Chapter 5 Electrostatic Potential and Capacitance (Electrostatics Part 5) – Physics free study material by TEACHING CARE online tuition and coaching classes
Chapter 5 Electrostatic Potential and Capacitance (Electrostatics Part 5) – Physics free study material by TEACHING CARE online tuition and coaching classes
Capacitance.
 Definition : We know that charge given to a conductor increases it’s potential e.,
Q µ V
Þ Q = CV
Where C is a proportionality constant, called capacity or capacitance of conductor. Hence capacitance is the ability of conductor to hold the charge.
 Unit and dimensional formula : I. unit is
Coulomb = Farad (F) Volt
Smaller S.I. units are mF, mF, nF and pF ( 1mF = 10 ^{3} F ,
1pF = 1mmF = 10 ^{12} F )
1mF = 10 ^{6} F ,
1nF = 10 ^{9} F ,
C.G.S. unit is Stat Farad 1F = 9 ´ 10^{11} Stat Farad . Dimension : [C] = [M ^{1} L^{2}T ^{4} A ^{2} ].
 Capacity of an isolated spherical conductor : When charge Q is given to a spherical conductor of
radius R, then potential at the surface of sphere is
+ + + Q
+ +
V = k . Q
ì = 1 ü
+ R +


R ík 4pe ý
+ +
+ O +
+ +
Hence it’s capacity C = Q = 4pe R
V 0
Þ C = 4πε_{0} R =
1 .R
9 ´ 10^{9}
+ + + + +
in C.G.S. C = R
Note : @ If earth is assumed to be spherical having radius
R = 6400 km.
It’s theortical capacitance
C = 1
9 ´ 10^{9}
 6400 ´ 10^{3} = 711 mF . But for all practical purpose capacitance of earth is taken
infinity.
 Energy of a charged conductor : When a conductor is charged it’s potential increases from 0 to V as shown in the graph; and work is done against repulsion, between charge stored in the conductor and charge coming from the source (battery). This work is stored as “electrostatic potential energy”
From graph : Work done = Area of graph
= 1 QV
2
Hence potential energy U = 1 QV ; By using Q = CV,
2
U = 1 QV = 1 CV ^{2} = Q2
we can write
2 2 2C
 Combination of drops : Suppose we have n identical drops each having – Radius – r, Capacitance – c, Charge – q, Potential – v and Energy – u.
If these drops are combined to form a big drop of – Radius – R, Capacitance – C, Charge – Q, Potential – V
and Energy – U then –
 Charge on big drop : Q = nq
 Radius of big drop : Volume of big drop = n ´ volume of a single drop e.,
4 pR ^{3} = n ´ 4 pr ^{3} ,
R = n^{1}^{/}^{3}r
 Capacitance of big drop :
 Potential of big drop :
C = n^{1}^{/}^{3}c
V = Q = nq
3 3
V = n^{2}^{/}^{3}v
C n1 / 3 c
(v) Energy of big drop :
U = 1 CV ^{2} = 1 (n^{1/} ^{3}c)(n^{2} ^{/} ^{3}v)^{2}
U = n^{5}^{/}^{3}u
2 2
 Sharing of charge : When two conductors joined together through a conducting wire, charge begins to flow from one conductor to another till both have the same potential, due to flow of charge, loss of energy also takes place in the form of
Suppose there are two spherical conductors of radii r_{1} and r_{2} , having charge Q_{1}
and
Q2 ,
potential V_{1}
and
V_{2} , energies U_{1}
and U 2
and capacitance C_{1}
and C2
respectively, as shown in figure. If these two spheres are
connected through a conducting wire, then alteration of charge, potential and energy takes place.
 New charge : According to the conservation of charge
Q1 + Q2 = Q^{‘}
 Q^{‘} = Q
(say), also


Q^{‘} C V 4pe r Q^{‘} r
Q^{‘} r
Q^{‘} + Q^{‘} r + r



1 = 1 = 0 1 ,
1 = 1
Þ 1 + ^{ } ^{1} = 1 + ^{1} Þ
1 2 = 1 2




‘ C2 V 4pe 0 r2
‘ r2
‘ r2
‘ r2





_{‘} = Q é r2 ù
r + r
and similarly
_{‘} = Q é r1 ù



r + r

ë 1 2 û ë 1 2 û
 Common potential : Common potential (V ) = Total charge
= Q1 + Q2
= Q‘
 Q^{‘}
Þ V = C1 V1 + C2 V2
Total capacity
C1 + C2
C1 + C2
C1 + C2
(iii) Energy loss : As electrical energy stored in the system before and after connecting the spheres is
1 1 1
1 æ C V
 C V ö 2
Ui =
C V ^{2} +
C V ^{2}
and U =
(C + C
). V ^{2} =
(C + C
)ç 1 1 2 2 ÷

2 1 1
2 2 2
f 2 1 2
2 1 2 è
C1 + C2 ø
so energy loss
ΔU = U – U =
C1C2
(V – V )^{2}


2(C1 + C2 )
+ + + +
+ + +
+ + +
++ + + +
Example: 95 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 [MP PMT 2002, 1990; MNR 1999, 87; DCE 1998]
(a) 16 times (b) 8 times (c) 4 times (d) 2 times
Solution: (d) By using relation C = n^{1/} ^{3} . c Þ C = (8)^{1/} ^{3} . c = 2c
Example: 96 Two spheres A and B of radius 4 cm and 6 cm are given charges of
80mC
and 40mC respectively. If they are
connected by a fine wire, the amount of charge flowing from one to the other is [MP PET 1991]
(a)
20 m C
from A to B (b)
16 m C
from A to B (c) 32mC from B to A (d)
32 m C from A to B
Solution: (d) Total charge
Q = 80 + 40 = 120 m C . By using the formula
Q ‘ = Q é r1 ù . New charge on sphere A is
1 ê r1 + r2 ú
ë û
Q^{‘} = Q é rA ù = 120 é 4 ù = 48 m C . Initially it was 80 m C, i.e., 32 m C charge flows from A to B.
A ê r_{A} + r_{B} ú êë 4 + 6 úû
ë û
Example: 97 Two insulated metallic spheres of 3mF
and 5mF
capacitances are charged to 300V
and 500V
respectively.
The energy loss, when they are connected by a wire, is [Pb PMT 1999; CPMT 1999; KCET (Engg.) 2000]
(a)
0.012 J
(b)
0.0218 J
(c)
0.0375 J
(d)
3.75 J
Solution: (c) By using ΔU =
C1C2
2(C1 + C2 )
(V1
 V2
)^{2} ; ΔU = 0.375 J
Example: 98 64 small drops of mercury, each of radius r and charge q coalesce to form a big drop. The ratio of the surface density of charge of each small drop with that of the big drop is [KCET 2002]
(a) 1 : 64 (b) 64 : 1 (c) 4 : 1 (d) 1 : 4
s Small
q / 4pr ^{2}
æ q öæ R ö ^{2}
Solution: (d) =
= ç ÷ç ÷ ; since R = n^{1/3}r and Q = nq
s Big
Q / 4pR ^{2}
è Q øè r ø
So s Small = 1 Þ s Small = 1
s Big
n1 / 3
s Big 4
Capacitor.
(1) Definition : A capacitor is a device that stores electric energy. It is also named condenser.
or
A capacitor is a pair of two conductors of any shape, which are close to each other and have equal and opposite charge.
 Symbol : The symbol of capacitor are shown below
or variable capacitor
 Capacitance : The capacitance of a capacitor is defined as the magnitude of the charge Q on the positive
plate divided by the magnitude of the potential difference V between the plates i.e.,
C = Q
V
(4) Charging : A capacitor get’s charged when a battery is connected across the plates. The plate attached to the positive terminal of the battery get’s positively charged and the one joined to the negative terminal get’s negatively charged. Once capacitor get’s fully charged, flow of charge carriers stops in the circuit and in this condition potential difference across the plates of capacitor is same as the potential difference across the terminals of battery (say V).
 Charge on capacitor : Net charge on a capacitor is always zero, but when we speaks of the charge Q on a capacitor, we are referring to the magnitude of the charge on each plate. Charge distribution in making of parallel plate capacitor can easily be understand by reading carefully the following sequence of figures –
(6) Energy stored : When a capacitor is charged by a voltage source (say battery) it stores the electric energy. If C = Capacitance of capacitor; Q = Charge on capacitor and V = Potential difference across
capacitor then energy stored in capacitor
U = 1 CV ^{2} = 1 QV = Q2
2 2 2C
Note : @ In charging capacitor by battery half the energy supplied is stored in the capacitor and remaining half energy (1/2 QV) is lost in the form of heat.
 Types of capacitors : Capacitors are of mainly three types as described in given table
Parallel Plate Capacitor Spherical Capacitor Cylindrical Capacitor
It consists of two parallel metallic plates (may be circular, rectangular, square) separated by a small distance
+Q – Q
It consists of two concentric conducting spheres of radii a and b (a < b). Inner sphere is given charge
+Q, while outer sphere is earthed
It consists of two concentric cylinders of radii a and b (a < b), inner cylinder is given charge +Q while outer cylinder is earthed. Common length of the cylinders is l then
+ – _{–} – –



– –
–Q
– + + –
b a q – q

+Q


A – +
a


– +
– +
+ – – –
d
+ b – _{l}
+ –
–
–
A = Effective overlaping area of
Capacitance C = 4πε . ab

b – a
Capacitance
each plate
d = Separation between the plates
Q = Magnitude of charge on the inner side of each plate
s = Surface density of charge of
in C.G.S. C = ab . In the presence of
b – a
dielectric medium (dielectric constant K)

between the spheres C‘ = 4pe K ab
b – a
Special Case :
In the presence of dielectric medium (dielectric constant K) capacitance increases by K times and
each plate æ= Q ö
If outer sphere is given a charge +Q
C‘ =
2pe0 Kl


ç ÷ while inner sphere is earthed

log
æ b ö
V = Potential difference across the
Q¢ – +Q
e ç a ÷
plates
E = Electric field between the plates
– – – – è ø
– –
– b
– – –
æ s ö

ç= ÷
è 0 ø
Induced charge on the inner sphere
□ A
Capacitance :
in C.G.S. :
C = 0
d
C = A
4πd
,
This arrangement is not a capacitor. But it’s capacitance is equivalent to the sum of capacitance of spherical
If a dielectric medium of dielectric constant K is filled completely
capacitor and spherical conductor
i.e.
between the plates then capacitance b 2 ab
increases by K times C’ = KC
4pe _{0} . b – a = 4pe _{0} b – a + 4pe _{0} b
Concepts
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.
Two plates of unequal area can also form a capacitor because effective overlapping area is considered.
A
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 comparision to that between the plates which is the main capacitance.
Capacitance of a parallel plate capacitor depends upon the effective overlapping area of plates (C µ A) , separation between the
plates æ C µ 1 ö and dielectric medium filled between the plates. While it is independent of charge given, potential raised or

ç ÷
è ø
nature of metals and thickness of plates.
The distance between the plates is kept small to avoid fringing or edge effect (nonuniformity of the field) at the bounderies of the plates.
+ –
+ –
Spherical conductor is equivalent to a spherical capacitor with it’s outer sphere of infinite radius.
A spherical capacitor behaves as a parallel plate capacitor if it’s spherical surfaces have large radii and are close to each other.
The intensity of electric field between the plates of a parallel plate capacitor (E = s/e_{0}) does not depends upon the distance between them.
The plates of a parallel plate capacitor are being moved away with some velocity. If the plate separation at any instant of time is
‘d’ then the rate of change of capacitance with time is proportional to 1 .
d 2
Radial and nonuniform electric field exists between the spherical surfaces of spherical capacitor.
Two large conducting plates X and Y kept close to each other. The plate X is given a charge Q_{1} while plate Y is given a charge
Q_{2} (Q_{1} > Q_{2} ) , the distribution of charge on the four faces a, b, c, d will be as shown in the following figure.

Q1 Q_{2}
– æ Q_{1} – Q_{2} ö
b d Þ
ç ÷


è 2 ø
æ Q_{1} + Q_{2} ö
æ Q_{1} – Q_{2} ö
æ Q_{1} – Q_{2} ö


a c è 2 ÷
ç ÷
è 2 ø
ç ÷
è 2 ø
+ –
Example: 99 The capacity of pure capacitor is 1 farad. In D.C. circuit, its effective resistance will be [MP PMT 2000]
 Zero (b) Infinite (c) 1 ohm (d)
1 ohm
2
Solution: (b) Capacitor does not work in D.C. for D.C. it’s effective resistance is infinite i.e. it blocks the current to flow in the circuit.
Example: 100 A light bulb, a capacitor and a battery are connected together as shown here, with switch S initially open. When the switch S is closed, which one of the following is true [MP PMT 1995]
 The bulb will light up for an instant when the capacitor starts charging
 The bulb will light up when the capacitor is fully charged
 The bulb will not light up at all
 The bulb will light up and go off at regular intervals
Solution: (a) Current through the circuit can flow only for the small time of charging, once capacitor get’s charged it blocks the current through the circuit and bulb will go off.
Example: 101 Capacity of a parallel plate condenser is 10mF when the distance between its plates is between the plates is reduced to 4cm, its capacity will be
8 cm . If the distance
[CBSE 2001; Similar to CPMT 1997; AFMC 2000]
(a) 10mF (b) 15mF (c) 20mF (d) 40mF
Solution: (c)
C = e _{0} A µ 1
\ C1 = d2
or C
= d1 ´ C
= 8 ´ 10 = 20m F
d d C2 d1
^{2} d2 1 4
Example: 102 What is the area of the plates of a 3F
parallel plate capacitor, if the separation between the plates is 5 mm
[BHU 2002; AIIMS 1998]
(a)
1.694 ´ 10^{9} m^{2}
(b)
4.529 ´ 10^{9} m^{2}
(c)
9.281 ´ 10^{9} m^{2}
(d)
12.981 ´ 10^{9} m^{2}
Solution: (a) By using the relation C = e 0 A
d
Þ A = Cd =
e _{0}
3 ´ 5 ´ 10 ^{3}
8.85 ´ 10 ^{12}
= 1.694 ´ 10^{9} m^{2} .
Example: 103 If potential difference of a condenser (6 m F) is changed from 10 V to 20 V then increase in energy is
[CPMT 1997, 87]
(a)
2 ´ 10 ^{4} J
(b)
4 ´ 10^{4} J
(c)
3 ´ 10 ^{4} J
(d)
9 ´ 10^{4} J
Solution: (d) Initial energy Ui = 1 CV ^{2} ; Final energy U = 1 CV ^{2}
2 1 f 2 2
\ Increase in energy DU = U f – Ui = 1 C (V ^{2} – V ^{2} ) = 1 ´ 6 ´ 10 ^{–}^{6} (20^{2} – 10^{2} ) = 9 ´ 10^{–}^{4} J.
2 2 1 2
Example: 104 A spherical capacitor consists of two concentric spherical conductors. The inner one of radius
R_{1} maintained
at potential V_{1} and the outer conductor of radius R_{2}
at potential
V_{2} . The potential at a point P at a distance
x from the centre (where R_{2} > x > R_{1} ) is [MP PMT 1997]
(a)
V1 – V2 (x – R )
(b)
V1 R1 (R2 – x) + V2 R2 (x – R1 )
R2 – R1 1
(R2 – R1 ) x
(c)
V1 +
V2 x
(R2 – R1 )
(d)
(V1 + V2 ) x
(R1 + R2 )
Solution: (b) Let Q_{1}
and Q2
be the charges on the inner and the outer sphere respectively. Now
V_{1} is the total potential
on the sphere of radius R_{1},
So, V
= Q1 + Q2
…….. (i)
 R1 R2
and V_{2} is the total potential on the surface of sphere of radius R_{2} ,
So, V
= Q2 + Q1
…….. (ii)
 R2 R2
If V be the potential at point P which lies at a distance x from the common centre then

V =
2 1 è 1
 V1
……..(iii)
Substracting (ii) from (i)
V – V
= Q1 – Q2 Þ (V
 V )R R
= R Q
 R Q
Þ Q = (V1 – V2 )R1 R2
1 2 R1 R2
1 2 1 2
2 1 1 1 1
R2 – R1
Now substituting it in equation (iii), we have
V = (R1 – x)(V1 – V2 )R1 R2 + V Þ V = V_{1} R_{1}(R_{2} – x) + V_{2} R_{2} (x – R_{1})
xR1 (R2 – R1 ) ^{1}
x(R2 – R1 )
Example: 105 The diameter of each plate of an air capacitor is 4 cm. To make the capacity of this plate capacitor equal to that of 20 cm diameter sphere, the distance between the plates will be [MP PET 1996]
(a)
4 ´ 10 ^{3} m
(b)
1 ´ 10^{3} m
 1 cm (d)
1 ´ 10 ^{–}^{3} cm
Solution: (b) According to the question
e 0 A = 4pe R Þ
d 0
d = A =
4pR
p (2 ´ 10^{2} )^{2} 4p ´ 10 ´ 10^{2}
= 1 ´ 10^{3} m.
Example: 106 A spherical condenser has inner and outer spheres of radii a and b respectively. The space between the two is filled with air. The difference between the capacities of two condensers formed when outer sphere is earthed and when inner sphere is earthed will be [MP PET 1996]
 Zero (b)
4pe _{0} a
(c)
4pe _{0}b
(d)
4pe _{0}
æ b ö

ç b – a ÷
è ø
Solution: (c) Capacitance when outer sphere is earthed
C1 = 4pe _{0} . ab

b a
and capacitance when inner sphere is earthed
C2 = 4pe _{0}
. b 2
b – a
. Hence C_{2}
 C1
= 4pe _{0} .b
Example: 107 After charging a capacitor of capacitance
4 m F
upto a potential 400 V, its plates are connected with a
resistance of 1kW . The heat produced in the resistance will be [CBSE PMT 1994]
(a) 0.16 J (b) 1.28 J (c) 0.64 J (d) 0.32 J
Solution: (d) This is the discharging condition of capacitor and in this condition energy released
U = 1 CV ^{2}
2
= 1 ´ 4 ´ 10 ^{6} ´ (400)^{2} = 0.32J 2
= 0.32 J.
Example: 108 The amount of work done in increasing the voltage across the plates of a capacitor from 5V to 10V is W. The work done in increasing it from 10V to 15V will be
(a) 0.6 W (b) W (c) 1.25 W (d) 1.67 W
Solution: (d) As we know that work done = U _{final} – U_{initial} = 1 C(V ^{2} – V ^{2} )
2 2 1
When potential difference increases from 5V to 10V then
W = 1 C(10^{2} – 5^{2} )
2
……..(i)
When potential difference increases from 10V to 15V then
W‘ = 1 C(15^{2} – 10^{2} )
2
On solving equation (i) and (ii) we get
W‘ = 1.67 W.
……..(ii)
Dielectric.
Dielectrics are insulating (nonconducting) 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.
 Type of Dielectrics : Dielectrics are of two types –
 Polar dielectrics : Like water, Alcohol, CO_{2} , NH_{3} ,
(ii) Non polar dielectric : Like
N 2 ,
O_{2} , Benzene,
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
Methane etc. are made of nonpolar atoms/molecules. In nonpolar molecules, when no electric field is applied the centre of positive charge coincides with the centre of
of negative charges. O– –
105^{o}
negative charge in the molecule. Each molecule has zero dipole moment in its normal state.
H+ r H+
P
– + +
r
– P = 0
A polar molecule has permanent electric dipole
moment (p) in the absence of electric field also. But a
When electric field is applied, positive charge
polar dielectric has net dipole moment is zero in the absence of electric field because polar molecules are randomly oriented as shown in figure.
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
In the presence of electric field polar molecules tends to line up in the direction of electric field, and the
electric dipole. r
– E
+ +
–
substance has finite dipole moment.
 In general, any nonconducting, material can be called
as a dielectric but broadly non conducting material having
– + – + – +
– +
– + – + – +
P
non polar molecules referred to as dielectric because induced dipole moment is created in the non polar molecule.
 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
Suppose a dielectric slab is inserted between the plates of a capacitor.
As shown in the figure.
Induced electric field inside the dielectric is
Ei , 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 E‘
where K is called dielectric constant
K is also known as relative permittivity (e _{r} ) 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
 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 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
V but practical unit is
m
kV .
mm
Variation of Different Variables (Q, C, V, E and U) of Parallel Plate Capacitor.
Suppose we have an air filled charged parallel plate capacitor having variables are as follows :
Charge – Q, Surface charge density – s =
Q , Capacitance – C =
A
e _{0} A d
Potential difference across the plates – V = E . d
Electric field between the plates –
E = s
e _{0}
= Q Ae _{0}
Energy stored – U = 1 CV 2 = Q 2
= 1 QV
2 2C 2
 When dielectric is completely filled between plates : If a dielectric slab is fills completely the gap
between the plates, capacitance increases by K times i.e., C‘ = Ke 0 A Þ
d
C‘ = KC
The effect of dielectric on other variables such as charge. Potential difference field and energy associated with a capacitor depends on the fact that whether the charged capacitor is disconnected from the battery or battery is still connected.
Note : @ If nothing is said it is to be assumed that battery is disconnected.
 When dielectric is partially filled between the plates : If a dielectric slab of thickness t (t < d) is
inserted between the plates as shown below, then
E = Main electric field between the plates,
Ei = Induced electric

field in dielectric.
E‘ = (E – Ei ) = The reduced value of electric field in the dielectric. Potential difference between the
two plates of capacitor is given by
V ‘ = E (d – t) + E‘ t = E (d – t) +
E . t K
Þ æ t ö s æ
t ö Q æ t ö
V ‘ = E ç d – t + K ÷ = e ç d – t + K ÷ = Ae ç d – t + K ÷
è ø _{0} è ø _{0} è ø
Now capacitance of the capacitor
C‘ = Q
V ‘
Þ C‘ =
e _{0} A d – t + t
K
C‘ = e_{0} A
C‘ =
e _{0} A
æ t1 t2 t3 ö
æ t1 t 2 t 3 t 4 ö
d – (t1 + t2 + t3 + ……..) + ç K + K + K +…….. ÷
ç + + + ÷
K K K K
è 1 2 3 ø è 1 2 3 4 ø
 When a metallic slab is inserted between the plates :
Capacitance
C‘ = e 0 A
(d – t)
C‘ = ¥
(In this case capacitor is said to be short circuited)
 When separation between the plates is changing : If separation between the plates changes then it’s
capacitance also changes according to
C µ 1 . The effect on other variables depends on the fact that whether the
d
charged capacitor is disconnected from the battery or battery is still connected.
(i) Separation is increasing
Quantity  Battery is removed
A
d¢  Battery remains connected
A
d¢
V 
Capacity  Decreases because C µ 1 i.e., C‘ < C d  Decreases i.e., C‘ < C 
Charge  Remains constant because a battery is not present i.e., Q‘ = Q  Decreases because battery is present i.e., Q‘ < Q Remaining charge (Q – Q‘ ) goes back to the battery. 
Potential difference  Increases because V = Q Þ V µ 1 i.e., V ‘ > V C C  V¢ = V (Since Battery maintains the potential difference) 
Electric field  Remains constant because E = s = Q e _{0} Ae _{0} i.e., E‘ = E  Decrease because E = Q Þ E µ Q Ae _{0} i.e., E‘ < E 
Energy  Increases because U = Q 2 Þ U µ 1 2C C i.e., U‘ > U  Decreases because U = 1 CV ^{2} Þ U µ C 2 i.e., U‘ < U 
 Separation is decreasing
Quantity  Battery is removed  Battery remains connected  
Capacity  Increases because C µ 1 i.e., C‘ > C d  Increases i.e., C‘ > C  
Charge  Remains constant because battery is not present i.e., Q‘ = Q  Increases because battery is present i.e., Q‘ > Q Remaining charge (Q‘ – Q) supplied from the battery.  
Potential difference  Decreases because V = Q Þ V µ 1 i.e., V ‘ < V C C  V¢ = V (Since difference)  Battery  maintains  the  potential 
Electric field  Remains constant because E = s = Q e _{0} Ae _{0} i.e., E‘ = E  Increases because E = Q Þ E µ Q Ae _{0} i.e., E‘ > E  
Energy  Decreases because U = Q 2 Þ U µ 1 2C C i.e., U‘ < U  Increases because U = 1 CV ^{2} Þ U µ C 2 i.e., U‘ > U 
Force Between the Plates of a Parallel Plate Capacitor.
Field due to charge on one plate on the other is
E = s , hence the force
2e _{0}
F = QE
æ s ö s ^{2}
F = –sA ´ ç
è 2e
Þ  F = s 2 A =
÷ = – A
0 ø 2e 0
Q2
2e _{0} 2e _{0} A
Energy Density Between the Plates of a Parallel Plate Capacitor.
The energy stored in a capacitor is not localised on the charges or the plates but is distributed in the field. And as in case of a parallel plate capacitor field is only between the plates i.e. in a volume (A× d), the so called energy density.
Energy
1 CV ^{2}
1 ée
A ù V ^{2}
1 æ V ö ^{2} 1
Hence Energy density =
= 2 = ê ú
= e _{0} ç ÷ =
e _{0} E ^{2} .

Volume
Ad 2 ë d û Ad
2 è d ø 2
Example: 109 The mean electric energy density between the plates of a charged capacitor is (here Q = Charge on the capacitor and A = Area of the capacitor plate) [MP PET 2002]
(a)
Q 2
2e _{0} A ^{2}
Q
2e _{0} A ^{2}
Q2
2e _{0} A
 None of these
Solution: (a) Energy density =
1 e _{0} E ^{2} =
æ ö Q^{2}




e _{0} ç ÷ .
2 2 è Ae _{0} ø
2e _{0} A^{2}
Example: 110 Plate separation of a 15m F capacitor is 2 mm. A dielectric slab (K = 2) of thickness 1 mm is inserted between
the plates. Then new capacitance is given by [BHU 1994, Similar to BHU 2000]
(a)
15m F
(b)
20m F
(c)
30m F
(d)
25m F
Solution: (b) Given C = e 0 A = 15m F
d
……..(i)
Then by using C‘ = e0 A = e0 A
= 2 ´ e_{0} A ´ 10^{3} ; From equation (i) C‘ = 20m F.
d – t + t
K
2 ´ 10^{–}^{3} – 10^{–}^{3} + 103 3
2
Example: 111 There is an air filled 1 pF parallel plate capacitor. When the plate separation is doubled and the space is filled
with wax, the capacitance increases to 2 pF. The dielectric constant of wax is [MNR 1998]
(a) 2 (b) 4 (c) 6 (d) 8
Solution: (b) Given that capacitance C = 1 pF
After doubling the separation between the plates C‘ = C
2
and when dielectric medium of dielectric constant k filled between the plates then C‘ = K C
2
According to the question, C‘ = K C = 2
2
Þ K = 4.
Example: 112 If a slab of insulating material 4 ´ 10 ^{5} m thick is introduced between the plate of a parallel plate capacitor, the distance between the plates has to be increased by 3.5 ´ 10^{5} m to restore the capacity to original value. Then the dielectric constant of the material of slab is [AMU 1999] (a) 10 (b) 12 (c) 6 (d) 8
Solution: (d) By using
K = t ; here t = 4 ´ 10^{–}^{5} m ; d‘ = 3.5 ´ 10^{–}^{5} m Þ
t – d‘
K = 4 ´ 10 ^{5} = 8
4 ´ 10 ^{5} – 3.5 ´ 10 ^{5}
Example: 113 The force between the plates of a parallel plate capacitor of capacitance C and distance of separation of the plates d with a potential difference V between the plates, is [MP PMT 1999]
(a)
CV 2 (b)
2d
C 2 V 2
2d ^{2}
(c)
C 2 V 2
d 2
(d)
V ^{2}d C
Solution: (a) Since F =
Q2
2e _{0} A
Þ F = C 2 V 2
2e _{0} A
= CV ^{2} .
2d
Example: 114 A capacitor when filled with a dielectric
K = 3
has charge
Q_{0} , voltage V_{0}
and field
E_{0} . If the dielectric is
replaced with another one having K = 9, the new values of charge, voltage and field will be respectively
(a)
3Q , 3V , 3E
(b)
Q , 3V , 3E
(c)
Q , V0 , 3E
(d)
Q , V0 , E0
0 0 0
0 0 0
0 3 0
^{0} 3 3
Solution: (d) Suppose, charge, potential difference and electric field for capacitor without dielectric medium are Q, V and E
respectively
With dielectric medium of K = 3
Charge Q_{0} = Q
With dielectric medium of K = 9
Charge Q‘ = Q = Q_{0}
Potential difference V0 = V
3
Potential difference V ‘ = V
9
= V0
3
Electric field
E = E
Electric field E‘ = E = E0 .
^{0} 3 9 3
Example: 115 A slab of copper of thickness b is inserted in between the plates of parallel plate capacitor as shown in the
figure. The separation between the plates is d. If
b = d
2
then the ratio of capacities of the capacitor after and
before inserting the slab will be [IITJEE 1976; Similar to Orissa JEE 2002, KCET 2001, MP PMT 1994]
(a) : 1 (b) 2 : 1 (c) 1 : 1 (d) 1 :
Solution: (b) Capacitance before inserting the slab C = e 0 A
d
and capacitance after inserting the slab C‘ = e 0 A .
d – t
Where t = b = d
so C‘ = 2e 0 A
hence,
C‘ = 2 .
2 d C 1
Example: 116 The capacity of a parallel plate condenser is C_{0} . If a dielectric of relative permitivity e _{r}
and thickness equal to
one fourth the plate separation is placed between the plates, then its capacity becomes C. The value of C
C0
will be
(a)
5e _{r}
4e _{r} + 1
(b)
4e _{r}
3e _{r} + 1
(c)
3e _{r}
2e _{r} + 1
(d)
2e _{r}
e _{r} + 1
Solution: (b) Initially capacitance C0
= e _{0} A d
……..(i) Finally capacitance C = e 0 A
d – d + d / 4
……..(ii)
By dividing equation (ii) by equation (i)
C = 4e _{r}
4 e _{r}
C0 3e _{r} + 1
Grouping of Capacitors.
Series grouping Parallel grouping
(1) Charge on each capacitor remains same and equals to the main charge supplied by the battery


C_{1} C_{2} C_{3}
 otential difference across each capacitor remains same and equal to the applied
+Q1
+

+
–Q1
–
–
+Q –Q +Q –Q +Q –Q
+ – + – + –
potential difference ^{+} ^{–}
+ – + – + –
+Q2 –Q2
V = V
+ – + – + – + –

+ V + V + – + – + – + –
1 2 3
V_{1} V_{2} V_{3}
+ –
V
Q = Q_{1}
+ Q_{2}
+ Q_{3}
Q2
Q Q3
+
+
+Q3
+
+
–
–
–Q3
–
–
 Equivalent capacitance + –
+ –
1
Ceq
= 1 + 1 + 1 or
C1 C2 C3 V
 In series combination potential difference and energy distribution in the reverse ratio of capacitance e.,
 C_{eq} = C_{1} + C_{2} + C_{3}
V µ 1
C
and U µ 1 .
C
 In parallel combination charge and energy distributes in
the ratio of capacitance i.e. Q µ C and U µ C
 If two capacitors having capacitances C_{1} and C_{2} are
connected in series then Ceq = C1C2 = Multiplication
 If two capacitors having capacitance C_{1} and C_{2}
respectively are connected in parallel then
C_{1} + C_{2} Addition
Ceq = C1 + C2
æ C1 ö
and
æ C2 ö
V1 = ç C + C ÷. V V2 = ç C + C ÷. V
and
è 1 2 ø è 1 2 ø
 If n identical capacitors each having capacitances C are connected in series with supply voltage V then
(5) If n identical capacitors are connected in parallel

Equivalent capacitance Ceq = C and Potential
Equivalent capacitance Ceq = nC and Charge on each
Q
difference across each capacitor V ‘ = V .
n
capacitor Q‘ = n
Redistribution of Charge Between Two Capacitors.
When a charged capacitor is connected across an uncharged capacitor, then redistribution of charge occur to equalize the potential difference across each capacitor. Some energy is also
wasted in the form of heat.
Suppose we have two charged capacitors C_{1}
and C2
after disconnecting
these two from their respective batteries. These two capacitors are connected to each other as shown below (positive plate of one capacitor is connected to positive plate of other while negative plate of one is connected to negative plate of other)
_{‘} æ C1 ö _{‘} æ C2 ö
Charge on capacitors redistributed and new charge on them will be Q1 = Q ç C + C ÷ , Q2 = Q ç C + C ÷
The common potential V = Q1 + Q2
= C1 V1 + C2 V2
è 1
and loss of energy DU =
2 ø
C1C2
è 1 2 ø


(V – V )^{2}
C1 + C2
C1 + C2
2(C1
+ C2 )
Note : @Two capacitors of capacitances C_{1} and C_{2} are charged to potential of V_{1} and V_{2} respectively. After disconnecting from batteries they are again connected to each other with reverse polarity i.e., positive plate
of a capacitor connected to negative plate of other. So common potential V = Q1 – Q2
C1 + C2
= C1 V1 – C2 V2 .
C1 + C2