0 r
?
? ?
[ ? = ?
0
 ?
r 
]
II step : Finding potential difference V = Ed = 
0 r
?
? ?
d= 
0 r
qd
A ? ?
V q
( E and )
d A
? ? ? ?
?III step : Finding capacitance
r 0
A q
C
V d
? ?
? ?
If medium between the plates is air or vacuum, then ?
r
= 1 ? ?
0
0
A
C
d
?
?
so
r 0 0
C C KC ? ? ? (where ?
r 
= K = dielectric constant)
JEEMAIN.GURU
J E E - P h y s i c s
E
5
NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit-07\Capacitance\English\theory.p65
(ii) Force between the plates
The two plates of capacitor attract each other because they are oppositely charged.
Electric field due to positive plate E = 
0
2
?
?
 = 
0
Q
2 A ?
Force on negative charge –Q is F = –Q E = – 
2
0
Q
2 A ?
Magnitude of force F = 
2
0
Q
2 A ?
 = 
1
2
 ?
0
 A E
2
Force per unit area or energy density or electrostatic pressure 
2
0
F 1
u p E
A 2
? ? ? ? ?
SPHERICAL CAPACITOR
(i) Outer sphere is earthed
?
?
?
?
?
?
?
?
?
?
R
2
R
1
? ?
?
+
+
+ +
+
+
+
+
+ +
When a charge Q is given to inner sphere it is uniformly  distributed on its
surface A charge –Q is induced on inner surface of outer sphere. The charge
+Q induced on outer surface of outer sphere flows to earth as it is grounded.
E = 0 for   r < R
1
 and E = 0 for   r > R
2
Potential of inner sphere V
1
 = 
0 1
Q
4 R ? ?
 + 
0 2
Q
4 R
?
? ?
 ?
0
Q
4 ? ?
 
2 1
1 2
R R
R R
? ? ?
? ?
? ?
As outer surface is earthed so potential  V
2
 = 0
Potential difference between plates    V = V
1
 – V
2
 = 
0
Q
4 ? ?
 
2 1
1 2
(R R )
R R
?
So C = 
Q
V
 = 4 ? ? ?
0
 
1 2
2 1
R R
R R ?
 (in air or vacuum)
In presence of medium between plate  C = 4 ? ? ?
r
 ?
0
 
1 2
2 1
R R
R R ?
(ii) Inner sphere is earthed 
Inside +ve and 
–ve charges 
are equal
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
(Q–Q') on outer
Surface
(+Q')
(–Q')
q
in
q
out
Here the system is equivalent to a spherical capacitor of inner and
outer radii R
1
 and R
2
 respectively and a spherical conductor of radius
R
2
 in parallel. This is because charge Q given to outer sphere
distributes in such a way that for the outer sphere.
Charge on the inner side is Q'=
1
2
R
R
Q    and
Charge on the outer side is Q – 
1
2
R
R
 Q = 
2 1
2
(R R )
R
?
Q
Q ' 
–Q' 
Q – Q' 
+
R
2
So total capacity of the system. C = 4 ? ? ?
0
 
1 2
2 1
R R
R R ?
 + 4 ? ? ?
0
 R
2
 =
2
0 2
2 1
4 R
R R
? ?
?
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