1.

ELECTROSTATICS
Two and Three Mark Question

1) What is meant by quantization of charges?

The charge ‘q’ on any object is equal to an integral multiple of the fundamental unit of charge ‘e’. That
is q = ne ; here n-is an integer

2) Write down Coulomb’s law in vector form and mention what each term represents
The Coulomb’s law in vector form = ̂ ; Here, -is force exerted on the second charge by
the first charge; - first charge ; - second charge; k- proportionality constant; r- distance between the
two charge; ̂ - is the unit vector pointing from charge to .

3) What are the difference between Coulomb force and gravitational force?

Coulomb’s force Gravitational force

It is act between two charges It is act between two masses
It is either attractive force or repulsive force Only attractive force
It is always greater in magnitude It is always smaller in magnitude for small
objects
Proportionality constant k = 9×109 N m2 C-2 Gravitational constant G = 6.67× 10-11 N m2 kg-2
Coulomb’s force depends on the medium Gravitational force independent of the medium
When the charges are in motion, yet another The gravitational force between two point
force (Lorentz force) comes into play in addition masses is the same whether two masses are at
to Coulomb’s force rest or in motion.

4) Write short notes on superposition principle.

According to the superposition principle, the total force acting on a given charge is equal to the vector
sum of forces exerted on it by all the other charges.
= + + + ……
5) Define “Electric field”.
The electric field at the point P at a distance’ r’ from the point charge ‘q’ is the force experienced by a
unit charge and is given by =
6) What is mean by ‘Electric field lines’?
Electric field lines are the visual representation of the electric field in some region of space. They form a
set of continuous lines.

7) The electric field lines never intersect. Justify.

If a charge is placed in the intersection point, then it has to move in two different directions at the same
time, which is physically impossible. Hence, electric field lines do not intersect.

8) Define ‘Electric dipole’.

Two equal and opposite charges separated by a small distance constitute an electric dipole. E.g. CO,
water, ammonia, HCl etc.,

9) What is the general definition of electric dipole moment?

The magnitude of the electric dipole moment is equal to the product of the magnitude of one of the
charges and the distance between them. p = q2a; Here, p-electric dipole moment; q- one of the charge; 2a -
distance between the two charge.

10) Define “Electrostatic potential”.

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 1
 The electrostatic potential at a point ‘P’ is equal to the work done by an external force to bring a unit
positive charge with constant velocity from infinity to the point ‘P’ in the region of the external electric
field (E). =

11) What is an Equi-potential Surface?

An equipotential surface is a surface on which all the points are at the same potential.

12) What are the properties of an equipotential surface?

 If two points A and B are lies on the equipotential surface, then work done to move the charge ‘q’ from A
to B is zero. Since VA = VB
 The electric field must always be normal to equipotential surface.

13) Give the relation between electric field and electric potential.
Electric field is the negative gradient of the electric potential. E = -

14) Define “electrostatic potential energy”.

If the charge ‘q2’ is brought from infinity to that point at a distance ‘r’ from ‘q1’, then the work done is
the product of q2 and the electric potential (V)at that point. Therefore W = q2V. Actually this work done is
stored as the electrostatic potential energy ‘U’ of a system of charges q1 and q2 separated by a distance ‘r’.
U=

15) Define ‘ Electric Flux’.

The number of electric field lines crossing a given area kept normal to the electric field lines is called
electric flux. Unit: N m2 C-1

16) What is meant by electrostatic energy density?

The energy stored per unit volume of space is defined as energy density (uE). uE =

17) Write short on ‘electrostatic shielding’.

Using Gauss’s law we can able to show that the electric field inside both hollow and solid conductor is
zero. Whatever the charge at the surfaces and whatever the electrical disturbances outside, the electric field
inside is zero.
A sensitive electrical instrument which is to be protected from external disturbance can kept inside the
cavity of conductor. This is called as electrostatic shielding.

18) What is Polarisation?

Polarisation is defined as the total dipole moment (induced) per unit volume of the dielectric. In
general, the polarisation is directly proportional to the strength of the external electric field.

= . Here, -electric susceptibility

19) What is dielectric field strength?

The maximum electric field the dielectric can withstand before it breakdowns is called dielectric strength.
E.g. dielectric field strength of air is 3×106 V m-1.

20) What is dielectric break down?

When the external electric field applied to a dielectric is very large, it tears the atoms apart so that the bound
charges become free charges. Then the dielectric starts to conduct electricity. This is called dielectric
breakdown.
A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 2
21) Define capacitance. Give its unit.
The capacitance of a capacitor is defined as the ratio of the magnitude of charge on either of the
conductor plates to the potential difference existing between the conductors. C =

Unit: coulomb volt-1 or farad.

22) What is action at points or corona discharge?

Leakage of electric charges from the sharp end of the charged conductor is called action at points. Or
corona discharge. This is used in lightning arrester and Van Dee Graff generator.

23) Write the use of capacitor in computer key board?

Computer keyboard keys are constructed using capacitors with a dielectric. When the key is pressed,
the separation between the plates decreases leading to an increase in the capacitance. This in turn triggers
the electronic circuits in the computer to identify which key is pressed.

24) Write the application of capacitor. Mention its disadvantages.

֍ Application
• The flash which comes from the camera when we take photographs is due to the energy released from the
capacitor, called a flash capacitor.
• During cardiac arrest, a device called heart defibrillator is used to give a sudden surge of a large amount
of electrical energy to the patient’s chest to retrieve the normal heart function. This defibrillator uses a
capacitor of 175 charged to a high voltage of around 2000 volt.
• Capacitors are used in the ignition system of automobile engines to eliminate sparking.
• Capacitors are used to reduce power fluctuations in power supplies and to increase the efficiency of power
transmission.
֍ Disadvantages
• Even after the battery or power supply is removed, the capacitor stores charges and energy for some time.
For example, if the TV is switched off, it is always advisable to not to touch the back side of the TV panel.

25) Write the application of capacitor in Ceiling fan.

For the ceiling fan, the initial torque is given by the capacitor widely known as a condenser. If the
condenser is faulty, it will not give sufficient initial torque to rotate the blades when the fan is switched on.

26) What is fringing field in capacitor?

In the capacitor with finite size plates, the electric field is not strictly uniform between the plates. At both
edges, the electric field is bend outwards. This is called “fringing field”.
However under the condition (d2 << A), this effect can be ignored.

27) Define polar and non-polar molecules.

Polar molecules Non-polar molecules
In polar molecules, the centers of the positive and A non-polar molecule is one in which centers of
negative charges are separated even in the absence of positive and negative charges coincide. As a result,
an external electric field. They have a permanent it has no permanent dipole moment.
dipole moment.
Examples, H2O, N2O, HCl, NH3 Examples, H2, O2 and CO2.

28) Write a short note on Microwave Oven.

• It is a device which is used to cook and heat the food very quickly.
֍ Principle
Torque acting on the electric dipole.
֍ Working

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 3
 When we switch on the oven it produces oscillating electromagnetic fields and produce torque on the
water molecules present in food. Due to this torque on each water molecule, it begins to rotate very fast and
produce thermal energy. Thus, heat generated is used to heat the food.

29) State the Gauss law.

The electric flux(Φ ) of the electric field(E) over any closed surface is equal to times the net
charge(Qencl) enclosed by the surface.

Φ =

COMPARISON OF ELECTRIC AND MAGNETIC FIELD

Electric Field Magnetic Field

1) Source: Charge (q) 1) Source: Magnetic charge or Pole strength(qm)
2) Electric field made up of electric field lines. 2) Magnetic field made up of magnetic field lines.
3) The test charge (q0) is used to measure the 3) Unit North pole is used to measure the magnetic
electric field at a point. (q0 = + 1 coulomb) field at a point. (Unit north pole = 1 ampere meter)
4) Electric field lines start from the +Ve charge 4) Magnetic field lines emerge from North pole and
and terminated at the -Ve charge. converged at South pole. Inside the bar magnet its
direction is from south pole to north pole.
5) Electric dipole 5) Magnetic dipole

6) Electric dipole moment (p). 6) Magnetic dipole moment (pm).

P = q2a Pm = qm2l
2a – distance between two charges. 2l – distance between two magnetic poles.
In vector notation, = 2 ̂ In vector notation = 2 ̂
Direction of electric dipole moment is Direction of magnetic dipole moment is from North
from -q to +q. pole to South pole.
Unit: coulomb meter. Unit: ampere meter2
7) Coulomb’s Law in electric field, 7) Coulomb’s law in magnetic field,

= =
k= = 9 × 10 = = 10 ℎ
4
– Permittivity of free space or vacuum. – Permeability of free space or vacuum.
r- Distance between the two charged particles. r- Distance between two magnetic poles.
− 8.854 × 10 − 4 × 10 ℎ

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 4
8) Electric field due to an electric dipole at a point 8) Magnetic field due to magnetic dipole at a point on
on the axial line. the axial line.
1 2 2
= =
4 4
9) Electric field due to an electric dipole at a point 9) Magnetic field due to a magnetic dipole at a point
on the equatorial line. on the equatorial line.
1
=− = −
4 4
10) Torque experienced by an electric dipole in 10) Torque experienced magnetic dipole in uniform
uniform electric field , magnetic field,
= × = ×
= sin = sin
11) Electro static potential energy of a dipole in a 11) Potential energy of the bar magnet in a uniform,
Uniform electric field. magnetic field.
= − ∙ = − ∙
= − cos = − cos

12) The force (FE) experienced by electric charge 12) The force (FB) experienced by magnetic charge
(q) in uniform electric field (E). (qm) in uniform magnetic field (B).
= =

FIVE MARKS QUESTIONS

1) Discuss the basic properties of electric charges.

֍ Electric Charge
Electric charge is intrinsic and fundamental property of particles like ‘masses’. The basic electric charge
is = 1.6 × 10 coulomb.
֍ Conservation of charges
The total charge in the universe is constant and charge can neither be created nor be destroyed. In any
physical process, the net change in charge will always be zero.
֍ Quantization of charges
The total charge ‘q’ of any object is equal to an integral multiple of this fundamental unit of charge ‘e’.
=
Here, n is an integer.( 0, ±1 , ±2, ±3, ±4 … … . )

2) Explain in detail Coulomb’s law and its various aspects.

֍ Coulomb’s law
The force of attraction or repulsion between two charges (q1, q2) is directly proportional to product of
charges and is inversely proportional to square the distance(r) between them.
=

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 5
 = Force acting on the second charge by the first charge. = unit vector directed from the first charge
(q1) to second charge (q2). = permittivity of free space or vacuum = 8.854 × 10 . The
value of = 9 × 10 N .

֍ Important aspects of Coulomb’s law

• The force of attraction or repulsion between two charges (q1, q2) is directly proportional to product of
charges and is inversely proportional to square the distance(r) between them.
• = Force acting on the second charge by the first charge, = unit vector directed from the first charge
(q1) to second charge (q2). Similarly = Force acting on the first charge (q1) by the second charge (q2),
its direction is from second charge to first charge. That is - .
Therefore the Coulomb’s law can be written as = −
From the above equation we came to conclusion the electrostatic force obeys Newton’s Third law.
• In SI units, = permittivity of free space or vacuum = 8.854 × 10 . The value of
= 9 × 10 N .
• The force between two point charges in a medium other than vacuum is always less than that in vacuum.
= , = permittivity of free space or vacuum; = Relative permittivity of the medium;
= permittivity of the medium. (For vacuum = 1, for all other medium its value is greater than one)
• Coulomb’s law has same structure as Newton’s law of gravitation. But there is some important difference
between these two laws.
Coulomb’s force Gravitational force
It is act between two charges It is act between two masses
It is either attractive force or repulsive force Only attractive force
It is always greater in magnitude It is always smaller in magnitude for small
objects
Proportionality constant k = 9×10 N m C9 2 -2
Gravitational constant G = 6.67× 10-11 N m2 kg-2
Coulomb’s force depends on the medium Gravitational force independent of the medium
When the charges are in motion, yet another The gravitational force between two point
force (Lorentz force) comes into play in addition masses is the same whether two masses are at
to Coulomb’s force rest or in motion.

• The expression for Coulomb’s force is true only for point charges. But the point charge is an ideal
concept. We can apply Coulomb’s law for two charged objects whose sizes are very much smaller than the
distance between them.

3) Define ‘Electric Field’ and discuss its various aspects.

The electric field at the point P at a distance’ r’ from the point charge ‘q’ is the force experienced by a
unit charge(q0) and is given by = = ̂= ̂
Electric field is a vector quantity. Unit = N C or V m
-1 -1

֍ Important aspects of electric field

• If the charge q is positive then the electric field points radially away from the source charge and if q is
negative, the electric field points radially towards the source charge.
• If the electric field at a point P is , then the force experienced by the test charge q0 placed at the point P
is, = . This is Coulomb’s law in terms of electric field.
• From the above equation the electric field is independent of the test charge and it is depends on source
charge.
• When the distance from the source charge increases then the electric field is decreases in magnitude. The
length of the electric field vector is depends on the magnitude of electric field.

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 6
• The test charge is made sufficiently small such that it will not modify the electric field of the source
charge.
• The expression = is valid only for point charges. For continuous and finite size charge
distributions, integration techniques must be used.
• There are two kinds of electric fields. Uniform and non-uniform electric fields.
Uniform electric field Non-uniform electric field
• Uniform electric field will have the same direction • Non-uniform electric field will have different
and constant magnitude at all points in space. directions or different magnitudes or both at different
points in space.
• If two oppositely charged plates are placed very • The electric field created by the point charge is
close to each other, may produce uniform electric basically a non-uniform electric field.
field.

4) a) Calculate the electric field due to a dipole on its axial line.

b) Calculate the magnetic field at a point on the axial line of a bar magnet.

ELECTRIC FIELD MAGNETIC FIELD

• Consider an electric dipole placed on the x – • Consider a bar magnet. N be the North pole and S be
axis. the south pole of the bar magnet, each of pole strength
A point C is located at a distance of ‘r’ from the qm and separated by a distance of 2l.
midpoint ‘O’ of the dipole along the axial line.
•

• The electric field at a point C due to +q is,

• In order to find out the magnetic field at ‘C’ we have
= ( ) to keep unit north pole at that point.
• Since the electric dipole moment vector is
from • The magnetic field at C due to North pole,
-q to +q and is directed along BC, the above = ̂ →→→ (1)
( )
equation is rewritten as
Here, (r – l) is the distance between north pole to the
1
= ̂ →→→ ( 1 ) considering point C
4 ( − )
Here, ̂ is the electric dipole moment unit vector • The magnetic field at C due to South pole,
from -q to +q
= − ( ) ̂ →→→ (2)
• The electric field at a point C due to –q
1 Here, (r + l) is the distance between south pole to the
= − considering point C
4 ( + )
• Since the electric dipole moment vector is The resultant magnetic field at C,
from = + →→→ (3)
-q to +q, which is opposite to the direction of CA,
the above equation is rewritten as, Substitute (1) and (2) in (3),
1 = ̂ + − ̂
= − ̂ →→→ ( 2 ) 4 ( − ) 4 ( + )
4 ( + )
• The total electric field at point C is calculated = ( )
− ( )
̂ →→→ ( 4 )
using the superposition principle of the electric By simplifying the equation (4)
field

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 7
 axial = + →→→ ( 3 ) =
∙
̂ →→→ (5)
( )
Substitute (1) and (2) in (3).
axial = ( )
̂ − ( )
̂ • Magnetic dipole moment, from the definition can be
written as,
= − ̂
| |= = ∙ 2 →→→ (6)
axial ( ) ( )
1 1 1 Substitute (6) in (5),
= − ̂
4 ( − ) ( + ) = ̂ →→→ (7)
= ̂ →→→ (4)
( )
( ) If the distance between two poles in a bar magnet
The total electric field is along , since +q is are small compared to the distance between
closer to C than –q. geometrical centre O of bar magnet and the location of
If the point C is very far away from the dipole point C, i.e., ( r >> l). Then the above equation can be
then ( r >> a), then the above equation can be written as,
written as, = ̂ →→→ (8)
1 4
= ̂ Since, (( − ) ) ≈
4 We know that ,
Since the electric dipole moment = 2 ̂ = ̂
1 2 By using the above equation to (8) ,
=
4 2
Note =
4
The direction of resultant electric field is in the Note
direction of electric dipole moment. The direction of resultant magnetic field is in the
direction of dipole moment.

5) a) Calculate the electric field due to a dipole on its equatorial line.

b) Calculate the magnetic field at a point on the equatorial line of a bar magnet.

Electric Field Magnetic Field

• Consider a point C at a distance ‘r’ from the • Consider a bar magnet. N be the North pole and S
midpoint ‘O’ of the dipole on the equatorial plane. be the south pole of the bar magnet, each of pole
• The point ‘C’ is equi-distant from +q and –q, the strength qm and separated by a distance of 2l.
Magnitude of the electric fields of +q and –Q are • In order to find out the magnetic field at ‘C’ we
same. have to keep unit north pole ( = 1 A m) at that
• is the electric field due to +q charge and acts point.
along the direction BC.
• is the electric field due to –q charge and acts
along the direction CA.

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 8
•
• The magnetic field at ‘C’ due to north pole is,
• and are resolved into two components, one
component parallel to the dipole axis and the other = − cos ̂ + sin ̂ →→→ (1)
perpendicular to it. Here, = ′
and ′ = ( + )
• The perpendicular components sin and
sin are oppositely directed and cancel each • The magnetic field at C due to south pole is,
other. = − cos ̂ − sin ̂ →→→ (2)
• The magnitude of the total electric field at point C Here, = and ′ = ( + )
is the sum of the parallel components of and
′

Its direction is along - ̂ .

• The magnetic field at C due to the dipole,
• =- +
= + →→→ (3)
• =- cos ̂ − cos ̂ → ( 1 ) Substitute (1) and (2) in (3)
= −( + ) cos ̂
• The magnitudes and are same and are given Since = =B
by
= −2 cos ̂ →→→ (4)
= = →→→ ( 2 )
( ) But, = and ′ = ( + ) →→→(5)
Substitute (2) in (1) ′

=- cos ̂– cos ̂
( ) ( ) Substitute (5) in (4),
= -2 ( )
cos ̂ = −2 cos ̂
′

But ′ = ( + )
=- ( )
̂ →→→ ( 3 ) The above equation can be rewritten as,
= − ( )
cos ̂ →→→ (6)

From the first figure,

cos = = ′= →→→ (7)
( )
Substitute (7) in (6),
×( )
= − ̂ →→→ (8)
( )
But the magnetic dipole moment,
| |= = ∙ 2 →→→ (9)
Substitute (9) in (8)

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 9
 = − ̂
4 ( + )
We know that ( r >> l ) then ( + ) ≈
Rewrite the above equation,
= − ̂
4
We know that ,
= ̂

Therefore the above equation can be written as,

= −
4
Note
֍ From the above equation the direction of
From the first diagram, cos can be written as, magnetic field is opposite to the direction of
cos = →→→ ( 4 ) magnetic dipole moment.
( )
֍ =2
Substitute (4) in (3),
=- ̂ →→→ (5)
( )
The electric dipole moment in vector notation,
=2 ̂
At very large distance ( r >> a ) By using this in the
above equation,
=-
Note
The direction of resultant electric field is opposite
to the direction of electric dipole moment.

6) a) Derive and expression for the torque experienced by a dipole due to a uniform electric field.
b) Derive an expression for the torque experienced by a magnetic dipole due to uniform magnetic
field.

Electric Field Magnetic Field

• Consider an electric dipole of dipole moment placed • Consider a bar magnet of length 2l of pole
in a uniform electric field . strength qm kept in a uniform magnetic field .
• The electric field is uniform since the field lines are • Each pole experiences a force of magnitude qmB
equally spaced and pointed in the same direction. But acts in opposite direction. Therefore, the net
• The force acting on +q charge is = force exerted on the magnet is zero.
And the force acting on the –q charge is = − • From the below diagram the two forces are not
Since the external electric field is uniform from the passing through the same point, so it will
above equation the resultant force acting on the constitute a couple.
dipole is zero.

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 10
 • Due to the couple, the magnet experience a
torque, this torque tends to align (rotate) the
magnet in the direction of external magnetic field.

• From the above diagram the two forces are not • The force experienced by the north pole,
passing through the same point, so it will constitute a = →→→ ( 1 )
couple. The force experienced by the south pole,
• Due to the couple, the dipole experience a torque, this
=− →→→(2)
torque tends to align (rotate) the dipole in the direction
The resultant force acting on the magnetic
of external electric field.
dipole,
• The torque on the dipole about the point ‘O’
= + = 0→→→ ( 3 )
=( × − )+ ( × )→→→ ( 1 )
The moment of force or torque experienced by
Using right hand corkscrew rule we can find the north and south pole about point O is
direction of torque which is perpendicular to the plane
of the paper and is directed into it.
=( × )+( × )→→→ ( 4 )
• The magnitude of the total torque, Substitute (1) and (2) in (4)
= − sin + sin →→→ (2) = × + × − →( 5 )
(Since, = = , = − , × = By using right hand cork screw rule, we
) conclude that the total torque is pointing into the
By using this let us rewrite the equation (2) paper.
= sin + sin
• The magnitude of the total torque,
=2 sin
But, the electric dipole moment p = q 2a = × sin + × sin →→→(6)
By using this, (since, = = , = − ,
= sin × = )
Where, = torque experienced by the electric dipole in By simplify the equation (6),
uniform electric field, = 2 × sin →→→ (7)
In vector notation the above equation can be written
as, • But the magnetic dipole moment,
= × 2 →→→ (8)
= ×
Substitute (8) in (7),
=
In vector notation,
= ×

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 11
7. a) Derive an expression for electrostatic potential energy of the dipole in a uniform electric field.
b) Derive an expression for potential energy of a bar magnet in a uniform magnetic field
Electric Field Magnetic Field
• Consider an electric dipole of dipole moment • Consider a bar magnet of length 2l of pole strength
placed in a uniform electric field . qm kept in a uniform magnetic field
• The electric field is uniform since the field lines • The magnitude of the torque acting on the dipole is
are equally spaced and pointed in the same | | = | | sin →→→ ( 1 )
direction. • If the dipole is rotated through a very small angular
• A dipole experience a torque when kept in an displacement gainst the torque
uniform electric field . at constant angular velocity, then the workdone
• This torque( ) tend to align the dipole along the by external torque ( ) for this small angular
electric field . Now let us rotate the dipole with displacement is given by
uniform angular velocity opposite to the torque = |τ | →→→ ( 2 )
produced by the Electric field . From the angle
′
, by using external torque( ).

• The bar magnet has to be moved at constant angular

• The workdone by the external torque to rotate the velocity, which implies that
dipole from angle ′ at constant angular | | = |τ |
velocity, Substitute (1) in (2)
= sin dθ→→→ ( 3 )
= ′ →→→ (1)
Total work done in rotating the dipole from
We know that, | |= | | ,
But, | | = × = pE sin →→→ (2)
Substitute (2) in (1), = τdθ = sin = (− cos )
W= ′ = (− cos ) ′
= − (cos − cos ′ )
֍ According Work – Energy principle, This work = − _ (cos − cos )
is stored as potential energy in the system. ֍ According to work-energy principle, this work
= − (cos − cos ) ′ done is stored as potential energy.
If the initial angle ′
= 90 and is taken as
° = − ( cos − cos )
reference point, then If the initial angle = 90° and is taken as reference
= − cos point, then
When we write the above equation in dot = − cos
product, When we write the above equation in dot product,
= − ∙
• If = 0 °
U = -pE
If = 180° ℎ
U = +pE.

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 12
8) Derive an expression for electrostatic potential due to a point charge.
Consider a positive charge q kept fixed at the origin. Let ‘P’ be a point at a distance r from the charge q.
֍ from the relation between electric field and electric potential
= − ∙ = − ∙ →→→ ( 1 )
From the definition electric field due to positive point charge at a distance ‘r’ is
= ̂ →→→ ( 2 )
Substitute (2) in (1),
= − ̂∙
= − ̂∙ →→→( 3 )
We know that mathematically, the infinitesimal displacement vector,
= ̂ and (( ̂ ∙ ̂ ) = 1) →→→( 4)
Substitute (4) in (3),
= − ̂∙ ̂
= − →→→ ( 5 )
By integrating the equation (5),
= − − =
Therefore the electric potential due to a point charge q at a distance r is

9) Derive an expression for electrostatic potential due to an electric dipole.

• Two equal and opposite charges separated by a small distance constitute and electric dipole. The electric
dipole moment p = q 2a. This is shown in the
figure.
• The point P is located at a distance ‘r’ from the
midpoint of the dipole. Let be the angle between
the line OP and dipole axis AB.
• Let r1 be the distance of point P from +q and r2 be
the distance of point P from –q.
•Potential at P due to charge +q is
V1 = →→→( 1 )
Potential at P due to charge –q is
V2 = − →→→ ( 2 )
The total potential at the point P is,
= + →→→ ( 3 )

Substitute (1) and (2) in (3),

= + −
= − →→→ ( 4 )
Rewrite the equation (4) in terms of ‘r’, Apply the cosine law for triangle BOP,
= + − 2 cos
2
= 1+ − cos

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 13
 Since the P is very far from dipole then r > > a. As a result the term is very small and can be
neglected. Therefore,
2
= 1− cos

2
= 1− cos

2
= 1− cos
Find the reciprocal of the above equation,
1 1 2
= 1− cos
Since a/r << 1, we can use binomial theorem and retain the terms up to the first order,
= + →→→ ( 5 )

Similarly applying the cosine law for triangle AOP,

= + − 2 (180 − )
But, cos(180 − ) = − cos
= + + 2 cos

2
= 1+ + cos

Since the P is very far from dipole then r > > a. As a result the term is very small and can be neglected.
Therefore,
2
= 1+ cos

2
= 1+ cos

2
= 1+ cos
Find the reciprocal of the above equation,
1 1 2
= 1+ cos
Since a/r << 1, we can use binomial theorem and retain the terms up to the first order,
= − →→→ ( 6 )

Substitute (5) and (6) in (4),

1 1 1
= 1 + cos − 1 − cos
4
q 1 a
= 1 + cos θ − 1 + cos
4πε r r
1 2
= cos
4

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 14
 1 2
= cos
4
But, the electric dipole moment p = q 2a and we get,

Note
In vector notation the above equation can be written as,
1 ∙ ̂
=
4
This equation is valid for distances very large compared to the size of the dipole.
SPECIAL CASES
Case (1)
If the point P lies on the axial line of the dipole on the side of +q (then θ = 0 then cos 0 = 1 )Now the
electric potential becomes
1
=
4
Case (2)
If the point P lies on the axial line of the dipole on the side of –q, ( then = 180° and hence cos 180° =
−1) Now the electric potential becomes,
1
= −
4
Case (3)
If the point P lies on the equatorial line of the dipole, ( then = 90° and hence cos 90° = 0 )
Now the electric potential becomes,
=0

10) Obtain the expression for electric field due to an infinitely long charged wire.
• Consider an infinitely long straight wire having uniform linear charge density . Let P be a point located at
a perpendicular distance ‘r’ from the wire.

• The electric field at the point P can be found using Gauss law. We choose two small charge elements
A1 and A2 on the wire which are at equal distances from the point P
• The resultant electric field due to these two charge elements points radially away from the charged wire.
• Let us choose a cylindrical Gaussian surface of radius r and length L as show in the in the figure.
• The total electric flux in this closed surface is calculated as follows.
Φ = ∙

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 15
Φ = ∙ + ∙ + ∙
֍ From the adjacent diagram,
• On the curved surface and are parallel to each other .
Therefore ( θ = 0 then cos 0 = 1 ) Hence the equation
∙ = .
• For the top and bottom surfaces, and are perpendicular to
each other. Therefore
( = 90° and hence cos 90° = 0 ) Hence the equation
∙ =0

• substituting these values in the above equation,

Φ =

From the definition of Gauss’s law,

Φ =
By comparing the above two equations,
= →→→ (1)

The linear charge density (Charge per unit length) =

→→→ (2)
The magnitude of the electric field for the entire curved surface is constant, hence E is taken out of the
integration.
The curved surface area = 2

• The equation can be rewritten as,

=
=
E2 =
1
=
2

The above equation in vector notation, = ̂

11) Obtain the expression for electric field due to a

charged infinite plane sheet.
Consider an infinite plane sheet of charges with
uniform surface charge density = .
That is charge per unit area. Let P be a point at a
distance of r from the sheet as shown in the figure.
• Since, the plane is infinitely large; the electric field
should be same at all points equidistant from the plane
and radially directed at all points.

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 16
• A cylindrical shaped Gaussian surface of length 2r and area A of the flat surfaces is chosen such that
infinite plane sheet passes perpendicularly through the middle part of the Gaussian surface.
• The electric field is perpendicular to the area element at all points on the curved surface and is parallel to
the surface areas at P and
Φ = ∮ ∙
Φ = ∙ + ∙ + ∙ →→→ (1)
Apply the Gauss’s for this cylindrical surface,

Φ = + =

Since the magnitude of the electric field at these two equal surfaces is uniform, E is taken out of the
integration and = , we get

2 =

The total area of surface either at P or

Hence,
2 =

=
2
In vector notation,
=
2

12) Obtain an expression for the Electric Field due to two parallel charged infinite sheets.
Consider two infinitely large charged plane sheets with equal and opposite charge densities + and –
which are placed parallel to each other as shown in the following figure.
֍ The electric field between the plates and outside the plates is found using Gauss law.
• The magnitude of the electric field due to an
infinite charged plane sheet is =
The direction of electric field is
perpendicularly outward if > 0.
The direction of electric field is
perpendicularly in ward if 0.
At the points P2 and P3, the electric field due
to both plates are equal in magnitude and
opposite in direction. As a result, electric field
at a point outside the plates is zero.
But, inside the plate, electric fields are in same direction. i.e., towards the right, the total electric field at a
point P1, is
= + =
Note
The direction of the electric field inside the plates is directed from positively charged plate to negatively
charged plate and is uniform everywhere inside the plate.

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 17
13) Obtain the expression for electric field due to a uniformly
charged spherical shell.
Consider a uniformly charged spherical shell or radius R and total
charge Q as shown in the figure.
Case 1 At a point outside the shell ( r > R )
Let us choose a point P outside the shell at a distance r from the
center as shown in the following figure.
• The charge is uniformly distributed on the surface of the sphere.
Hence the electric field must point radially outward if Q > 0 and
point radially inward if Q < 0.
• So we choose a spherical Gaussian surface of radius r
and the total charge enclosed by this Gaussian surface is Q.
• Applying Gauss law,

But ∮ = total area of Gaussian surface

=4
By using this above equation can be rewritten as,
∙4 =

In vector notation, = ̂

Case 2 At a point on the surface of the spherical shell

(r=R)
The electrical field at points on the spherical shell ( r = R ) is given by,
= ̂
Case 3 At a point inside the spherical shell ( r < R )
Consider a point P inside the shell at a distance r from the center. A Gaussian sphere of radius r is
constructed as show in
the figure.
• Applying Gauss law,

The surface area of the Gaussian surface is A = 4

∙4 =
Since Gaussian surface encloses no charge, so Q = 0.
Now the above equation becomes
=0
Therefore the electric field due to the uniformly charged spherical shell
is zero at all points inside the shell.

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 18
14) Obtain the expression for capacitance for a parallel plate
capacitor.
֍ Capacitor
Capacitor is device used to store electric charge and electrical energy.
• Structure
A simple capacitor consists of two parallel metal plates of area (A)
separated by a small distance (d) as shown in the figure. Among the two
plates one plate consists of positive charge with uniform surface charge
density = , another plate has uniform negative charge .
• Electric field between two plates is =
By substituting the surface charge density value,
= →→→ (1)
Since, the electric field is uniform, the electric potential between the plates
having separation d is given by,

V =Ed →→→ (2)

Substitute (1) in (2),
= →→→ (3)
The general formula to find out the capacitance of parallel capacitor is
= →→→ (4)
Substitute (3) in (4),
= =

Therefore the capacitance of the parallel plate capacitor is =

15) Obtain the expression for energy stored in the parallel plate capacitor.
֍ Capacitor
Capacitor is device used to store electric charge and electrical energy.
֍ Electrostatic potential energy
When a battery is connected to the capacitor, electrons of total charge Q – are transferred from one plate to
the other plate. To transfer the charge work is done by the battery. This workdone is stored as electrostatic
potential energy in the capacitor.

֍ To derive the Expression

To transfer an infinitesimal charge dQ for a potential difference V, the workdone is given by,
=
Where =
Substitute this in the above equation,

The total workdone to charge a capacitor is,

=

1 1
= = =
2

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 19
 =
2
֍ According to work-energy principle, this workdone is stored as electrostatic potential energy ( ) in the
capacitor.
=
2
To express the above equation in terms of electric potential,
= since, =

16) Explain in detail how charges are distributed in a conductor and the principle behind the
lightning conductor.
Consider two conducting spheres A and B of radii r1 and r2 respectively connected to each other by a thin
conducting wire as shown in the following figure. The distance between the spheres is much greater than
the radii of either sphere.
• If a charge Q is introduced into any one of the spheres, this charge Q is redistributed into both the spheres
such that the electrostatic potential is same in both the spheres. They are now uniformly charged and attain
electrostatic equilibrium.
• Let q1 be the charge residing on the surface of sphere A and q2 is the charge residing on the surface of
sphere B such that ,
= +
• The charges are distributed only on the surface and there is no net charge inside the conductor.
• The electrostatic potential at the surface of the sphere A is given by,
= →→→ ( 1 )
• The electrostatic potential at the surface of the sphere A is given by,
= →→→ ( 2 )
֍ Since the spheres are connected by the conducting wire, the surfaces of both the spheres together form an
equipotential surface.
= →→→ ( 3 )

Substitute (1) and (2) in (3)

1 1
=
4 4

= →→→ ( 4 )
From the definition surface charge density, =
That is =
The total charge on the surface area of the sphere A,
=4 →→→ ( 5 )
The total charge on the surface area of the sphere B,
=4 →→→ ( 6 )

Substitute (4) and (5) in (6),

4 4
=

=
From which we conclude that,

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 20
 =
Note
From the above equation the surface charge density is inversely
proportional to the radius of the sphere. For a smaller radius, the charge
density will be larger and vice versa.

LIGHTNING ARRESTER OR LIGHTNING CONDUCTOR

This is a device used to protect tall buildings from lightning strikes.
֍ Principle
It works on the principle of action at points or corona discharge.
֍ Structure
This device consists of a long thick copper rod passing from top of the
building to the ground. The upper end of the rod has a sharp spike or a
sharp needle. The lower end of the rod is connected to the copper plate
which is buried deep into the ground.
֍ Working
When a negatively charged cloud is passing above the building, it
induces a positive charge on the spike.
Since the induced charge density on thin sharp spike is large, it results in
a corona discharge. This positive charge ionizes the surrounding air which in turn neutralizes the negative
charge in the cloud.
The negative charge pushed to the spikes passes through the copper rod and is safely diverted to the
earth.
Note
The lightning arrester does not stop the lightning; rather it diverts the lightning to the ground safely.

17) Explain in detail the construction and working of a Van de Graaff generator.

֍ Van de Graaff generator

It is a machine which is designed to produces a large amount of electrostatic potential difference, up to
( 107 volt). It was designed by Robert Van De Graaff in 1929.

֍ Principle
Electrostatic induction and Action at points

֍ Input = 104 volt

Output = 107 volt
֍ Structure
A large hollow spherical conductor is fixed on
the insulating stand as shown in the figure. A
pulley B is mounted at the center of the hollow
sphere and another pulley C is fixed at the bottom.
• A belt made up of insulating materials like silk
or rubber runs over both pulleys. The pulley C s
driven continuously by the electric motor.
• Two comb shaped metallic conductors E and D
are fixed near the pulleys. The comb D is
maintained at a positive potential of 104 volt by a
power supply. The upper comb E is connected to
the inner side of the hollow metal sphere.

֍ Working

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 21
 Due to action at points, a high electric field near comb D produced. So, the air between the belt and
comb D gets ionized, now the positive charges are pushed towards the belt and negative charges are
attracted towards comb D.
• The positive charges stick to the belt and move up. When the positive charges reach the comb E, a large
amount of negative and positive charges are induced on either side of comb E due to electrostatic
induction.
• As a result, the positive charges are pushed away from the comb E and they reach the outer surface of the
sphere. Since the sphere is a conductor, the positive charges are distributed uniformly on the outer surface
of the hollow sphere. At the same time, the negative charges nullify the positive charges in the belt due to
action at points before it passes over the pulley.
• When the belt descends, it has almost no net charge. At the bottom, it again gains a large positive charge.
The belt goes up and delivers the positive charges to the outer surface of the sphere. This process
continuous until the outer surface produces the potential difference of the order of 107 volt which is the
limiting value.

֍ Leakage of electric charges and reducing

We cannot store charges beyond this limit since the extra charge starts leaking to the surroundings due to
ionization of air. The leakage of charges can be reduced by enclosing the machine in a gas filled steel
chamber at very high pressure.

֍ Application
The high potential is used to accelerate the positive ions (protons and deuterons) for nuclear
disintegrations and other applications.

18) Write down the properties of electric field lines and magnetic field lines.
Electric field Magnetic Field
• The electric field lines start from a positive charge • Magnetic field lines are continuous closed curves.
and end at negative charges or at infinity. The direction of magnetic field lines is from North
• The electric field lines are never intersect with each pole to South pole outside the magnet and South
other. pole to North pole inside the magnet.
• The electric field lines are denser in a region where •
the electric field has larger magnitude and less dense The magnetic field lines are never intersect with
in a region where the electric field is of smaller each other.
magnitude. • The magnetic field is strong where magnetic field
• The electric field vector at a point on the field line lines crowd and weak where magnetic field lines thin
is the direction of electric field at that point. out.
• For a positive point charge the electric field lines • The direction of magnetic field at any point on the
point radially outward and for a negative point curve is known by drawing tangent to the magnetic
charge, the electric field lines points radially inward. field lines at that point.
• The number of electric field lines that emanate
from the positive charge or end at a negative charge
is directly proportional to the magnitude of the
charges.
(No. field lines ∝ Q )
• The magnitude of the electric field for a point
charge decreases as the distance increases.
1
∝

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 22
19) a)Derive an expression for resultant capacitance, when capacitors are connected in series.
b) Derive an expression for resultant resistance, when resistors are connected in series.
c) Derive an expression for resultant emf and current, when cells are connected in series.
Capacitor Resistor Cell

• Consider three capacitors of • When two or more

resistors are connected end • Several cells can be connected to form a
capacitance ,
to end, they are said to be battery.
Connected in series with a
in series. •In series connection, the negative terminal
battery of voltage V as shown in
• Consider three resistor of of one cell is connected to the positive
the above figure.
resistance , terminal of the second cell; the negative
• As soon as the battery is
• The amount of charge terminal of second cell is connected to the
connected to the capacitors in
passing through resistor R1 positive terminal of the third cell and so on.
series, the electrons of charge –Q
must also pass through • The free positive terminal of the first cell
are transferred from negative
resistors R2 and R3 since and the free negative terminal of the last
terminal to the right plate of C3
the charges cannot cell become the terminals of the battery.
which pushes the electrons of
accumulate anywhere in • Suppose ‘n’ cells, each of emf volts and
same amount –Q from left plate
the circuit. internal resistance of ‘r’ ohms are
of C3 to the right plate of C2 due
• Due to this reason, the connected in series with an external
to electrostatic induction.
current I passing through resistance R as shown in the above figure.
• Similarly, the left plate of C2
all the three resistors is the • The total emf of the battery = n
pushes the charges of –Q to the
same. The total resistance in the circuit = nr + R
right plate of C1 which induces
the positive charge +Q on the • According to Ohm’s law
(V= IR), If same current By Ohm’s law, the current in the circuit
left plate of C1.
pass through different is,
• At the same time, electrons of
resistors of different total emf nε
charge –Q are transferred from I= =
left plate of C1 to positive values, then the potential total resistance nr + R
terminal of the battery. difference across each
resistor must be different. Case (1)
• By these processes, each
• Let , be the If (r<<R), then
capacitor stores the same amount
of charge Q. The capacitances potential difference = ≈
of the capacitors are in general (voltage) across each of
Where, I1 is the current due to a single cell
different, so that the voltage the resistors ,
across each capacitor is also Respectively. =
different and are denoted as • The total voltage V is If ‘r’ is negligible when compared to R
, Respectively. equal to the sum of the current supplied by the battery is ‘n’
• The total voltage across each voltages across each times that supplied by a single cell.
capacitor must be equal to the resistor.
voltage of the battery. = + + Case(2)
= + + Here, If (r>>R), then
From the definition, = ;
= ; = ≈
= = It is the current due to single cell. That
Therefore, s, current due to the whole battery is the

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 23
 Substitute these in the above same as that due to a single cell and hence
= ; = ; = equation, there is no advantage in connecting several
Substitute these in the above = + + cells.
equation, = ( + + ) Note
= ∙ Thus series connection of cells is
= + +
advantageous only when the effective
1 1 1 Where RS is the internal resistance of the cell is negligibly
= + + equivalent resistance. small compared with R.
That is,
But,
1
= = = + +

Here, CS is the resultant of three

capacitor.
= + +

20) a)Derive an expression for resultant capacitance, when capacitors are connected in parallel.
b) Derive an expression for resultant resistance, when resistors are connected in parallel.
c) Derive an expression for resultant emf and current, when cells are connected in parallel.

Capacitor Resistor Cell

CAPACITORS IN PARALLEL
CONNECTION
• Consider three capacitors of
capacitance ,
Connected in parallel with a
battery of voltage V as shown in
• Resistors are in parallel when • In parallel connection all the
the above figure.
they are connected across the positive terminals of the cells are
same potential difference as connected to one point and all the
• Since corresponding sides of
negative terminals to a second point.
the capacitors are connected to shown in the above figure.
These two points form the positive
the same positive and negative
and negative terminals of the
terminals of the battery, the • In this case, the total current I
battery.
voltage across each capacitor is that leaves the battery is split into
three separate paths. • Let ‘n’ cells be connected in
equal to the battery’s voltage.
parallel between the points A and B
• Let , be the current as shown in the above figure.
• But, the capacitance of the
through the resistors, • Let be the emf and r be the
capacitors is different; the charge
, . internal resistance of each cell.
stored in each capacitor is not the
Due to the conservation of • the equivalent internal resistance
same. Let the charge stored in
charges, total current in the of the battery is
the three capacitors be
, respectively. circuit I is equal to sum of the
currents through each of three
A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 24
 resistors. 1 1 1 1
• According to the law of = + + +(
conservation of total charge, = + +
− )….
The sum of these three charges is
equal to the charge Q transferred Since the voltage across each From the equation the equivalent
by the battery. resistor is the same, applying Ohm’s internal resistance,
law(V=IR) to each resistor,
=
= + +
we have Now the total resistance in the
From the definition,
circuit becomes,
= ; = ; = = +
=
• Net current flowing the circuit is,
= + +
Substitute these in the above =
equation, +
If these three capacitors are
considered to form a single = + + Rewrite the above equation,
equivalent capacitance CP which =
stores the total charge Q . 1 1 1 +
= + +
Case (1)
We can write, = = = If r >> R , then = = nI1
Where I1 is the current due to a
Therefore we can rewrite the single cell and is equal to when R
Here, RP is the resultant
above equation as,
resistance in parallel connection. is negligible. Thus, the current
through the external resistance due
= + +
to the whole battery is N times the
current due to a single cell.
= + +
= + +
Case(2)
If r << R, then =
The above equation implies that
current due to the whole battery is
the same as that due to a single cell.
Hence it is advantageous to connect
cells in parallel when the external
resistance is very small compared to
the internal resistance of the cells.

21) Discuss the various properties of conductors in electrostatic equilibrium.

֍ Electrostatic equilibrium
When there is no external field, the free electrons are in continuous random motion in all directions. As a
result, there is no net motion of electrons along any particular direction which implies that the conductor is
in electrostatic equilibrium. The time taken by a conductor to reach electrostatic equilibrium is in the
order of 10-16 second.( that is almost instantaneous)

֍ Properties
• The electric field is zero everywhere inside the conductor. This is true regardless of whether the
conductor is solid or hollow.
• There is no net charge inside the conductors. The charges must reside only on the surface of the
conductors.

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 25
• The electric field outside the conductor is perpendicular to the surface of the conductor and has a
magnitude of where is the surface charge density at that point.
• The electrostatic potential has the same value on the surface and inside of the conductor.

22) Explain detail the effect of a dielectric placed in a parallel plate capacitor

When the capacitor is disconnected from the When the capacitor is connected to the battery
battery

Consider a capacitor with two parallel plates each

of cross-sectional area A and are separated by a Let us now consider what happens when the
distance‘d’. The capacitor is charged by a battery of battery of voltage V0 remains connected to the
voltage V0 and the charge stored is Q0. The capacitor when the dielectric is inserted into the
capacitance of the capacitor without dielectric is capacitor. This is shown in the above figure.
The potential difference V0 across the plates
= remains constant. But it is found experimentally that
The battery is then disconnected from the when dielectric is inserted, the charge stored in the
capacitor and the dielectric is inserted between the capacitor is increased by a factor .
plates. This is shown in the above figure. =
The introduction of dielectric between the plates Due to this increased charge, the capacitance is
will decrease the electric field. Experimentally it is also increased.
found that the modified electric field is given by, The new capacitance is

= = = =
Here, E0 is the electric field inside the capacitors Note
when there is no dielectric and is the relative The reason for the increase in capacitance in this
Permeability of the dielectric or dielectric constant. case when the battery remains connected is different
Since, > 1, the electric field E < E0. from the case when the battery is disconnected
As a result, the electrostatic potential difference before introducing the dielectric.
between the plates (V = Ed) is also reduced. But, at
the same time, the charge Q0 will remain constant CHANGE IN ENERGY
once the battery is disconnected.
Here the new potential difference is The energy stored in the capacitor before the
insertion of a dielectric is given by,
= = = 1
=
We know that capacitance is inversely 2
proportional to the potential difference. Therefore Here, V 0 is constant so for this formula has been
As ‘V’ decreases, ‘C’ increases. used.
Thus new capacitance in the presence of a After the dielectric is inserted, the capacitance is
dielectric is increased; hence the stored energy is also increased.

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 26
 = = = = = =
From the above equation we know that
Since, > 1, we have C > C0. Thus insertion of
the dielectric constant increases the capacitance. energy will increase.

CHANGE IN ENERGY
The energy stored in the capacitor before the
insertion of a dielectric is given by
1
=
2
After the dielectric is inserted, the charge Q0
remains constant but the capacitance is increases. As
a result, the stored energy is decreased.

= = =
Since >1
We get U < U0. There is a decrease in energy
because, when the dielectric is inserted,. The
capacitor spends some energy in pulling the
dielectric inside.

23) Obtain Gauss’s law from Coulomb’s law.

Strictly speaking, Gauss’s law cannot be derived from Coulomb’s law alone. Since, Coulomb’s law
gives the electric field due to an individual point charge only. However, Gauss’s law can be proven from
Coulomb’s law if it assumed, in addition that the electric field obeys the super position principle.
֍ Gauss’s law from Coulomb’s law

SPHERICAL SURFACE ENCLOSING SINGLE POINT CHARGE

Consider a point charge ‘+q’. From this point charge at a distance ‘r’, we consider a point c.
The electric field at this point, from the definition E =
q0 is the test charge q0 = 1 C ; F coulomb force = ℎ
Therefore, = ̂
From the definition, the electric flux
Φ = ∙
1
Φ = ̂∙
4
1
Φ = ̂
4
Φ = 4 ( ̂ = 1; ∮ =4 )
Φ = This is Gauss’s law.
So we can able to derive Gauss’s law from Coulomb’s law.

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 27
24) Obtain an expression for potential energy
due to a collection of three point charges which
are separated by finite distances.
Three charges are arranged in the following
configuration as shown in the following.
• Bring the charge q1 from infinity to the point A
requires no work, because there are no other
charges already present in the vicinity of charge
q1.
• To bring the second charge q2 to the point B,
work must be done against the electric field
created by the charge q1. So the workdone on the
charge q2 is
=
Here, V1B is the electrostatic potential due to the charge q1 at point B.
= = →→→ (1)
Note that the expression is same when q2 is brought first and then q1 later.
• To bring the charge q3 to the point C, work has to be done against the total electric field due to both
charges q1 and q2.
The electrostatic potential energy,
= + →→→ (2)
By adding equation (1) and (2) the total electrostatic potential energy for the system of three charges
, is
1
= + +
4
Note
The stored potential energy U is equal to the total external work done to assemble the three charges at the
given locations.

25) Explain the process of electrostatic induction.

֍ Electrostatic induction
Charging without actual contact is called electrostatic induction
• Consider an uncharged conducting sphere at rest on an insulating stand. Suppose a negatively charged rod
is brought near the conductor without touching it.
In the figure ‘a’. The negative charge of the rod
repels the electrons in the conductor to the
opposite side. As a result, positive charges are
induced near the region of the charged rod while
negative charges on the farther side.
• Now the conducting sphere is connected to the
ground through a conducting wire. This is called
grounding. Grounding removes the electron
from the conducting sphere. Note that positive
charges will not flow to the ground because they
are attracted by the negative charges of the rod.
(Observe the figure ‘b’)
• When the grounding wire is removed from the
conductor, the positive charges remain near the
charged rod (Observe the figure ‘c’)
• Now the charged rod is taken away from the
conductor. As soon as the charged rod is
A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 28
removed, the positive charge gets distributed uniformly on the surface of the conductor. By this process, the
neutral conducting sphere becomes positively charged.

26) Explain dielectrics in detail and how an electric field is induced inside a dielectric.
֍ Dielectric
A dielectric is non-conducting material and has no free electrons. The electrons in a dielectric are bound
within the atoms. Examples Ebonite, glass, mica, wood, rubber, plastic etc., A dielectric is made up of
either polar molecules or non-polar molecules.
֍ Inducing of electric field inside the dielectric
In the case of conductor, when an external electric field is applied on a conductor, the charges are
aligned in such a way that an internal electric field is created which cancels the external electric field. But,
in the case of a dielectric, which has no free electrons, the external electric field only realigns the charges so
that an internal electric field is produced.
The magnitude of the internal electric field is
smaller than that of external electric field.
Therefore the net electric field inside the
dielectric is not zero but is parallel to an external
electric field with magnitude less than that of the
external electric field.
֍ Experimental setup
Let us consider a rectangular dielectric slab
placed between two oppositely charged plates a
shown in the following figure.
The uniform electric field between the plates
acts as an external electric field which
polarizes the dielectric placed between plates.
The positive charges are induced on one side
surface and negative charges are induced on the
other side of surface.
Inside the dielectric, the net charge is zero
even in small volume. From the diagram the
dielectric in the external field is equivalent to two
Oppositely charged sheets with the surface
charge densities + and - . These charges are
called bound charges. They are not free to
move like free electrons in conductors.
For example, the charged balloon after
rubbing sticks onto a wall. The reason is that the
negatively charged balloon is brought near the
wall, it polarizes opposite charges on the surface
of the wall, which attracts the balloon.

A.SENTHIL KUMAR M.Sc.,M.Phil.,B.Ed.,PGDCA GBHSS, MECHERI, METTUR TK, SALEM DT. 636453 Page 29