Physics - Grade XII - Capacitors and Dielectrics.pptx

Capacitors and Dielectrics
Class XII |Physics
Copyright Abiona Education

Overview
Capacitance of a Conductor
Capacitance of an Isolated Spherical Conductor
Potential Energy of a Charged Conductor
Redistribution of Charges : Common Potential
(Optional)
Capacitor
Capacitance of a Parallel Plate Capacitor
Copyright Abiona Educationv

Electrostatics Of Conductors
3
We have talked about –
Conductors
Metallic (electrons)
Electrolytic (ions : positive and negative)
Insulators
What we are going to learn –
Important results regarding electrostatics of conductors.

Important results regarding electrostatics of conductors –
A good electric conductor such as Copper, contains charges (free electron)
that are not bound to anty atom and are free to move within the material
1. Inside a conductor, electrostatic field is zero
2. At the surface of a charged conductor, electrostatic field must be normal to the surface at every point
3. Interior of a conductor can have no excess charge in the static situation
4. Electrostatic potential is constant throughout the volume of the conductor and has the same value
(as inside) on its surface
5. Electric field at the surface of a charged conductor
6. Electrostatic Shielding
4
𝑬=
𝝈
𝝐𝟎
^
𝒏

1. Inside a conductor, electrostatic field is zero
5
The electric field inside the conductor is zero. (Any net electric field in the
conductor would cause charge to move since it is abundant and mobile.
This violates the condition of equilibrium: net force = 0.)

2. At the surface of a charged conductor, electrostatic field must be normal
to the surface at every point.
If there were a field component parallel to the surface, it would cause
mobile charge to move along the surface, in violation of the assumption of
equilibrium.
6

3. Interior of a conductor can have no excess charge in the static situation.
There is no net charge at any point inside the conductor, and any excess
charge must reside at the surface.
The mutual repulsion of like charges from Coulomb's Law demands that
the charges be as far apart as possible, hence on the surface of the
conductor.
7

8
4. Electrostatic potential is constant throughout the volume of the
conductor and has the same value (as inside) on its surface.
Since E = 0 inside the conductor and has no tangential component on the
surface, no work is done in moving a small test charge within the
conductor and on its surface.
That is, there is no potential difference between any two points inside or
on the surface of the conductor. If the conductor is charged, electric field
normal to the surface exists; this means potential will be different for the
surface and a point just outside the surface.
In a system of conductors of arbitrary size, shape and charge
configuration, each conductor is characterised by a constant value of
potential, but this constant may differ from one conductor to the other.

5. Electric field at the surface of a
charged conductors
• For > 0, electric field is normal to surface
outward
• For < 0, electric field is normal to surface
inward
06/08/2025 Electric Charges and Fields 9
Here σ is the surface charge density and n is the unit vector normal to the
surface in the outward direction.

6. Electrostatic Shielding
If a conductor have cavity with no charges inside the cavity, the
electric field inside the cavity is zero.
10
Whatever be the charge and field
configuration outside, any cavity in a
conductor remains shielded from
outside electric influence: the field
inside the cavity is always zero. This is
known as electrostatic shielding.
The effect can be made use of in
protecting sensitive instruments from
outside electrical influence.

Summary: Electrostatics Of
Conductors
11
Below figure gives a summary of the important electrostatic properties of
a conductor –

Electrostatics of Dielectrics
gives a
summary of the important electrostatic properties
of a conductor.

Polarisation of Dielectrics
The dipole moment per unit volume is called polarisation and
is denoted by P. For linear isotropic dielectrics,
Non-polar molecules Polar molecules

Polarisation of Dielectrics
Polarised dielectric is
equivalent to two charged
surfaces with induced surface
charge densities, say σp and –
σp

Capacitors
What is a Capacitor?
A capacitor is a device in which electrical energy can be stored.
Electrical energy is stored in battery also. How are capacitors different from Battery? Where do we use a
capacitor?
The batteries can supply energy at only a modest rate, too slowly for the photoflash unit to emit a flash of
light. However, once the capacitor is charged, it can supply energy at a much greater rate when the
photoflash unit is
triggered—enough energy to allow the unit to emit a burst of bright light.
Batteries in a camera store energy in the photoflash unit by charging a capacitor.

Capacitors
Capacit0r (condenser) is a device that stores (or condenses) electrostatic field energy. Capacitors
provides temporary storage of energy in circuit and can be made to release the energy when required.
The property of a capacitor that characterizes its ability to store energy is called as its capacitance.
When energy is stored in a capacitor, the electric field exist within the capacitor and the stored energy
within the capacitor can be associated with the electric field.
Capacitance can be explained for –
1. An isolated conductor
2. A system of conductors (called as plates or electrodes)

Capacitance of an isolated
conductor
18
A single or isolated conductor can act as Capacitor.
[Note: In this case the other conductor is supposed to be at infinity, whose potential will be taken as zero.]
An electrical conductor (such as metal) holds its charge at its surface. When an isolated spherical
conductor of radius R is given with a charge Q, potential on it is given as –
V =
Hence, Potential
This means that the ratio of charge given to an isolated conductor and its potential rise, that is Q/V is a
constant, which is defined as the Capacitance of the conductor, denoted by C.
C = Q/V = R
Hence, capacitance of a spherical conductor, C = Q/V =

Checkpoint
Two uniformly charged spherical drops at potential V coalesce to form a larger drop. If capacity of each
smaller drop is C, then find the capacitance and potential of the larger drop.

Checkpoint
20
Two uniformly charged spherical drops at potential V coalesce to form a larger drop. If capacity of each
smaller drop is C, then find the capacitance and potential of the larger drop.

Capacitance of system of Conductor
How are capacitors made using system of conductors?
To make a capacitor, just insulate two conductors from each other.
To store energy in this device, transfer charge from one conductor to
the other so that one has a negative charge and the other has an equal
amount of positive charge.
Work must be done to move the charges through the resulting
potential difference between the conductors, and the work done is
stored as electric potential energy.
Two conductors, isolated electrically from each other and from their
surroundings, form a capacitor. When the capacitor is charged, the charges
on the conductors, or plates as they are called, have the same magnitude q
but opposite signs.

A simplest form of capacitor
A parallel-plate capacitor, consisting of two parallel conducting plates of area A separated by a distance d
is the simplest form of Capacitor.
The symbol we use to represent a capacitor is based on the structure of a parallel-plate capacitor but is
used for capacitors of all geometries.
When a capacitor is charged, its plates have charges of equal magnitudes but opposite signs: +q and - q.
However, we refer to the charge of a capacitor as being q, the absolute value of these charges on the
plates.
Note:
q is not the net charge on the capacitor.
Net charge on the capacitor is (+q) + (-q)
= 0

Electric Charges and Fields 23
A simplest form of capacitor
Because the plates are conductors, they are equipotential surfaces; all points on a plate are at the same
electric potential. Moreover, there is a potential difference between the two plates.
The charge q and the potential difference V for a capacitor are proportional to each other. Which means
–
Its value depends only on the
geometry of the plates and not on
their charge or potential difference.
The capacitance is a measure of how
much charge must be put on the
plates to produce a certain potential
difference between them. The greater
the capacitance, the more charge is
required.
The proportionality constant C is called the capacitance of the capacitor.

1. A capacitor is a system of two conductors separated by an insulator
2. Conductors have charges, say Q1 and Q2, and potentials V1 and V2
• Usually, in practice, the two conductors have charges Q and – Q, with potential
difference V = V1 – V2 between them
• Total charge of the capacitor is zero
3. V is also proportional to Q, and the ratio Q/V is a constant called as Capacitance

Capacitance
• Capacitance depends on
1. Geometrical configuration (shape, size, separation) of conductors
2. Nature of dielectric (insulator) between conductors
• SI unit is Farad
• Large capacitance of a capacitor means that it can hold
large amount of charge Q at relatively small V
• Maximum electric field that a dielectric can withstand
without break-down (of its insulating property) is called
dielectric strength

The capacitance of a conductor is a measure of its ability to hold
electric charge. When a conductor is given a charge, its potential rises in
proportion to the charge given. Thus, if a charge Q raise the potential of a
conductor by V, then
Q ∝ V
or Q = C V,
where C is a constant depending upon the size and shape of the
conductor, the surrounding medium and he presence of other conductors
nearby. This constant C is known as the ‘capacitance’ of the conductor.
From the above equation,
C = Q/V.
Thus, the capacitance of a conductor is defined as the ratio of the charge
given to the rise in the potential of the conductor.

Unit of Capacitance : The SI unit of charge Q is ‘coulomb’ and that of
potential is ‘volt’.
As per the definition of capacitance , the SI unit of capacitance C will be
‘coulomb/volt’.
This is called ‘farad’ and is denoted by F. Thus,
1 farad = 1 coulomb / volt or, 1 F = 1 C V–1
.
The capacitance of a conductor is 1 farad if a charge of 1 coulomb raises
the potential of the conductor by 1 volt.
The farad is too large a unit. (A 1 farad spherical conductor must have a
radius of 9 x 109
meter). Hence, in practice smaller units ‘microfarad’,
‘nanofarad’ and ‘picofarad’ are used:
1 microfarad (1 μF) = 10–6
F
1 nanofarad (1 nF) = 10–9
F
–12

Dimensions of Capacitance :
farad =
So, dimensions of capacitance are [AT]2
/ [MLT–2
] [L] = [M–1
L–2
T4
A2
].

Capacitance of an Isolated Spherical
Conductor
Suppose an isolated spherical conductor of radius a meter is placed in
vacuum (or air) and a charge of + Q coulomb is given to this sphere.
The charge spreads on the outer surface of the sphere uniformly so that the
potential at every point on the surface is ‘same’. As a result, the lines of force
emerging from the sphere are everywhere normal to the surface, that is, they
appear diverging radially from the centre O of the sphere.
Therefore, we can assume the charge +
Q as concentrated at the centre O of the
sphere.
Then, the potential at the surface of the
sphere is
V= 1 / 4 πε0 (Q / a) volt

Capacitance of an Isolated Spherical
Conductor
where 1/4 π ε0 = 9.0 x 109
N-m2
C–2
. If the capacitance of the sphere be C, then
C = Q / V.
Substituting the value of V from the last expression, we get
C = Q / (Q / 4πε0a)
or
Thus, the capacitance of a spherical conductor is directly proportional to its
radius. The larger the radius, the greater the ability to hold a given amount of
charge without running up too high a voltage. ε0 is the permittivity of free
space. According to the above formula
ε0 = C /4 π a.
Hence, the unit of ε0 will be ‘farad / meter’ (F m–1
). Earlier, we have read the
unit of ε0 as coulomb2
/ newton – meter2
(C2
N–1
m–2
). Thus F m–1
and C 2
N–1
m–2
are the units of the same physical quantity (ε0 ).
C = 4πε0a
farad.

Potential Energy of a Charged
Conductor
When a conductor of capacitance C is given a charge Q, it acquires a potential
give by V = Q / C.
The work done in charging the conductor is stored as potential energy in the
field in the vicinity of the conductor.
Suppose we bring from infinity infinitesimally small charge dQ to the
conductor. The work done will be (potential x charge)
dW = V dQ = (Q / C) dQ.
Therefore, the work done in increasing the charge on the conductor from 0 to
Q is

Potential Energy of a Charged
Conductor
This work is the electric potential energy U stored in the conductor. Thus,
But Q = CV,
If C is in farad and V in volt, then the energy U will be in joule.

Suppose two insulated conductors A and B of capacitances C1 and C2 are
given charges Q1 and Q2 raised to potentials V1 and V2 respectively.
Then,
Q1 = C1 V1 and Q2 = C2 V2.

If the two conductors are joined by a thin wire, then the positive charge
begins to flow from the conductor at potential (say V1) to that at lower
potential (say V2) till their potentials become equal. That is, on joining the
conductors, the charges on them are redistributed although the total quantity
of charge remains still Q1 + Q2.
If the conductors are at a sufficient distance apart
so that they do not exert any electrical effect on
each other and the capacitance of the connecting
‘thin’ wire is negligible, then their combined
capacitance will be C1 + C2, Now, if after the
redistribution of charge, the common potential of
the conductors is V, then

Suppose, after redistribution, the charge on A is Q1´ and on Bi is Q2´. Then.
Q1´ = C1 V and Q2´ = C2 V.
∴ Q´1 / Q´2 = C1V / C2V = C1 / C2.
On connecting two charged conductors, the redistributed charged on
them are in the ratio of their capacitances.

(Optional)
Quantity of Transferred Charge : Before joining, the conductor A had a
charge Q1 which becomes Q1´ after A is joined to B. Thus, the charge
transferred from A to B is
Q1 – Q1´ = C1V1 – C1V
= C1 (V1 – C1 V1 + C2 V2 / C1 + C2)
= C1 C2 (V1 – V2) / C1 + C2.

(Optional)
Loss of Energy in Redistribution of Charges : When charge flows from a
conductor at higher potential to that at lower potential, the total potential
energy decreases. Before connecting, the total potential energy of the
two conductors is
U = 1 / 2 (C1 V1
2
+ C1 V2
2
).
After connecting the two conductors, their conbined capacitancce
becomes (C1 + C2) and common potential V. Hence, after connecting, the
total potential energy is
U´ = 1 / 2 (C1 + C2) V2.
Substituting the value of V from Eq. (i), we get
U´= 1/ 2 (C1 + C2) (C1 V1 + C2 V2 / C1 + C2)2
.

(Optional)
Subtracting Eq. (iii) from Eq. (ii), the loss in energy is
U – U´= 1 / 2 (C1 V1
2
+ C2 V2
2
) – 1 / 2 (C1 V1 + C2 V2)2
/ C1 + C2
= 1 / 2 (C1 + C2) [(C1 + C2) (C1 V1
2
+ C2V2
2
) (C1V1 + C2V2)2
]
= 1 / 2 (C1 + C2) [C1
2
V1
2
+ C1 C2 V2
2
+ C1 C2 V1
2
+ C2
2
V2
2
– C1
2
V1
2
– C2
2
V2
2
–
2C1 C2 V1V2]
= C1C2 / 2 (C1 + C2) [V1
2
+ V2
2
– 2V1 V2]
= C1C2 / 2 (C1 + C2) (V1 – V2)2
.
In this expression, C1 and C2 are both positive and (V1 – V2)2
, being a square
term, is also positive. Thus, in redistribution of charges there is always a
loss of energy. The balance of energy (U – U´) appears partly as heat in the
connecting wire and partly as light and sound if sparking occurs.

Capacitor -
A capacitor is a pair of two conductors of any shape which are close to
each other and have equal and opposite charges.

Capacitor
Capacitance of a Capacitor : If the charges on the plates of a capacitor
are + Q and – Q and the potential difference between them be V, then the
capacitance of the capacitor is given by
C = Q/V.
The capacitance of a capacitor is defined as the ratio of the charge given
to a plate of the capacitor to the potential difference produced between
the plates.

Capacitor
The capacitance of a capacitor depends upon the following factors:
1. On the Area of the Plates : the capacitance of a capacitor depends
upon the area (A) of the plates, and is directly proportional to it –
(C ∝ A).

Capacitor
2. On the Distance between the Plates : the capacitance of a capacitor
depends upon the distance (d) between the plates and is inversely
proportional to it
(C ∝ 1 / d).
3. On the Medium between the Plates : the capacitance of a capacitor
depends upon the medium between he two plates and increases for a
dielectric medium. It is proportional to the dielectric constant K of the
medium.
(C ∝ K).

Parallel Plate Capacitor
𝑪=
𝜺𝒐 𝑨
𝒅

Effect of Dielectric on Capacitance

Capacitance of a Parallel Plate
Capacitor
A parallel plate capacitor consists of two
long, plane, metallic plates mounted on two
insulating stand and placed at a small
distance apart. The plates are exactly parallel
to each other and the space between them
is filled with air or some dielectric material.
In Fig –
X and Y are two plane metallic plates, each
of area A meter2
and having a distance of d
meter between them. The space between
the plates is completely filled with a
dielectric material of dielectric constant K.

Capacitor
Suppose the plate X is given a charge of + Q coulomb.
By induction, – Q coulomb of charge is produced on the
inner surface of the plate Y and + Q coulomb on the
outer surface. Since, the plate Y is connected to the
earth, the + Q charge on the other surface follows to the
earth. Thus, the plates X and Y have equal and opposite
charges. All the lines of force starting from the plate X
reach the plate Y and except near the edges, the electric
field between the plates is uniform everywhere.
Suppose, the surface density of charge on each plate is
σ. We know that the intensity of electric field at a point
between two plane parallel sheets of equal and
opposite charges is σ/, where is the permittivity of free
space.
If the space between the sheets is filled with a dietetic
material of dielectric constant K, then intensity of
electric field will be given by –

Capacitor
The charge on each plate is Q and the area of each plate is A. thus σ = Q / A,
and so
Now, let the potential difference between the two plates be V volt. Then the
electric field between the plates is given by
or
Substituting the value of E, we get
So, capacitance of the capacitor is
C = Q/V =
or C = farad Where = 8.85 x 10–12
farad / meter.

Capacitor
Note:
In order to obtain high capacitance -
i) A should be large, that is, the plates of large area should be taken,
ii) d should be small, that is, that is, the plates should be kept close to each
other
iii) space between the plates should be filled with a dielectric of high
dielectric constant K.
If these is vacuum (or air) between the plates, then K = 1 and capacitance of
the parallel plate (air) capacitor is
C0 = farad.

with Dielectric Slab between Plates
Let us consider a parallel plate capacitor
having plate charge Q, plate area A and
plate separation d. Suppose a slab of
some dielectric material of dielectric
constant K and thickness t (< d) is
introduced between the plates.
Then, the distance between the plates is t
in the dielectric and (d – t) in vacuum (or
air).

If σ be the surface charge density on the plates, then the electric field in air
between the plates is
[ σ = Q / A]
and that in the dielectric slab (of dielectric constant k) is
The field E0 between the plates is in the distance (d – t) and E in the distance t.
Hence, if the potential difference between the plates be V, then
V = E0 (d – t) + E t
=

So, capacitance of the capacitor is
=
Note:
Since, K > 1, the ‘effective’ distance between the plates becomes less than d and so
the capacitance increases.

Special Cases :
(a) If the dielectric slab fills the entire space between the plates, then t = d, and we
have –
b) If there be vacuum (or air) in the whole space between the plates (that is, t = 0),
then the capacitance will be
c) If there be a slab of metal (K = ∞) of thickness t between the plates, then the
capacitance will be
If the metallic slab fills the entire space (t = d), then the capacitance will become infinite.
However if a metal foils is introduced, then since, t ≈ 0, the capacity will reaction unaltered.

d) If slabs of dielectric constants K1, K2, K3, … and respective thicknesses t1, t2,
t3, … be placed in the entire space between the two plates then, from Eq.
(i), the capacitance will be
But d = t1 + t2 + t3 …
∴

Capacitors in Series
𝟏
𝑪𝒆𝒒
=
𝟏
𝑪𝟏
+
𝟏
𝑪𝟐

Capacitors in Parallel
𝑪𝒆𝒒=𝑪𝟏+𝑪𝟐

Energy Stored in a Charged Capacitor
When a capacitor is charged, say by a battery, work is done by the charging
battery (at the cost of its chemical energy).
As the capacitor charges, the potential difference across its plates rises. More
and more work has to be done by the battery in delivering the same amount
of charge to the capacitor due to the rising potential difference across its
plates.
The total amount of work in charging the capacitor is stored up in the
capacitor in the form of electric potential energy.
This energy is recovered as heat when the capacitor is discharged through a
resistance.

Le us consider a capacitor of capacitance C farad, with a potential difference of
V volt between the plates.
The charge Q is equal to C V coulomb. There is a charge + Q on one plate and –
Q on the other.
In the process of charging, electrons are transferred from the positive to the
negative plate, until each plate acquires an amount of charge Q.
Suppose during the process of charging, we increase the charge from Q´ to Q´
+ dQ´ by transferring an amount of negative charge dQ´ from the positive to
the negative plate.
The potential at that instant will be V´ = Q´ / C. The work that has to be done is
dW = V´dQ´ = (Q´/ C) dQ´

Therefore, the total work done in charging the capacitor from the uncharged
state to the final charge Q will be
But Q = C V.
∴ W = 1 / 2 Q2
/ C = 1 / 2 C V2
.
This is the energy U which is “stored” in the capacitor. Thus,
U = 1/2 (Q2
/C )= 1/2 CV2
joule

This energy resides in the electric field created between the plates of the
charged capacitor.
The above expression can lead to the energy of a charged conducting
sphere for with C = 4πε0a, where a is radius of the sphere.
Thus, its energy is
U = 1/2 (Q2
/C) = Q2
/ (8πε0a)

Energy Stored in a Combination of
Capacitors
If many capacitors are combined in series, or in parallel, the total energy
stored in either combination is equal to the sum of the energies stored in
individual capacitors.
This follows from the combination formulae of capacitors:
The energy stored in a series combination (Q constant) is
U = 1 / 2 Q2
/ C = 1 / 2 Q2
(1 / C1 + 1 / C2 + 1 / C3 + ……)
= 1 / 2 Q2
/ C1 + 1 / 2 Q2
/ C2 + 1 / 2 Q2
/ C3 + ….
= U1 + U2 + U3 + ….
The energy stored in a parallel combination (V constant) is
U = 1 / 2 C1 V2
= 1 / 2 (C1 + C2 + C3 + …) V2
= 1 / 2 C1V2
+ 1 / 2 C2V2
+ 1 / 2 C3V2
+ …

Force between the Plates of a Charged Parallel Plate
Capacitor
Imagine a parallel plate capacitor with a charge
+Q on one of its plate and –Q on the other plate.
Let initially the plates of capacitor are almost, but
not quite touching. Due to opposite polarity
there is an attractive force F between the plates.
Now, if we gradually pull the plates apart to a
distance d, in such a way that d is still small
compared to the linear dimension of the plates,
then the approximation of uniform field E =
(σ/ε0) between the plates is maintained and thus
the force remains constant at F. Now, the work
done in separating the plates from near 0 to d,
W = F d

Force between the Plates of a Charged Parallel Plate
Capacitor
This work done is stored as electrostatic potential energy between the
plates,
U = 1 / 2 Q V
But V = E d,
∴ U = 1 / 2 Q E d
From above equations W = F d and U = 1 / 2 Q E d, we have
F d = 1 / 2 Q E d
or F = 1 / 2 Q E
The factor 1 / 2 arises because just outside the conducting plates the field
is E and inside the plates the field is zero. So, the average value of E/2
contributes to the force.

Charges Induced on the Surfaces of Dielectric
Slab Placed between the Plates of Parallel
Plate Capacitor
Suppose, a slab of some dielectric material (dielectric constant K) is placed
between the plates of a parallel plate capacitor (Fig). The charges on the plates
are + Q and – Q. Let the charges induced on the surfaces of the dielectric slab be
– Q ´ and + Q ´ (Q ´ < Q). Then, by Gauss’ law, the magnitude of the electric field
in the air between the plates is
E0 = Q / ε0 A
and that within the dielectric slab is
E = (Q – Q´) / ε0 A.
But E = E0 / K = Q / K ε0 A.
∴ Q / K ε0 A = Q / ε0 A – Q´ / ε0 A
or Q´ = Q – Q / K = Q (1 – 1 / K).
Thus, the charges induced on the surfaces of the dielectric slab will be – Q (1 – 1 /

Checkpoint
1. Assuming the earth as an insulated spherical conductor of radius 6400
km. Calculate its capacitance.
Answer : 711 μF.
2. Each of the two charged metallic spheres of radii 15 cm and 10 cm has
+ 100 μ C of charge. They are connected by a wire. Find the common
potential and final charge on each sphere. What is the amount of
charge transferred through the wire?
Answer: 7.2 x 106
v 120 μC 80 μC.

Checkpoint
3. Two insulated metallic spheres of capacitances 3.0 and 5.0 μF are
charged to potentials of 300 and 500 volt respectively. They are
connected by a wire. Calculate the common potential, charge on each
sphere and the loss of energy.
Answer: 425 V. 1.275 x 10–3
C 2.125 x 10 – 3
C.
0.0375J
4. Two isolated metallic solid spheres of radii R and 2R are charged such
that both of these have same charge density σ. The spheres are
located far away from each other and connected by a thin conduction
wire. Find the new charge density on the bigger sphere.
Answer: 5 / 6 σ.

Checkpoint
5. What is the area of the plates of 2 F parallel plate capacitor with plate
separation of 0.5 cm? Why do ordinary capacitors have capacitances of
the order of microfarads?
Answer: 1130 km2
6. 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 material
of dielectric constant 6?
Answer: 96 pF.

Checkpoint
7. The area of each plate of a parallel plate capacitor is 100 cm2
and the
intensity of electric field between the plates is 100 N-C–1
. Find charge
on each plate. ε0 = 8.85 x 10–12
C2
N–1
m–2
.
Answer: 8.85 x 10–12
C.
8. A parallel plate capacitor of plate area A = 600 cm2
and plate separation
d = 2.0 mm is connected to a DC source of 200 V. Calculate in SI unit: (i)
the magnitude of the uniform electric field between the plates, (ii)
the charge density σ on any plate. Given : ε0 = 8.85 x 10–12
F m–1
.
Answer : 105
volt / meter. 8.85 x 10–7
C-m–2
.

Checkpoint
9. The area of the parallel plates of an air capacitor is 0.20 m2
and the
distance between them is 0.01 m. The potential difference between its
plates is 3000 V. When a 0.01 m thick sheet of an insulating material is
placed between the plates, the potential difference decreases to 1000
V. Determine :
(i) capacitance of the capacitor before placing the sheet,
(ii) charge on each plate,
(iii) dielectric constant of the material,
(iv) capacitance of the capacitor after placing the dielectric and
(v) permittivity of the dielectric ε. (ε0 = 8.85 x 10–12
F m–1
).
Answer : (i)1.77 x 10–10
F. (ii) 5.31 x 10–7
C.
(iii) 3.0. (iv) 5.31 x 10–10
E.
–11 –1

Checkpoint
10. An isolated 16 μF parallel plate air capacitor has a potential difference of 100
V [Fig]. A dielectric slab having relative permittivity (i.e., dielectric constant) =
5 is introduced to fill the space between the two plates completely [Fig].
Calculate:
i) the new capacitance of the capacitor.
ii) the new potential difference between the two plates of capacitor.
Answer : 20 V.

Checkpoint
11.A parallel plate capacitor is to be designed with a voltage rating 1 kV,
using a material of dielectric constant 3 and dielectric strength 107
V-
m–1
. If the field is not to exceed 10% of the dielectric strength, find the
minimum area of the plate required to have a capacitance of 50 pF.
Answer : 19 cm2
.
12.A 1.0 μF capacitor C1 and 2.0 μF capacitor C2 can separately withstand
maximum voltages V1 = 6.0 kV and V2 = 4.0 kV respectively. What
maximum voltage will the system C1 and C2 withstand if they are
constant in series?
Answer : 9.0 kV.

Checkpoint
13.A parallel plate capacitor contains one mica sheet of thickness d1 = 1.0
x 10–3
m and one fibre sheet of thickness d2 = 0.5 x 10–3
m. The
dielectric constants of mica and fibre are 8.0 and 2.5 respectively. Fibre
breaks down in an electric field of 6.4 x 106 v-m–1
. What maximum
voltage can be applied to the capacitor?
Answer: 5200 V.
14.What should be the capacitance of a capacitor capable of storing 1 J of
energy at 100 V DC supply?
Answer: 200 μF.

Checkpoint
15.(a) A 900 pF capacitor is charged by a 100 V battery. How much
electrostatic energy is stored by the capacitor?
(b) The capacitor is disconnected from the battery and connected in
parallel to another 900 pF capacitor. What is the energy stored by the
system?
Answer: (a) 4.5 x 10–6
J. (b) 2.25 x 10–6
J.
16.The plates of a parallel plate capacitor have an area of 30 cm2
each and
are separated by 2.5 mm. The capacitor is charged to 400 V. How much
electrostatic energy is stored in it? How much, when it is filled with a
dielectric medium of K = 3 and then charged? If it is first charged as an
air capacitor and then filled with the dielectric, then? (ε0 = 8.85 x 10–12
F-m–1
)
Answer: 2.55 x 10–6
J. 7.65 x 10–6
J. 0.85 x 10–6
J.

Checkpoint
17.A 2 μF parallel plate capacitor with a dielectric slab (K = 5) between the
plates is charged to 100 V and then isolated. What will be the PD. If the
dielectric be removed? How much work would be done in removing
the dielectric?
Answer: 500 V. 0.20 J.
18.The capacitance of a parallel plate capacitor is 50 pF and the distance
between the plates is 4 mm. It is charged to 200 V and the charging
battery is removed. Now, a dielectric slab (K = 4) of thickness 2 mm is
placed between the plates. Determine : (i) final charge on each plate,
(ii) final potential difference between the plates, (iii) final energy in the
capacitor (iv) energy loss.
Answer: (i) the charge 10–8
C will remain as such.
(ii) 125 V.
(iii) 6.25 x 10–7
J.

Checkpoint
19.A parallel plate capacitor of capacitance 100 μF is charged to 200 V.
After disconnecting it from the battery, using a insulating handle, the
distance between the plates is doubled. Find (i) potential difference
between the plates and (ii) energy stored in the capacitor after the
separation between the plates has been increased.
Answer: (i) 400 V. (ii) 4 J.
20.A 10 μF capacitor is charged by a 30 V DC supply and then connected
across an uncharged 50 μF capacitor. Calculate (i) the final potential
difference across the combination and (ii) the initial and the final
energies. How will you account for the difference in energy?
Answer: (i) 5.0 V. (ii) 4.5 x 10–3
J. (iii) 0.75 x 10–3
J.

Checkpoint
21.An electric field = E0 x 104
v-m–1
is established between the plates, 0.05
m apart, of a parallel plate capacitor. After removing the charging
battery, an uncharged metal plate of thickness t = 0.01 m is inserted
between the capacitor plates. Find the PD across the capacitor (i)
before, (ii) after the introduction of the metal plate. (iii) If a dielectric
slab (K = 2) were introduced in place of metal plate.
Answer: (i) 1500 V. (ii) 1200 V. (iii) 1350 V.
22.A parallel plate is charged to a certain potential difference. When a 3.0
mm thick slab is slipped between the capacitor plates, then to
maintain the same PD between the plates, the plate separation is to be
increased by 2.4 mm. Find the dielectric constant of the slab.
Answer: 5.

Checkpoint
23.How would you combine 8, 12 and 24 μF capacitors to obtain (i)
minimum capacitance, (ii) maximum capacitance? (ii) if a PD of 100 V
be applied across the system, what would be the charges on the
capacitors in each case?
Answer : (i) 4 μF. (ii) 44 μF. (iii) 400 μC.
24.Connect three capacitors of 3 μF, 3 μF and 6 μF such that their
equivalent capacitance is 5 μF.
25.Three capacitors C1 = 3μF, C2 = 6μF and C3 = 10μF are connected to a 50
V Battery as shown in figure. Calculate (i) the equivalent capacitance of
the circuit between points A and B, (ii) the charge on C1.
Answer: (i) 12 μF. (ii) 100 μC.

Checkpoint
26.You are provided with 8 μF capacitors. Show with the help of a diagram
how you will arrange minimum number of them to get a resultant
capacitance of 20 μF.

Checkpoint
27.An 8 μF capacitor C1 is charged to a potential difference V0 = 120 volt.
The charging battery is than removed and the capacitor is connected
in parallel to an uncharged 4 μF capacitor C2 as shown. (a) What will be
the final PD across the combination? (b) What will be the stored energy
before and after the switch K is pressed? What happens to the energy
difference?
Answer: (b) 5.76 x 10–2
J. 3.84 x 10–2
J.

Checkpoint
28.A battery of 10 V is connected to a capacitor of capacity 0.1 F. The
battery is now removed and this capacitor is connected to a second
uncharged capacitor. If the charge distributes equally on these two
capacitors, find the total energy stored in the two capacitors. Further,
compare this energy with the initial energy stored in the first capacitor.
Answer: 2.5 J. 1/2.
29. A parallel plate capacitor has plates each of area A and separation d.
Two dielectrics of dielectric constants K1 and K2 are filled between the
plates in two arrangements as shown. Find the out the capacitance of
the capacitor in each of the arrangements (a) and (b).
Answer: (a) (b)

Checkpoint
30.A capacitor constant of 7 square plates of side 2.0 cm placed one
above the other, with 1.0 mm thick mica (K = 6) foils between them. The
first, third, fifth and seventh plates are connected to one point; and the
second, fourth and sixth plates to another point. If the potential
difference between these points be 300 V, then calculated the stored
energy in the capacitor.
Answer: 5.71 x 10–6
J.
31.In the combination of four identical capacitors shown, the equivalent
capacitance between points P and Q is 1 μF. Find the value of each
separate capacitance.
Answer: 4 μF.

Checkpoint
32.Four capacitors are connected, as shown. Calculate the equivalent
capacitance between the points P and Q.
Answer: 5 μF.
33.In the network of capacitors given in adjacent figure, find the effective
capacitance between the points P and Q. Given : C1 = C2 = C3 = C4 = 4 μF
and C5 = 5 μF. If a 10 V battery be connected across P and Q, what will
be the charges on the capacitors?
Answer: 20 μC. Zero.

Checkpoint
34.Find the equivalent capacitance of the adjoining circuit between the
junctions A and B; given C1 C2 = C3 = C4 = 10 μF and C5 = 5 μF.
Answer : 10 μF

Checkpoint
35.Three capacitors C1 = 6 μF, C2 = 12 μF and C3 = 20 μF are connected to a
100 V battery as shown in figure. Calculate:
i) charge on each plate of capacitor C1.
ii) electrostatic potential energy stored in capacitor C3.
Answer: 0.1 J.

Checkpoint
36.In the given arrangement of capacitors, C1 = 2.0 μF, C2 = 6.0 μF and C3 =
2.5 μF. Calculate : (i) total capacitance, charge and energy of the
system, (ii) charges on separate capacitors and (iii) PD’s across
separate capacitors.
Answer : (i) 4.0 μF. 800 μC. 0.08 J.
(ii) 300 μC. 500 μC.
(iii) 1500 V, 50 V 200 V.

Checkpoint
37.In the given network, two identical parallel plate capacitors are
connected to a battery with switch S closed. The switch is opened and
the free space between the plates of the capacitors is filled with a
dielectric K = 3. Find the ratio of the total electrostatic energy stored in
both thee capacitors before and after the introduction of the dielectric.
Answer: 0.6.

Checkpoint
38.Three capacitors each of capacitance C are connected in series. their
equivalent capacitance is Cs. The same three capacitors are now
connected in parallel. Their equivalent capacitance becomes Cp. Find
the ratio (Cp/Cs).
Answer: Cp : Cs = 9 : 1.
39.Determine the equivalent capacitance between A and B in the shown
below.
Answer : 8 / 3 μF.

Checkpoint
40.In the given network, each capacitor is of 1 F. Determine the
equivalent capacitance between points A and B.
Answer: 2 F.

Van De Graaff Generator
• Machine that can build up high
voltages
• Resulting high electric field
used to accelerate charged
particles like electrons, protons
for various experiments of
Nuclear physics
• Based on two principles
1. Corona Discharge
2. Movement of charge from
smaller sphere to larger sphere

Physics - Grade XII - Capacitors and Dielectrics.pptx

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