electric charges and fields class 12 study material pdf download
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About This Presentation
Have you ever experienced a crackling sound or witnessed a spark while removing synthetic clothes or a sweater, especially in dry weather? This phenomenon occurs due to the discharge of electric charges accumulated through the rubbing of insulating surfaces. Another example of electric discharge is ...
Slide Content
ELECTRIC CHARGES AND FIELDS
Introduction
The Greek word for amber is “elektron”; this is the origin of the terms electricity and
electron. Electrostatic is a branch of physics that deals with the phenomena and properties of
stationary or slow-moving electric charges with no acceleration.
While it’s hard to see the electric charges that are responsible for electricity, it’s easy to
see their effects. They’re all around us: in the sparks and shocks of a cold winter day, the
imaging process of a xerographic copier, and the illumination of a flashlight when you turn on
its switch. Although we often take electricity for granted, it clearly underlies many aspects of
our modern world.
Just imagine what life would be like if there were no electric charges and no electricity.
For starters, we’d probably be sitting around campfires at night, trying to think of things to do
without television, cell phones, or computer games. But before you remark on just how
peaceful such a pre-electronic-age existence would be, let me add one more sobering thought:
we wouldn’t exist either. Whether it’s motionless as static charge or moving as electric current,
electricity really does make the world go ‘round.
Electricity may be difficult to see, but you can easily observe its effects. How often have
you found socks clinging to a shirt as you remove them from a hot dryer or struggled to throw
away a piece of plastic packaging that just won’t leave your hand or stay in the trash can? The
forces behind these familiar effects are electric in nature and stem from what we commonly
call “static electricity.” Static electricity does more than just push things around, however, as
you’ve probably noticed while reaching for a doorknob or a friend’s hand on a cold, dry day.
In this section, we’ll examine static electricity and the physics behind its intriguing forces and
often painful shocks.
When a plastic comb is rubbed with your hairs, it acquires the property of attracting light
objects such as paper pieces.
What we study under electrostatics is
static electricity.
The charges at rest develop due
to friction when we rub two insulating bodies against each other.
Some industrial applications of electrostatics are:
1.In designing electrostatics generators like Van de Graff generator
2.In electrostatic spraying of paints, powders etc. All right copy reserved. No part of the material can be produced without prior permission
Introduction
The Greek word for amber is “elektron”; this is the origin of the terms electricity and
electron. Electrostatic is a branch of physics that deals with the phenomena and properties of
stationary or slow-moving electric charges with no acceleration.
While it’s hard to see the electric charges that are responsible for electricity, it’s easy to
see their effects. They’re all around us: in the sparks and shocks of a cold winter day, the
imaging process of a xerographic copier, and the illumination of a flashlight when you turn on
its switch. Although we often take electricity for granted, it clearly underlies many aspects of
our modern world.
Just imagine what life would be like if there were no electric charges and no electricity.
For starters, we’d probably be sitting around campfires at night, trying to think of things to do
without television, cell phones, or computer games. But before you remark on just how
peaceful such a pre-electronic-age existence would be, let me add one more sobering thought:
we wouldn’t exist either. Whether it’s motionless as static charge or moving as electric current,
electricity really does make the world go ‘round.
Electricity may be difficult to see, but you can easily observe its effects. How often have
you found socks clinging to a shirt as you remove them from a hot dryer or struggled to throw
away a piece of plastic packaging that just won’t leave your hand or stay in the trash can? The
forces behind these familiar effects are electric in nature and stem from what we commonly
call “static electricity.” Static electricity does more than just push things around, however, as
you’ve probably noticed while reaching for a doorknob or a friend’s hand on a cold, dry day.
In this section, we’ll examine static electricity and the physics behind its intriguing forces and
often painful shocks.
When a plastic comb is rubbed with your hairs, it acquires the property of attracting light
objects such as paper pieces.
What we study under electrostatics is
static electricity.
The charges at rest develop due
to friction when we rub two insulating bodies against each other.
Some industrial applications of electrostatics are:
1.In designing electrostatics generators like Van de Graff generator
2.In electrostatic spraying of paints, powders etc. All right copy reserved. No part of the material can be produced without prior permission
3. In the design of cathode ray tubes for radar, television etc.
4. Ink-jet printing
5. Understanding lightning that strikes from the cloud base to the ground.
6. Adhesive forces of glue associated with surface tension, all are electric in nature
Methods of charging
There are three main methods for charging of a body
(i) Charging by rubbing and Frictional electricity:-
Rubbing as the term suggest is moving two things back and forth against each other. The
simplest way to experience electric charge is to rub certain bodies among each other. Rubbing
or friction makes electrons move. This gives one material a positive charge and the other a
negative charge. The charges stay on the surfaces of the materials until they can flow or they
discharge.
If we pass a comb through hairs, comb becomes charged and can attract small pieces of
paper. This is because the comb might have lost its electrons or acquired some electrons
when we rub it with hairs. Now, this comb is a charged body. The net charge on the comb
interacts with the net charge on small pieces of paper which results in attraction. Many such
solid materials are known which on rubbing attract light objects like a light feather, bits of
papers, straw etc.
Explanation of appearance of electric charge on rubbing is simple. Material bodies consist of
a large number of electrons and protons in equal number and hence is in neutral in their normal
state. But when a glass rod is rubbed with a silk cloth, electrons are transferred from glass rod
to silk cloth. The glass rod becomes positively charged and the silk cloth becomes negatively
charged as it receives extra electrons from the glass rod. In this case rod after rubbing, comb
after passing through dry hairs becomes electrified and these are the example of frictional
electricity.
(ii) Charging by induction (Electrostatic Induction):-
The temporary electrification of a conductor, when a charged body is brought near it is called
electrostatic induction.
All right copy reserved. No part of the material can be produced without prior permission
4. Ink-jet printing
5. Understanding lightning that strikes from the cloud base to the ground.
6. Adhesive forces of glue associated with surface tension, all are electric in nature
Methods of charging
There are three main methods for charging of a body
(i) Charging by rubbing and Frictional electricity:-
Rubbing as the term suggest is moving two things back and forth against each other. The
simplest way to experience electric charge is to rub certain bodies among each other. Rubbing
or friction makes electrons move. This gives one material a positive charge and the other a
negative charge. The charges stay on the surfaces of the materials until they can flow or they
discharge.
If we pass a comb through hairs, comb becomes charged and can attract small pieces of
paper. This is because the comb might have lost its electrons or acquired some electrons
when we rub it with hairs. Now, this comb is a charged body. The net charge on the comb
interacts with the net charge on small pieces of paper which results in attraction. Many such
solid materials are known which on rubbing attract light objects like a light feather, bits of
papers, straw etc.
Explanation of appearance of electric charge on rubbing is simple. Material bodies consist of
a large number of electrons and protons in equal number and hence is in neutral in their normal
state. But when a glass rod is rubbed with a silk cloth, electrons are transferred from glass rod
to silk cloth. The glass rod becomes positively charged and the silk cloth becomes negatively
charged as it receives extra electrons from the glass rod. In this case rod after rubbing, comb
after passing through dry hairs becomes electrified and these are the example of frictional
electricity.
(ii) Charging by induction (Electrostatic Induction):-
The temporary electrification of a conductor, when a charged body is brought near it is called
electrostatic induction.
All right copy reserved. No part of the material can be produced without prior permission
Electrostatic Induction: - Electrical charges in the conductor are redistributed
When a body is charged this way there is no transfer of electrons from one body to
other. This happens because there is no physical contact taking place between charging
body and conductor being charged.
If a charged body is brought near an uncharged body, then the neutral body becomes
oppositely charged. By induction method, we can charge any type of material body.
(iii) By conduction (by touch without rubbing):-
Because of having excess free electrons in metals they can be charged by conduction. When
we bring two conductors, one charged and other uncharged in contact, the same type of
charge will appear on both the conductors.
Basic properties of electric charge
Additivity of charges
T
system.
For example, the total charge of a system containing five charges +1, +2, –3, +4 and – 5, in some
arbitrary unit, is (+1) + (+2) + (–3) + (+4) + (–5) = –1 in the same unit.
Conservation of charge
The total charge of an isolated system is always conserved that means charge can neither be
created nor be destroyed but can be transferred from one body to another.
When bodies are charged by rubbing, there is transfer of electrons from one body to the other; no
new charges are either created or destroyed.
Quantisation of charge
A
experimental fact is called quantisation of charge.
q = ± ne, where n is an integerand e = 1.6 × 10
- 19
C
By convention, the charge on an electron is taken to be negative; therefore charge on an electron
is written as –e and that on a proton as +e.
The SI unit of charge is Coulomb and is denoted by the symbol C. All right copy reserved. No part of the material can be produced without prior permission
When a body is charged this way there is no transfer of electrons from one body to
other. This happens because there is no physical contact taking place between charging
body and conductor being charged.
If a charged body is brought near an uncharged body, then the neutral body becomes
oppositely charged. By induction method, we can charge any type of material body.
(iii) By conduction (by touch without rubbing):-
Because of having excess free electrons in metals they can be charged by conduction. When
we bring two conductors, one charged and other uncharged in contact, the same type of
charge will appear on both the conductors.
Basic properties of electric charge
Additivity of charges
T
system.
For example, the total charge of a system containing five charges +1, +2, –3, +4 and – 5, in some
arbitrary unit, is (+1) + (+2) + (–3) + (+4) + (–5) = –1 in the same unit.
Conservation of charge
The total charge of an isolated system is always conserved that means charge can neither be
created nor be destroyed but can be transferred from one body to another.
When bodies are charged by rubbing, there is transfer of electrons from one body to the other; no
new charges are either created or destroyed.
Quantisation of charge
A
experimental fact is called quantisation of charge.
q = ± ne, where n is an integerand e = 1.6 × 10
- 19
C
By convention, the charge on an electron is taken to be negative; therefore charge on an electron
is written as –e and that on a proton as +e.
The SI unit of charge is Coulomb and is denoted by the symbol C. All right copy reserved. No part of the material can be produced without prior permission
1�C=10
-6
C
At the macroscopic level, one deals with charges that are enormous compared to the magnitude
of charge e. A charge of magnitude, say 1 μC, contains something like 10
13
times the electronic charge.
At this scale, the fact that charge can increase or decrease only in units of e. Thus, at the macroscopic
level, the quantisation of charge has no practical consequence and can be ignored.
At the microscopic level, where the charges involved are of the order of a few tens or hundreds
of e, i.e., they can be counted and quantisation of charge cannot be ignored.
Coulomb's Law
In 1785 the French physicist
Charles Augustin Coulomb
measured the electric force between
small charged spheres using a torsion balance. He then formulated his observations in the form of
Coulomb's Law. Coulomb's Law is an electrical analog of Newton's Universal Law of Gravitation. It
states that
The force of attraction or repulsion between two stationary point charges is
(i) directly proportional to the product of the magnitude of two charges.
(ii) Inversely proportional to the square of the distance between them.
This force acts along the line joining the two charges.
To explain above statement consider the figure given below
Coulomb's Law
Above figure consists of two point charges q1 and q2. These two charges are separated by a
distance r. Then according to Coulomb's Law,
F α
�1�2
�
2
F= k
�1�2
�
2
k=
1
4�??????0
??????
0
is called permittivity of air or free space. (absolute permittivity)
F=
1
4�??????0
�1�2
�
2
Value of ??????
0= 8.85x10
-12
??????
2
??????�
2
All right copy reserved. No part of the material can be produced without prior permission
-6
C
At the macroscopic level, one deals with charges that are enormous compared to the magnitude
of charge e. A charge of magnitude, say 1 μC, contains something like 10
13
times the electronic charge.
At this scale, the fact that charge can increase or decrease only in units of e. Thus, at the macroscopic
level, the quantisation of charge has no practical consequence and can be ignored.
At the microscopic level, where the charges involved are of the order of a few tens or hundreds
of e, i.e., they can be counted and quantisation of charge cannot be ignored.
Coulomb's Law
In 1785 the French physicist
Charles Augustin Coulomb
measured the electric force between
small charged spheres using a torsion balance. He then formulated his observations in the form of
Coulomb's Law. Coulomb's Law is an electrical analog of Newton's Universal Law of Gravitation. It
states that
The force of attraction or repulsion between two stationary point charges is
(i) directly proportional to the product of the magnitude of two charges.
(ii) Inversely proportional to the square of the distance between them.
This force acts along the line joining the two charges.
To explain above statement consider the figure given below
Coulomb's Law
Above figure consists of two point charges q1 and q2. These two charges are separated by a
distance r. Then according to Coulomb's Law,
F α
�1�2
�
2
F= k
�1�2
�
2
k=
1
4�??????0
??????
0
is called permittivity of air or free space. (absolute permittivity)
F=
1
4�??????0
�1�2
�
2
Value of ??????
0= 8.85x10
-12
??????
2
??????�
2
All right copy reserved. No part of the material can be produced without prior permission
1
4�??????0
= 9x10
9
??????�
2
�
2
F=
1
4�??????0
�1�2
�
2
F = 9x10
9
�1�2
�
2
If the charges are placed in a medium of permittivity ??????
F=
1
4�??????
�1�2
�
2
Relative permittivity Or Dielectric constant (K Or ?????????r)
It is the ratio of permittivity of a medium to the permittivity of free space.
??????
r
=
??????
??????0
??????= ??????
0
??????
r
F=
1
4�??????
�1�2
�
2
F=
1
4�??????0????????????
�1�2
�
2
OR F=
1
4�??????0�
�1�2
�
2
UNIT OF CHARGE- COULOMB
F=
1
4�??????
0
�
1�
2
�
2
F = 9x10
9
�
1�
2
�
2
If q
1
=q
2
= 1C and r=1m
F = 9x10
9
N
1 Coulomb is that charge when placed in air or vacuum at a distance of 1m from an
equal and similar charge experiences a force of 9x10
9
N.
Coulomb’s law in vector form
Force on q
2
due to q
1,
�
21
̅̅̅̅̅
=
1
4�??????
0
�
1�
2
�
2
�
12̂
�
12̂ Is the unit vector pointing from q
1 to
q
2
All right copy reserved. No part of the material can be produced without prior permission
4�??????0
= 9x10
9
??????�
2
�
2
F=
1
4�??????0
�1�2
�
2
F = 9x10
9
�1�2
�
2
If the charges are placed in a medium of permittivity ??????
F=
1
4�??????
�1�2
�
2
Relative permittivity Or Dielectric constant (K Or ?????????r)
It is the ratio of permittivity of a medium to the permittivity of free space.
??????
r
=
??????
??????0
??????= ??????
0
??????
r
F=
1
4�??????
�1�2
�
2
F=
1
4�??????0????????????
�1�2
�
2
OR F=
1
4�??????0�
�1�2
�
2
UNIT OF CHARGE- COULOMB
F=
1
4�??????
0
�
1�
2
�
2
F = 9x10
9
�
1�
2
�
2
If q
1
=q
2
= 1C and r=1m
F = 9x10
9
N
1 Coulomb is that charge when placed in air or vacuum at a distance of 1m from an
equal and similar charge experiences a force of 9x10
9
N.
Coulomb’s law in vector form
Force on q
2
due to q
1,
�
21
̅̅̅̅̅
=
1
4�??????
0
�
1�
2
�
2
�
12̂
�
12̂ Is the unit vector pointing from q
1 to
q
2
All right copy reserved. No part of the material can be produced without prior permission
�
12
̂ =
�
12̅̅̅̅̅
�
�
21
̅̅̅̅̅
=
1
4�??????
0
�
1�
2
�
3
�
12
̅̅̅̅̅
Force on q
1
due to q
2,
�
12
̅̅̅̅̅
=
1
4�??????
0
�
1�
2
�
2
�
21̂
�
21̂ is the unit vector pointing from q
2
to q
1
�
21
̂ =
�
21̅̅̅̅̅
�
�
12
̅̅̅̅̅
=
1
4�??????
0
�
1�
2
�
3
�
21
̅̅̅̅̅
�
12̅̅̅̅= - �
21̅̅̅̅
�
12
̅̅̅̅= - �
21
̅̅̅̅
ie Coulomb’s law obey Newton’s third law
Principle of Superposition
Principle of superposition gives the method to find force on a charge when system consists of
large number of charges.
According to this principle when multiple charges are interacting the total force on a given
charge is vector sum of forces exerted on it by all other charges.
This principle makes use of the fact that the forces with which two charges attract or repel one
another are not affected by the presence of other charges.
If a system of charges has n number of charges say q1, q2,...................., qn, then total force on
charge q1 according to principle of superposition is
�
.
̅
=
�
12 ???
̅̅̅̅̅̅
+�
13
̅̅̅̅̅
+…………………. �
1�
̅̅̅̅̅
As per the principle of superposition, the force on any charge due to a number of
other charges is the vector sum of all the forces on that charge due to other charges,
taken one at a time. All right copy reserved. No part of the material can be produced without prior permission
12
̂ =
�
12̅̅̅̅̅
�
�
21
̅̅̅̅̅
=
1
4�??????
0
�
1�
2
�
3
�
12
̅̅̅̅̅
Force on q
1
due to q
2,
�
12
̅̅̅̅̅
=
1
4�??????
0
�
1�
2
�
2
�
21̂
�
21̂ is the unit vector pointing from q
2
to q
1
�
21
̂ =
�
21̅̅̅̅̅
�
�
12
̅̅̅̅̅
=
1
4�??????
0
�
1�
2
�
3
�
21
̅̅̅̅̅
�
12̅̅̅̅= - �
21̅̅̅̅
�
12
̅̅̅̅= - �
21
̅̅̅̅
ie Coulomb’s law obey Newton’s third law
Principle of Superposition
Principle of superposition gives the method to find force on a charge when system consists of
large number of charges.
According to this principle when multiple charges are interacting the total force on a given
charge is vector sum of forces exerted on it by all other charges.
This principle makes use of the fact that the forces with which two charges attract or repel one
another are not affected by the presence of other charges.
If a system of charges has n number of charges say q1, q2,...................., qn, then total force on
charge q1 according to principle of superposition is
�
.
̅
=
�
12 ???
̅̅̅̅̅̅
+�
13
̅̅̅̅̅
+…………………. �
1�
̅̅̅̅̅
As per the principle of superposition, the force on any charge due to a number of
other charges is the vector sum of all the forces on that charge due to other charges,
taken one at a time. All right copy reserved. No part of the material can be produced without prior permission
�
12
̅̅̅̅̅
=
1
4�??????
0
�
1�
2
�12
2
�
12̂
�
13
̅̅̅̅̅
=
1
4�??????
0
�
1�
2
�13
2
�
13̂
�
1
̅̅̅̅
=
�
12 ???
̅̅̅̅̅̅
+�
13
̅̅̅̅̅
In case of n charges
�
.
̅
=
�
12 ???
̅̅̅̅̅̅
+�
13
̅̅̅̅̅
+…………………. �
1�
̅̅̅̅̅
ELECTRIC FIELD
All right copy reserved. No part of the material can be produced without prior permission
12
̅̅̅̅̅
=
1
4�??????
0
�
1�
2
�12
2
�
12̂
�
13
̅̅̅̅̅
=
1
4�??????
0
�
1�
2
�13
2
�
13̂
�
1
̅̅̅̅
=
�
12 ???
̅̅̅̅̅̅
+�
13
̅̅̅̅̅
In case of n charges
�
.
̅
=
�
12 ???
̅̅̅̅̅̅
+�
13
̅̅̅̅̅
+…………………. �
1�
̅̅̅̅̅
ELECTRIC FIELD
All right copy reserved. No part of the material can be produced without prior permission
The electric field is defined as the region or space around a charge where an electric
force of attraction or repulsion can be experienced.
Electric field Intensity
Consider a charge Q placed in vacuum. If we place another point charge q at a point
P, then the charge Q will exert a force on q as per Coulomb’s law.
Let F
̅
be the force experienced by the charge q.
Force experienced by unit charge, E̅ =
F̅
q
The electric field or field intensity at a point is defined as the force experienced by unit
positive charge placed at that point.
E̅=lim
�→�
???
F̅
q
Significance of lim
�→�
???
F̅
q
is that the test charge q should be negligibly small so that the
source charge Q remain at its original position.
Unit Electric field Intensity is N/C or V/m
Electric field Intensity is a vector quantity.
The force acting on the charge q is �̅= q�̅
Physical Significance of electric field
It is very important concept in understanding various electrostatic phenomenon.
The space around every electric charge or electrically charged body is filled
with an electric field thereby altering the space around it. This is the reason why
electrostatic force like gravitational force is an action-at-a-distance force.
Electric field should not be thought of as a kind of matter filled in space
surrounding electric charge. It is a kind of aura or the distinctive atmosphere or
quality that seems to surround and be generated by an electric charge.
Electric field intensity due to a point charge
All right copy reserved. No part of the material can be produced without prior permission
force of attraction or repulsion can be experienced.
Electric field Intensity
Consider a charge Q placed in vacuum. If we place another point charge q at a point
P, then the charge Q will exert a force on q as per Coulomb’s law.
Let F
̅
be the force experienced by the charge q.
Force experienced by unit charge, E̅ =
F̅
q
The electric field or field intensity at a point is defined as the force experienced by unit
positive charge placed at that point.
E̅=lim
�→�
???
F̅
q
Significance of lim
�→�
???
F̅
q
is that the test charge q should be negligibly small so that the
source charge Q remain at its original position.
Unit Electric field Intensity is N/C or V/m
Electric field Intensity is a vector quantity.
The force acting on the charge q is �̅= q�̅
Physical Significance of electric field
It is very important concept in understanding various electrostatic phenomenon.
The space around every electric charge or electrically charged body is filled
with an electric field thereby altering the space around it. This is the reason why
electrostatic force like gravitational force is an action-at-a-distance force.
Electric field should not be thought of as a kind of matter filled in space
surrounding electric charge. It is a kind of aura or the distinctive atmosphere or
quality that seems to surround and be generated by an electric charge.
Electric field intensity due to a point charge
All right copy reserved. No part of the material can be produced without prior permission
Consider a point P at a distance r from a point charge +q. Electric field intensity at the
point P, Let us imagine a test charge q0 to be placed at P. Now we find force on charge q0 due
to q through Coulomb's law.
F
̅
=
1
4????????????
0
qq
0
r
2
r̂
Electric field at point P, E̅=
F̅
q
=
1
4????????????
0
q
r
2
r̂
E̅ =
1
4????????????
0
q
r
2
r̂
Magnitude of Electric Field, E =
1
4????????????
0
q
r
2
Electric Field in terms of position vectors
Electric dipole and Electric Dipole moment
Electric dipole is a pair of equal and opposite charges, +q and −q, separated by a very
small distance.
Total charge of the dipole is zero but electric field of the dipole is not zero as charges q
and -q are separated by some distance and electric field due to them when added is not zero.
Dipole moment is the product of one of the charges and distance between them.
??????̅ = q 2??????̅̅̅̅
The unit of electric dipole moment is coulomb – meter (C-m)
Electric dipole moment is a vector quantity and by convention, its direction is from –
q to +q. All right copy reserved. No part of the material can be produced without prior permission
point P, Let us imagine a test charge q0 to be placed at P. Now we find force on charge q0 due
to q through Coulomb's law.
F
̅
=
1
4????????????
0
0
r
2
r̂
Electric field at point P, E̅=
F̅
q
=
1
4????????????
0
q
r
2
r̂
E̅ =
1
4????????????
0
q
r
2
r̂
Magnitude of Electric Field, E =
1
4????????????
0
q
r
2
Electric Field in terms of position vectors
Electric dipole and Electric Dipole moment
Electric dipole is a pair of equal and opposite charges, +q and −q, separated by a very
small distance.
Total charge of the dipole is zero but electric field of the dipole is not zero as charges q
and -q are separated by some distance and electric field due to them when added is not zero.
Dipole moment is the product of one of the charges and distance between them.
??????̅ = q 2??????̅̅̅̅
The unit of electric dipole moment is coulomb – meter (C-m)
Electric dipole moment is a vector quantity and by convention, its direction is from –
q to +q. All right copy reserved. No part of the material can be produced without prior permission
Examples of electric dipole:-
Dipoles are common in nature. Molecules like H2O, HCl, and CH3COOH are electric
dipoles and have permanent dipole moments. They have permanent dipole moments because
the center of their positive charges does not fall exactly over the center of their negative
charges. Figure given below shows molecule of water.
Physical significance of Electric Dipole and dipole moment
Atoms as a whole are electrically neutral in their ground state. We know that atoms
have equal amount of positive and negative charge. Similar to atoms molecules are
also neutral but they also have equal amount of positive and negative charges.
Now when in a system, algebraic sum of all the charges is zero it does not necessarily
mean that electric field produced by the system is zero everywhere. This makes study
of electric dipoles important for electrical phenomenon in matter.
Matter which is made up of atoms and molecules is electrically neutral. If the center of
mass of positive charges coincides with that of negative charges then molecule
behaves as
non-polar molecule. On the other hand, if center of mass of positive
charges does not coincides with that of negative charges then molecule behaves
as
polar molecule. These polar molecules have permanent dipole moments. These
dipole moments are randomly oriented in the absence of external electric field. If we
place a material with polar molecules in external electric field then these molecules
align themselves in the direction of the field. This results in the development of a net
dipole moment. This particular piece of material is said to be
polarized.
So study of dipole and dipole moments gives a measure of the polarization of a net
neutral system. The study of dipole moments measures the tendency of a dipole to
align with an external electric field.
All right copy reserved. No part of the material can be produced without prior permission
Dipoles are common in nature. Molecules like H2O, HCl, and CH3COOH are electric
dipoles and have permanent dipole moments. They have permanent dipole moments because
the center of their positive charges does not fall exactly over the center of their negative
charges. Figure given below shows molecule of water.
Physical significance of Electric Dipole and dipole moment
Atoms as a whole are electrically neutral in their ground state. We know that atoms
have equal amount of positive and negative charge. Similar to atoms molecules are
also neutral but they also have equal amount of positive and negative charges.
Now when in a system, algebraic sum of all the charges is zero it does not necessarily
mean that electric field produced by the system is zero everywhere. This makes study
of electric dipoles important for electrical phenomenon in matter.
Matter which is made up of atoms and molecules is electrically neutral. If the center of
mass of positive charges coincides with that of negative charges then molecule
behaves as
non-polar molecule. On the other hand, if center of mass of positive
charges does not coincides with that of negative charges then molecule behaves
as
polar molecule. These polar molecules have permanent dipole moments. These
dipole moments are randomly oriented in the absence of external electric field. If we
place a material with polar molecules in external electric field then these molecules
align themselves in the direction of the field. This results in the development of a net
dipole moment. This particular piece of material is said to be
polarized.
So study of dipole and dipole moments gives a measure of the polarization of a net
neutral system. The study of dipole moments measures the tendency of a dipole to
align with an external electric field.
All right copy reserved. No part of the material can be produced without prior permission
Electric field of a dipole (On axial line)
Consider a dipole of charge q and length 2a. Let P be a point at a distance x from the
centre of the electric dipole.
Electric field intensity at the point P due to the charge +q,
E1=
1
4�??????
0
�
(??????−??????)
2
along BP
Electric field intensity at the point P due to the charge -q,
E2 =
1
4�??????
0
�
(??????+??????)
2
along PA
The total electric field at P due to the dipole is
E = E1 - E2
E =
1
4�??????
0
�
(??????−??????)
2
−
1
4�??????
0
�
(??????+??????)
2
E =
�
4�??????
0
(
1
(??????−??????)
2
-
1
(??????+??????)
2
)
E =
�
4�??????
0
(
(??????+??????)
2
−(??????−??????)
2
(??????
2
−??????
2
)
2
)
E =
�
4�??????
0
(
4????????????
(??????
2
−??????
2
)
2
a<<x (a
2
can be neglected)
E =
�
4�??????
0
4????????????
??????
4
E =
�
4�??????
0
4??????
??????
3
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Consider a dipole of charge q and length 2a. Let P be a point at a distance x from the
centre of the electric dipole.
Electric field intensity at the point P due to the charge +q,
E1=
1
4�??????
0
�
(??????−??????)
2
along BP
Electric field intensity at the point P due to the charge -q,
E2 =
1
4�??????
0
�
(??????+??????)
2
along PA
The total electric field at P due to the dipole is
E = E1 - E2
E =
1
4�??????
0
�
(??????−??????)
2
−
1
4�??????
0
�
(??????+??????)
2
E =
�
4�??????
0
(
1
(??????−??????)
2
-
1
(??????+??????)
2
)
E =
�
4�??????
0
(
(??????+??????)
2
−(??????−??????)
2
(??????
2
−??????
2
)
2
)
E =
�
4�??????
0
(
4????????????
(??????
2
−??????
2
)
2
a<<x (a
2
can be neglected)
E =
�
4�??????
0
4????????????
??????
4
E =
�
4�??????
0
4??????
??????
3
$OO???ULJKW???FRS\???UHVHUYHG???1R???SDUW???RI???WKH???PDWHULDO???FDQ???EH???SURGXFHG???ZLWKRXW???SULRU???SHUPLVVLRQ
E =
1
4�??????
0
2(2??????�)
??????
3
E =
1
4�??????
0
2p
??????
3
p=2??????�(Electric dipole moment)
Electric field of a dipole (On equatorial line line)
The electric field at P due to the charge +q,
E1=
1
4�??????
0
�
�
2
along BP
E
1
Can be resolved into two components E
1
Cos?????? and E
1
Sin??????.
The electric field at P due to the charge -q,
E2=
1
4�??????
0
�
�
2
along PA
E
2
Can also be resolved into two components E
2
Cos and E
2
Sin??????.
Here Sin?????? components are equal and opposite, therefore they cancel out. But Cos??????
components are in the same direction, they can be added up.
The total electric field at P due to the dipole,
E = E
1
Cos?????? + E
2
Cos??????
E =
1
4�??????
0
�
�
2
Cos?????? +
1
4�??????
0
�
�
2
Cos??????
E = 2
1
4�??????
0
�
�
2
Cos??????
Cos??????= a/r $OO???ULJKW???FRS\???UHVHUYHG???1R???SDUW???RI???WKH???PDWHULDO???FDQ???EH???SURGXFHG???ZLWKRXW???SULRU???SHUPLVVLRQ
1
4�??????
0
2(2??????�)
??????
3
E =
1
4�??????
0
2p
??????
3
p=2??????�(Electric dipole moment)
Electric field of a dipole (On equatorial line line)
The electric field at P due to the charge +q,
E1=
1
4�??????
0
�
�
2
along BP
E
1
Can be resolved into two components E
1
Cos?????? and E
1
Sin??????.
The electric field at P due to the charge -q,
E2=
1
4�??????
0
�
�
2
along PA
E
2
Can also be resolved into two components E
2
Cos and E
2
Sin??????.
Here Sin?????? components are equal and opposite, therefore they cancel out. But Cos??????
components are in the same direction, they can be added up.
The total electric field at P due to the dipole,
E = E
1
Cos?????? + E
2
Cos??????
E =
1
4�??????
0
�
�
2
Cos?????? +
1
4�??????
0
�
�
2
Cos??????
E = 2
1
4�??????
0
�
�
2
Cos??????
Cos??????= a/r $OO???ULJKW???FRS\???UHVHUYHG???1R???SDUW???RI???WKH???PDWHULDO???FDQ???EH???SURGXFHG???ZLWKRXW???SULRU???SHUPLVVLRQ
E = 2
1
4�??????
0
�
�
2
??????
�
E =
1
4�??????
0
2??????�
�
3
E =
1
4�??????
0
2??????�
�
3
2aq=p
E =
1
4�??????
0
p
�
3
� = (??????
2
+ ??????
2
)
1/2
E =
1
4�??????
0
p
(??????
2
+??????
2
)
3/2
if the length of the dipole is very small ??????
2
can be neglected
E =
1
4�??????
0
p
??????
3
Torque acting on a dipole in an electric field
Consider a dipole of charge q and length 2a placed in a uniform electric field E makes an
angle ?????? with the direction of electric field.
The charges +q and –
q experience forces + qE and – qE respectively.
These two equal and unlike forces constitute a couple.
The net force acting on the dipole + qE- qE = 0
Torque(τ) = force x Perpendicular distance
τ = qE x BN
Sin??????=
�??????
2??????
BN= 2a Sin??????
τ = qE x 2a Sin??????
τ = 2aqE Sin?????? 2aq=p
τ = PE Sin??????
In vector form, All right copy reserved. No part of the material can be produced without prior permission
1
4�??????
0
�
�
2
??????
�
E =
1
4�??????
0
2??????�
�
3
E =
1
4�??????
0
2??????�
�
3
2aq=p
E =
1
4�??????
0
p
�
3
� = (??????
2
+ ??????
2
)
1/2
E =
1
4�??????
0
p
(??????
2
+??????
2
)
3/2
if the length of the dipole is very small ??????
2
can be neglected
E =
1
4�??????
0
p
??????
3
Torque acting on a dipole in an electric field
Consider a dipole of charge q and length 2a placed in a uniform electric field E makes an
angle ?????? with the direction of electric field.
The charges +q and –
q experience forces + qE and – qE respectively.
These two equal and unlike forces constitute a couple.
The net force acting on the dipole + qE- qE = 0
Torque(τ) = force x Perpendicular distance
τ = qE x BN
Sin??????=
�??????
2??????
BN= 2a Sin??????
τ = qE x 2a Sin??????
τ = 2aqE Sin?????? 2aq=p
τ = PE Sin??????
In vector form, All right copy reserved. No part of the material can be produced without prior permission
??????τ̅ =P̅ X E̅
Direction of torque is perpendicular to the plane containing dipole axis and electric field.
Electric field lines
Electric field is a vector quantity and can be represented as we represent vectors. Let
us try to represent E due to a point charge pictorially. Let the point charge be placed at the
origin. Draw vectors pointing along the direction of the electric field with their lengths
proportional to the strength of the field at each point. Since the magnitude of electric field at a
point decreases inversely as the square of the distance of that point from the charge, the
vector gets shorter as one goes away from the origin, always pointing radially outward. Figure
shows such a picture. In this figure, each arrow indicates the electric field, i.e., the force acting
on a unit positive charge, placed at the tail of that arrow. Connect the arrows pointing in one
direction and the resulting figure represents a field line. Now the magnitude of the field is
represented by the density of field lines. E is strong near the charge, so the density of field
lines is more near the charge and the lines are closer. Away from the charge, the field gets
weaker and the density of field lines is less, resulting in well-separated lines.
$OO???ULJKW???FRS\???UHVHUYHG???1R???SDUW???RI???WKH???PDWHULDO???FDQ???EH???SURGXFHG???ZLWKRXW???SULRU???SHUPLVVLRQ
Direction of torque is perpendicular to the plane containing dipole axis and electric field.
Electric field lines
Electric field is a vector quantity and can be represented as we represent vectors. Let
us try to represent E due to a point charge pictorially. Let the point charge be placed at the
origin. Draw vectors pointing along the direction of the electric field with their lengths
proportional to the strength of the field at each point. Since the magnitude of electric field at a
point decreases inversely as the square of the distance of that point from the charge, the
vector gets shorter as one goes away from the origin, always pointing radially outward. Figure
shows such a picture. In this figure, each arrow indicates the electric field, i.e., the force acting
on a unit positive charge, placed at the tail of that arrow. Connect the arrows pointing in one
direction and the resulting figure represents a field line. Now the magnitude of the field is
represented by the density of field lines. E is strong near the charge, so the density of field
lines is more near the charge and the lines are closer. Away from the charge, the field gets
weaker and the density of field lines is less, resulting in well-separated lines.
$OO???ULJKW???FRS\???UHVHUYHG???1R???SDUW???RI???WKH???PDWHULDO???FDQ???EH???SURGXFHG???ZLWKRXW???SULRU???SHUPLVVLRQ
The electric field lines are imaginary lines drawn in such a way that the tangent to which at
any point gives the direction of the electric field at that point.
Electric field lines of a single positive Charge
Electric field lines of a single negative Charge
The field lines of a single positive charge and a single negative
The field lines of a single positive charge are radially outward while those of a single
negative charge are radially inward. $OO ULJKW FRS\ UHVHUYHG 1R SDUW RI WKH PDWHULDO FDQ EH SURGXFHG ZLWKRXW SULRU SHUPLVVLRQ
any point gives the direction of the electric field at that point.
Electric field lines of a single positive Charge
Electric field lines of a single negative Charge
The field lines of a single positive charge and a single negative
The field lines of a single positive charge are radially outward while those of a single
negative charge are radially inward. $OO ULJKW FRS\ UHVHUYHG 1R SDUW RI WKH PDWHULDO FDQ EH SURGXFHG ZLWKRXW SULRU SHUPLVVLRQ
Field lines around the system of two positive charges
Field lines around the system of two positive charges gives a different picture and
describe the mutual repulsion between them.
Field lines around a system of a positive and negative charge (Electric
dipole)
$OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ
Field lines around the system of two positive charges gives a different picture and
describe the mutual repulsion between them.
Field lines around a system of a positive and negative charge (Electric
dipole)
$OOULJKWFRS\UHVHUYHG1RSDUWRIWKHPDWHULDOFDQEHSURGXFHGZLWKRXWSULRUSHUPLVVLRQ
Field lines around a system of a positive and negative charge clearly shows the mutual
attraction between them.
Uniform electric field
Electric field corresponding to a negative charge is placed with in the
vicinity of a metal plate
Properties of Electric field lines.
1) Electric field lines start from +ve charge and end in –
ve charge.
2) Electric field lines do not form any closed loop.
3) Electric field lines never intersect each other.
If two lines intersect at a point, it means two directions for the field at that point which
is not at all possible and hence they never intersect each other. All right copy reserved. No part of the material can be produced without prior permission
attraction between them.
Uniform electric field
Electric field corresponding to a negative charge is placed with in the
vicinity of a metal plate
Properties of Electric field lines.
1) Electric field lines start from +ve charge and end in –
ve charge.
2) Electric field lines do not form any closed loop.
3) Electric field lines never intersect each other.
If two lines intersect at a point, it means two directions for the field at that point which
is not at all possible and hence they never intersect each other. All right copy reserved. No part of the material can be produced without prior permission
4) If the field lines are crowded, then the field is strong and if the field lines are not
crowded, then field is weak.
5) The electric field lines are always normal to the surface of the charge body.
6) In a charge-free region, electric field lines can be taken to be continuous curves
without any breaks.
Electric flux (ϕ)
The number of field lines crossing a unit area, placed normal to the field at a point is a
measure of the strength of electric field at that point. This means that if we place a small planar
element of area Δ
S normal to E at a point, the number of field lines crossing it is proportional*
to
E ΔS. Now suppose we tilt the area element by angle θ. Clearly, the number of field lines
crossing the area element will be smaller. The projection of the area element normal to
E is
Δ
S cosθ. Thus, the number of field lines crossing Δ S is proportional to E ΔS cosθ.
The electric flux is defined as the measure of total number of electric field lines passing
normally through a given surface.
If the surface is perpendicular to the field, then the flux through an area ΔS is
Δ ϕ = E ΔS
If the normal to the coil makes an angle ?????? with the electric field,
Flux through the surface
Δϕ = E ΔS Cos ??????
Δϕ = E. ΔS All right copy reserved. No part of the material can be produced without prior permission
crowded, then field is weak.
5) The electric field lines are always normal to the surface of the charge body.
6) In a charge-free region, electric field lines can be taken to be continuous curves
without any breaks.
Electric flux (ϕ)
The number of field lines crossing a unit area, placed normal to the field at a point is a
measure of the strength of electric field at that point. This means that if we place a small planar
element of area Δ
S normal to E at a point, the number of field lines crossing it is proportional*
to
E ΔS. Now suppose we tilt the area element by angle θ. Clearly, the number of field lines
crossing the area element will be smaller. The projection of the area element normal to
E is
Δ
S cosθ. Thus, the number of field lines crossing Δ S is proportional to E ΔS cosθ.
The electric flux is defined as the measure of total number of electric field lines passing
normally through a given surface.
If the surface is perpendicular to the field, then the flux through an area ΔS is
Δ ϕ = E ΔS
If the normal to the coil makes an angle ?????? with the electric field,
Flux through the surface
Δϕ = E ΔS Cos ??????
Δϕ = E. ΔS All right copy reserved. No part of the material can be produced without prior permission
Total Flux through a given surface
ϕ =?????? E . ΔS
OR
ϕ = E . S
Unit of electric flux is Nm
2
/C
Electric field due to continuous charge distributions
If a charge q is uniformly distributed along a line of length L, the linear charge density λ is
defined by
� =
�
�
and the unit of λ is Coulomb/meter(C/m).
For a charge q uniformly distributed over a surface of area A, the surface charge density σ is
?????? =
�
�
and unit of surface charge density is C/m
2
.
Similarly for uniform charge distributions volume charge density is
� =
�
??????
and unit of volume charge distribution is C/m
3
.
GAUSS’S LAW
Gauss's law was suggested by Karl Frederich Gauss (1777-1855) who was German
scientist and mathematician.
Gauss's law is basically the relation between the charge distribution producing the
electrostatic field to the behaviour of electrostatic field in space.
Gauss's law is based on the fact that flux through any closed surface is a measure of
total amount of charge inside that surface and any charge outside that surface would not
contribute anything to the total flux. All right copy reserved. No part of the material can be produced without prior permission
ϕ =?????? E . ΔS
OR
ϕ = E . S
Unit of electric flux is Nm
2
/C
Electric field due to continuous charge distributions
If a charge q is uniformly distributed along a line of length L, the linear charge density λ is
defined by
� =
�
�
and the unit of λ is Coulomb/meter(C/m).
For a charge q uniformly distributed over a surface of area A, the surface charge density σ is
?????? =
�
�
and unit of surface charge density is C/m
2
.
Similarly for uniform charge distributions volume charge density is
� =
�
??????
and unit of volume charge distribution is C/m
3
.
GAUSS’S LAW
Gauss's law was suggested by Karl Frederich Gauss (1777-1855) who was German
scientist and mathematician.
Gauss's law is basically the relation between the charge distribution producing the
electrostatic field to the behaviour of electrostatic field in space.
Gauss's law is based on the fact that flux through any closed surface is a measure of
total amount of charge inside that surface and any charge outside that surface would not
contribute anything to the total flux. All right copy reserved. No part of the material can be produced without prior permission
Gauss‘s law state that the total electric flux or total number of field lines passing through
any closed surface is equal to
1
??????
0
times the charges enclosed by the surface.
ϕ =
�
??????
0
E . S =
�
??????
0
The electric flux through the surface = E . S
The electric field intensity at P, E=
1
4�??????
0
�
�
2
Surface area of the spherical surface, S = 4�???r
2
The electric flux through the surface = E . S =
1
4�??????
0
�
�
2
x 4�???r
2
E S =
�
??????
0
ϕ =
�
??????
0
ie Gauss’s Law
All right copy reserved. No part of the material can be produced without prior permission
any closed surface is equal to
1
??????
0
times the charges enclosed by the surface.
ϕ =
�
??????
0
E . S =
�
??????
0
The electric flux through the surface = E . S
The electric field intensity at P, E=
1
4�??????
0
�
�
2
Surface area of the spherical surface, S = 4�???r
2
The electric flux through the surface = E . S =
1
4�??????
0
�
�
2
x 4�???r
2
E S =
�
??????
0
ϕ =
�
??????
0
ie Gauss’s Law
All right copy reserved. No part of the material can be produced without prior permission
APPLICATIONS OF GAUSS ’S LAW
Field due to an infinitely long straight uniformly charged wire.
Consider an infinitely long thin straight wire with uniform linear charge density λ. Let P
be a point at a distance r from the straight wire. The electric field lines are radially outward. To
find the electric field intensity at P, imagine a Gaussian surface of radius r. The electric flux
through two flat surfaces is zero because the electric field lines are radially outward.
The electric flux through though the curved surface,
E . S =
�
??????
0
E S =
�
??????
0
q = �???l
S = 2�???rl
E 2�???rl =
λl
??????
0
E =
λ
2πr??????
0
E̅ =
λ
2πr??????
0
??????̂
where ˆn is the radial unit vector plane normal to the wire.
Electric Flux
ϕ =
�
??????
0
All right copy reserved. No part of the material can be produced without prior permission
Field due to an infinitely long straight uniformly charged wire.
Consider an infinitely long thin straight wire with uniform linear charge density λ. Let P
be a point at a distance r from the straight wire. The electric field lines are radially outward. To
find the electric field intensity at P, imagine a Gaussian surface of radius r. The electric flux
through two flat surfaces is zero because the electric field lines are radially outward.
The electric flux through though the curved surface,
E . S =
�
??????
0
E S =
�
??????
0
q = �???l
S = 2�???rl
E 2�???rl =
λl
??????
0
E =
λ
2πr??????
0
E̅ =
λ
2πr??????
0
??????̂
where ˆn is the radial unit vector plane normal to the wire.
Electric Flux
ϕ =
�
??????
0
All right copy reserved. No part of the material can be produced without prior permission
ϕ =
λl
??????
0
Field due to a uniformly charged thin spherical shell
(i) Field outside the shell
Consider a spherical shell of radius R with uniform surface charge density ??????. Let P be a
point at a distance r from the centre of the spherical shell. Here the electric field lines are
radially outward. To find the electric field intensity at P imagine a Gaussian surface of radius
r.
Electric flux through the surface,
E . S =
�
??????
0
E S =
�
??????
0
q = 4�???R
2
?????????
S = 4�???r
2
E 4�???r
2
=
4πR
2
σ
??????
0
E =
σR
2
??????
0r
2
If the point is on the surface of the charged spherical shell, r =R
All right copy reserved. No part of the material can be produced without prior permission
λl
??????
0
Field due to a uniformly charged thin spherical shell
(i) Field outside the shell
Consider a spherical shell of radius R with uniform surface charge density ??????. Let P be a
point at a distance r from the centre of the spherical shell. Here the electric field lines are
radially outward. To find the electric field intensity at P imagine a Gaussian surface of radius
r.
Electric flux through the surface,
E . S =
�
??????
0
E S =
�
??????
0
q = 4�???R
2
?????????
S = 4�???r
2
E 4�???r
2
=
4πR
2
σ
??????
0
E =
σR
2
??????
0r
2
If the point is on the surface of the charged spherical shell, r =R
All right copy reserved. No part of the material can be produced without prior permission
E =
σ
??????
0
Field inside the shell
If the point P is inside the
Shell, the Gaussian surface is again a sphere through P centred at O.
The flux through the Gaussian surface,
E . S =
�
??????
0
Here the charge enclosed by the Gaussian surface is zero. (q = 0)
E . S = 0
E =0
The electric field inside a charged spherical shell is zero
Variation of electric field with distance from the centre of the spherical
shell
$OO???ULJKW???FRS\???UHVHUYHG???1R???SDUW???RI???WKH???PDWHULDO???FDQ???EH???SURGXFHG???ZLWKRXW???SULRU???SHUPLVVLRQ
σ
??????
0
Field inside the shell
If the point P is inside the
Shell, the Gaussian surface is again a sphere through P centred at O.
The flux through the Gaussian surface,
E . S =
�
??????
0
Here the charge enclosed by the Gaussian surface is zero. (q = 0)
E . S = 0
E =0
The electric field inside a charged spherical shell is zero
Variation of electric field with distance from the centre of the spherical
shell
$OO???ULJKW???FRS\???UHVHUYHG???1R???SDUW???RI???WKH???PDWHULDO???FDQ???EH???SURGXFHG???ZLWKRXW???SULRU???SHUPLVVLRQ
Field due to a uniformly charged infinite plane sheet
Consider an infinite plane sheet of charge with uniform charge density ??????. To find the electric
field intensity at P, imagine a Gaussian cylinder of cross sectional area A normal to the plane
of the sheet. Since the electric field lines are parallel to the curved surface, the flux through
this surface is zero.
The flux through two flat surfaces,
E . S =
�
??????0
E S =
�
??????0
S = 2ΔS
q =?????????
ΔS
E 2ΔS =
σ ΔS
??????0
E =
σ
2??????0
E =
σ
2??????0
??????̂
E is independent of x
All right copy reserved. No part of the material can be produced without prior permission
Consider an infinite plane sheet of charge with uniform charge density ??????. To find the electric
field intensity at P, imagine a Gaussian cylinder of cross sectional area A normal to the plane
of the sheet. Since the electric field lines are parallel to the curved surface, the flux through
this surface is zero.
The flux through two flat surfaces,
E . S =
�
??????0
E S =
�
??????0
S = 2ΔS
q =?????????
ΔS
E 2ΔS =
σ ΔS
??????0
E =
σ
2??????0
E =
σ
2??????0
??????̂
E is independent of x
All right copy reserved. No part of the material can be produced without prior permission
Electric field between two parallel plates
In region I
E = - (
σ
2??????0
+
−σ
2??????0
)=0
In region II
E = (
σ
2??????0
−
−σ
2??????0
)
E =
2σ
2??????0
E =
σ
??????0
Electric field intensity between two parallel plates E =
σ
??????0
In region III
E = (
σ
2??????0
+
−σ
2??????0
)=0
All right copy reserved. No part of the material can be produced without prior permission
In region I
E = - (
σ
2??????0
+
−σ
2??????0
)=0
In region II
E = (
σ
2??????0
−
−σ
2??????0
)
E =
2σ
2??????0
E =
σ
??????0
Electric field intensity between two parallel plates E =
σ
??????0
In region III
E = (
σ
2??????0
+
−σ
2??????0
)=0
All right copy reserved. No part of the material can be produced without prior permission
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