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Electricity EEE

Physics (Green University of Bangladesh)

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ELECTRICITY & MAGNETISM

Electrical Charge and Coulomb’s Law
Electromagnetic forces are responsible for the structure of atoms and for the binding of atoms in molecules
and solids. Many properties of materials that we have studied so far are electromagnetic in their nature. Such
as the elasticity of solids and the surface tension of liquids. The spring force, friction, and the normal force all
originate with the electromagnetic force between atoms.

Among the examples of electromagnetism that we shall study are the force between electric charges, such as
occurs between an electron and the nucleus in an atom; the motion of a charged body subject to an external
electric force. Such as an electron in an oscilloscope beam; the flow of electric charges through circuits and
the behavior of circuit elements; the force between permanent magnets and the properties of magnetic
materials; and electromagnetic radiation, which ultimately leads to the study of optics, the nature

and propagation of light.

In this chapter, we begin with a discussion of electric charge, some properties of charged bodies, and the
fundamental electric force between two charged bodies.

Electric Charge

Electric Charge and Electrical Forces

A body is said to be electrical neutral if it contains equal number of +ve and –ve charges. When two
bodies are rubbed together, their neutrality is distributed due to transfer of electrons from one body to the other.
The body which gives electrons becomes electrically positive and the body which gains electrons becomes
negative.
“Charges of the same signs repel each other and
Charges of the oppositely sign attract each other.”

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These attractive and repulsive forces among the charges are called electrical forces.

Point Charges:
The charge bodies whose sizes are much smaller than the distance between them are called point charges.

Coulomb’s Law

Charles Augustin Coulomb (l736- 1806) measured electrical attractions and repulsions quantitatively

and deduced the law that governs them. His apparatus, shown in Fig. consist of
spheres A and B.
If A and B are charged, the electric force on A tends to twist the suspension fiber. The
angle is then a relative measure of the electric force acting on charge
A. Experiments due to Coulomb and his contemporaries showed that

Statement

The magnitude of electrical force between two point charges is directly proportional
to the product of magnitude of charges and inversely proportional to the square of the
distance between their centers and the force acts along the line connecting the
charges.

Mathematical Form

Suppose two point charges 𝑞1 and 𝑞2 are separated by distance 𝑟. According

to the Coulomb’s law, the electrical force between two charges is:

𝐹 ∝ 𝑞1 𝑞2 − − − − − (1)

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1
𝐹∝ − − − − − −(2)
𝑟2

Combining (1) and (2) we have,

𝑞1 𝑞2
𝐹∝
𝑟2
𝑞1 𝑞2
𝑜𝑟, 𝐹 = 𝑘
𝑟2

Where k is called Coulomb’s constant. Its value depends upon the system of
units and medium between the charges. For free space and in system international
‘k’ is expressed as:
1
𝑘=
4𝜋𝜖0

Where 𝜖0 is the permittivity of free space and its value in SI unit is: 8.854 × 10−12 𝐶 2 𝑁 −1

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Example: What must be the distance between point charge 𝒒𝟏 = 𝟐𝟔. 𝟑 𝝁𝑪 and point charge
𝒒𝟐 = −𝟒𝟕. 𝟏 𝝁𝑪 for the attractive electrical force between them to have a magnitude of 5.66
N.

Solution:

𝑞1 = 26.3 𝜇𝐶 = 26.3 × 10−6 𝐶

𝑞2 = −47.1 𝜇𝐶 = −47.1 × 10−6 𝐶

𝐹 = 5.66 𝑁

From Coulomb’s law:

𝑞1 𝑞2
𝐹=𝑘
𝑟2

𝑘𝑞1 𝑞2 (9×109 )×(26.3×10−6 )×(47.1×10−6 )

Or,𝑟 = √ 𝐹
=√ 5.66
= 1.40 𝑚

Example : In the radioactive decay of 𝑼𝟐𝟑𝟖 , the center of the emerging 𝟒𝟐𝑯𝒆 particle is at a certain
distance 𝟏𝟐 × 𝟏𝟎−𝟏𝟓 𝒎 from the center of residual 𝟐𝟑𝟔𝑻𝒉 nucleus at that instant. (a) What is the
force on helium atom and (b) what is its acceleration?

Solution:

238
𝑈 → 42𝐻𝑒 + 234
90𝑇ℎ

𝑟 = 12 × 10−15 𝑚
𝐹 =?
𝑎 =?
𝑞1 𝑞2 9
(2𝑒) × (90𝑒)
𝐹=𝑘 = 9 × 10 × = 288 𝑁
𝑟2 (12 × 10−15 )2
Mass of helium atom 𝑚 = 4 𝑎𝑚𝑢 = 4 × 1.67 × 20−27 𝑘𝑔 = 6.67 × 10−27 𝑘𝑔
From Newton’s second law of motion:
𝐹 288
𝐹 = 𝑚𝑎 ⇒ 𝑎 = = = 43.18 × 1023 𝑚𝑠 −2
𝑚 6.67 × 10−27

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Vector Form of Coulomb’s Law

So far, we have considered only the magnitude of the force between two charges determined according

to Coulomb’s law. Force, being a vector, has directional properties as well. In case of coulomb’s law, the
direction of the force is determined by the relative sign of the two
electric charges.

Suppose we have the two point charges ′𝑞1 ′ and ′𝑞2 ′

separated by a distance 𝑟12 . For the moment, we assume the two
charges to have the same sign, so that they repel each other. Let us
consider the force exerted on 𝑞1 by 𝑞2 is denoted by⃗⃗⃗𝐹12 . The
position
vector that locates particle 1 relative to particle 2 is 𝑟12 ; that is, if we
were to define the origin of our coordinate system at the location of
𝑞2 , then 𝑟12 would be the position vector of 𝑞1.
If the charges have the same sign, then the force is repulsive

and 𝐹12 must be parallel to 𝑟12, as shown in figure a. If the charges are of opposite sign, as in figure b,
then 𝐹12 is attractive and anti-parallel to 𝑟12. In either case we can represent the force as:
𝑞1 𝑞2
𝐹12 = 𝑘 2 𝑟̂12
𝑟12
Coulomb force due to many point charges

Let 𝑞1 , 𝑞2 , 𝑞3 … … … 𝑞𝑛 are the ‘n’ point charges as

shown in the figure. We want to find out electrical force on
charge
𝑞1 by the assembly of n point charges. The point charges
𝑞2 , 𝑞3 , 𝑞4 … … … 𝑞𝑛 are at the distances 𝑟12 , 𝑟13 , 𝑟14 … … … 𝑟1𝑛
from charge 𝑞1 respectively.

Now if 𝐹12 , 𝐹13 , 𝐹14 … … … 𝐹1𝑛 be the electrostatic force

on charge 𝑞1 due to the point charges 𝑞2 , 𝑞3 , 𝑞4 … … … 𝑞𝑛

respectively. Thus, the total electrostatic force 𝐹1 on charge 𝑞1 is
given by;

𝐹1 = 𝐹12 + 𝐹13 + 𝐹14 + ⋯ … … + 𝐹1𝑛 − − − − − −(1)

Where,
1 𝑞1 𝑞2
𝐹12 = Force on charge ′𝑞1 ′ exerted by ′𝑞2 ′ = 2 𝑟̂12
4𝜋𝜖0 𝑟12

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1 𝑞1 𝑞2
𝐹13 = Force on charge ′𝑞1 ′ exerted by ′𝑞3 ′ = 2
4𝜋𝜖0 𝑟13
𝑟̂13

… … … …

… … … …
1 𝑞1 𝑞𝑛
𝐹1𝑛 = Force on charge ′𝑞1 ′ exerted by ′𝑞𝑛 ′ = 4𝜋𝜖 2 𝑟̂1𝑛
0 𝑟1𝑛

Putting values in equation (1) we get,

1 𝑞1 𝑞2 1 𝑞1 𝑞2 1 𝑞1 𝑞𝑛
𝐹1 = 2 𝑟̂12 + 2 𝑟̂13 + ⋯ … … + 2 𝑟̂1𝑛
4𝜋𝜖0 𝑟12 4𝜋𝜖0 𝑟13 4𝜋𝜖0 𝑟1𝑛

𝑞1 𝑞2 𝑞2 𝑞𝑛
= ( 2 𝑟̂12 + 2 𝑟̂13 + ⋯ … … + 2 𝑟̂1𝑛 )
4𝜋𝜖0 𝑟12 𝑟13 𝑟1𝑛
𝑛
𝑞1 𝑞𝑖
= ∑ 2 𝑟̂1𝑖
4𝜋𝜖0 𝑟1𝑖
𝑖=1

This expression gives the electrical force between on a point charge due to many point charges.

Quantization of Charges
When the two bodies are rubbed together, transfer of electrons from one body to the other takes place
and they are said to be electrified. The magnitude of charge q that can be detected and measured on any
object is given by
𝑞 = 𝑛𝑒 ---------------------- (1)

where 𝑛 = 0, ±1, ±2, … … … and 𝑒 is the elementary unit of charge called on unit charge, has
the experimentally determined value
𝑒 = 1.6 × 10−19 𝐶

When a physical quantity is discrete values, it is called quantized quantity.

Equation (1) shows that charge is also a quantized quantity like matter, energy, angular momentum
etc. It means that we can find a body that can have a charge of 10𝑒 or −5𝑒 but it is not possible to find a
body with fractional charge such as +3.57𝑒 or −2.35𝑒.

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Example- The electrostatic force between identical ions that are separated by a distance of 𝟓 ×
𝟏𝟎−𝟏𝟎 𝒎 is 𝟑. 𝟕 × 𝟏𝟎−𝟗 𝑵 (a) Find the charge on each ion? (b) How many electrons are missing from
each ion?

Solution: 𝐹 = 3.7 × 10−9 𝑁

𝑟 = 5 × 10−10 𝑚

(a) 𝑞1 = 𝑞2 = 𝑞 =?

𝑞1 𝑞2 𝑞2
𝐹=𝑘 = 𝑘
𝑟2 𝑟2

𝐹𝑟 2 3.7 × 10−9 × (5 × 10−10 )2

𝑜𝑟, 𝑞 2 = = = 10.27 × 10−38
𝑘 9 × 109

𝑞 = 3.20 × 10−19 𝐶
(b) 𝑛 = ?
𝑞 = 𝑛𝑒

𝑞 3.20 × 10−19
𝑛= = =2
𝑒 1.6 × 10−19

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THE ELECTRIC FIELD

Electric charges interact with each other over vast distances. Electrons or ionized atoms at the
furthest reaches of the known universe can exert forces that cause electrons to move on the Earth.
How can we explain these interactions? We do so in terms of electric field- the distant charge set up
an electric field, which exist throughout the space between the Earth and is the origin of the field.

In this chapter we consider only the static electric field due to charges at rest.
Electric Field Intensity

To describe the mechanism by which one charge particle exert the force the force on other charge particles,

Michael Faraday introduced the concept of electric field.

Electric Field

The region or space around a charge in which it can exert the force of attraction or repulsion on other
charged bodies is called electric field.
Electric Field Intensity

The electrostatic force on unit positive charge at a specific field point is called the electric field intensity.
In order to find out electric field intensity, a test charge 𝑞0 is placed in the electric field at a field point. The
electric field intensity 𝐸⃗ is expressed as,
𝐹
𝐸⃗ =
𝑞0

Where 𝐹 is the electrostatic force on test charge 𝑞0 .

The test charge 𝑞0 should be very small, so that it cannot disturb the field produced by source charge

𝑞 Therefore, the electric field intensity can be written as,

𝐹
𝐸⃗ = lim
𝑞0 →0 𝑞0

Electric Field Lines or Lines of Force

A visual representation of the electric field can be obtained in terms of electric field lines. Electric field
lines can be thought of a map that provides information about the direction and strength of the electric field at
various places. As electric field line provides the information about the electric force exerted on a charge, the
lines are commonly called “Lines of Force”.

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Properties of Electric Field Lines

i) Electric field lines originate from positive charges and end on negative charges.

ii) The tangent to a field line at any point gives the direction of the electric field intensity at that point.

iii) The lines are closer where the field is strong, the lines are farther apart where the field is weak.

iv) No two lines cross each other.

Electric Field Intensity Due to a Point Charge

Consider a test charge 𝑞0 placed at point P in the electric field of a point charge 𝑞 at a distance 𝑟 apart.

We want to find out electric field intensity at point 𝑃 due to a point charge 𝑞.

The electrostatic force 𝐹 between 𝑞 and 𝑞0 can be find out by using expression,

1 𝑞𝑞0
𝐹=
4𝜋𝜖0 𝑟 2

The electric field intensity 𝐸 due to a point charge 𝑞 can be obtained by putting the value of
electrostatic force in expression of electric field intensity:

1 𝑞𝑞0
(
)
4𝜋𝜖0 𝑟 2
𝐸=
𝑞0

1 𝑞
𝐸=
4𝜋𝜖0 𝑟 2

This expression gives the magnitude of electric field intensity due to a point charge 𝑞. In vector form,
the electric field intensity 𝐸⃗ will be:
1 𝑞
𝐸⃗ = 𝑟̂
4𝜋𝜖0 𝑟2

Where 𝑟̂ is the unit vector which gives the direction of electric field intensity.

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Example-. In an ionized helium atom (a helium atom in which one of the two electrons has been
removed) the electron and nucleus are separated by a distance of 26.5 pm. What electric field due to the
nucleus at the location of the electron.

Solution:

Total charge of helium nucleus 2𝑒 = 2 × 1.6 × 10−19 = 3.2 × 10−19 𝐶

Distance 𝑟 = 26.5 𝑝𝑚 = 26.5 × 10−12 𝑚

𝑞 9
3.2 × 10−19
𝐸=𝑘 = 9 × 10 × = 4.1 × 1012 𝑁𝐶 −1
𝑟2 (26.5 × 10−12 )2

Example- Two equal and opposite charges of magnitude 𝟏. 𝟖𝟖 × 𝟏𝟎−𝟕 𝑪 are held 15.2 cm apart.
What is the direction and magnitude of E at mid-point between the charges? What is the force act
on an electron placed here?

Solution:

𝑞1 = 1.88 × 10−7 𝐶

𝑞2 = 1.88 × 10−7 𝐶

Distance, 𝑟 = 15.2 𝑐𝑚 = 15.2 × 10−2 𝑚

𝑟
Mid-point, 𝑑 = 2 = 7.6 × 10−2 𝑚

Total electric field 𝐸 = |𝐸+ | + |𝐸− | = ?

Force on an electron placed at the same point 𝐹 = ?

𝑞1 9
1.88 × 10−7 𝑁
|𝐸+ | = 𝑘 2
= 9 × 10 × −2 2
= 1.28 × 106
𝑑 (7.6 × 10 ) 𝐶

𝑞2 9
1.88 × 10−7 6
𝑁
|𝐸− | = 𝑘 = 9 × 10 × = 1.28 × 10
𝑑2 (7.6 × 10−2 )2 𝐶
𝑁
Total electric field 𝐸 = |𝐸+ | + |𝐸− | = 2.56 × 106
𝐶

𝑁
Now, Force on an electron placed at the same point 𝐹 = 𝑞𝐸 = −1.6 × 10−19 × 2.56 × 106
𝐶

= −4.096 × 10−13 𝑁

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GAUSS’S LAW

Coulomb’s law can always be used to calculate the electric field intensity for any discrete or continuous charge
distribution of charges at rest. The sums or integrals might be complicated (and a computer might be needed to evaluate
them numerically), but resulting electric field intensity can always be found.

In this chapter, we discuss an alternative to Coulomb’s law, called Gauss’s law, that provides a more useful and
instructive approach to calculating the electric field in the situations having certain symmetries.

The number of situations that can directly be analyzed using Gauss’s law is small, but those cases can be done with
extraordinary ease. Although Gauss’s law and Coulomb’s law gives identical results in the cases in which both can be

used. Gauss’s law is considered a more fundamental equation than Coulomb’s law. It is fair to say that while Coulomb’s
law provides workhorse of electrostatics, Gauss’s law provides the insight.
Electric Flux

The number of electric lines of force passing normally through a certain area is called the electric

flux. It is measured by the product of area and the component of electric field intensity normal to the area. It is
denoted by the symbol 𝜙𝑒 .
Consider a surface ′𝑆′ placed in a uniform electric field of intensity 𝐸⃗ .
Let 𝐴 be the area of the surface. The component of 𝐸⃗ normal to the area 𝐴 is
𝐸𝑐𝑜𝑠𝜃 as shown in the figure below.
The electric flux through the surface 𝑆 is given by;

𝜙𝑒 = 𝐴(𝐸𝑐𝑜𝑠𝜃)

𝜙𝑒 = 𝐸𝐴𝑐𝑜𝑠𝜃

𝜙𝑒 = 𝐸⃗ . 𝐴

Thus, the electric flux is the scalar product of electric field intensity and the vector area. The SI unit of
𝑁𝑚2
the electric flux is
𝐶

Electric Flux through an Irregular Shaped Object

Consider an object of irregular shape placed in a non-uniform electric field. We want to find out the

expression of electric flux through this irregular shaped object.

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We divide the surface into n number of small
patches having area Δ𝐴1 , Δ𝐴2 , Δ𝐴3 , … … … , Δ𝐴𝑛 .
Let 𝐸⃗1 , 𝐸⃗2 , 𝐸⃗3 , … … … , 𝐸⃗𝑛 are the electric field intensities
which makes angle 𝜃1 , 𝜃2 , 𝜃3 , … … … , 𝜃𝑛 with the normal
to the area elements Δ𝐴1 , Δ𝐴2 , Δ𝐴3 , … … … , Δ𝐴𝑛
respectively. If 𝜙1 , 𝜙2 , 𝜙3 , … … … , 𝜙𝑛 be the electric
flux through Δ𝐴1 , Δ𝐴2 , Δ𝐴3 , … … … , Δ𝐴𝑛 , then the total
electric flux 𝜙𝑒 will be:
𝜙𝑒 = 𝜙1 + 𝜙2 + 𝜙3 + ⋯ … … + 𝜙𝑛
⇒ 𝜙𝑒 = 𝐸1 (Δ𝐴1 𝑐𝑜𝑠𝜃1 ) + 𝐸2 (Δ𝐴2 𝑐𝑜𝑠𝜃2 ) + ⋯ … … + 𝐸𝑛 (Δ𝐴𝑛 𝑐𝑜𝑠𝜃𝑛 )

⇒ 𝜙𝑒 = 𝐸1 Δ𝐴1 𝑐𝑜𝑠𝜃1 + 𝐸2 Δ𝐴2 𝑐𝑜𝑠𝜃2 + ⋯ … … + 𝐸𝑛 𝐴𝑛 𝑐𝑜𝑠𝜃𝑛

⇒ 𝜙𝑒 = 𝐸⃗1 . Δ𝐴1 + 𝐸⃗2 . Δ𝐴2 + ⋯ … … + 𝐸⃗𝑛 . Δ𝐴𝑛 .

Where Δ𝐴1 , Δ𝐴2 , … … … Δ𝐴𝑛 are the vector areas corresponding to area elements
Δ𝐴1 , Δ𝐴2 , … … … Δ𝐴𝑛 respectively

𝜙𝑒 = ∑ 𝐸⃗𝑖 . Δ𝐴𝑖
𝑖=1

When 𝑛 → ∞, 𝑜𝑟 Δ𝐴 → 0, then the sigma is replaced by the surface integral

i.e,

𝜙𝑒 = ∫ 𝐸⃗ . 𝑑𝐴
𝑆

By convention, the outward flux is taken as positive and inward flux is taken as negative

Example-. Consider a hypothetical closed cylinder of radius 𝑹 immersed in a uniform electric field ⃗𝑬, the
cylinder axis being parallel to the field What is 𝝓𝒆 for this closed surface?

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Solution:

The flux 𝜙𝑒 can be written as the sum of three terms, an

integral over (a) the left cylinder cap, (b) the cylindrical
surface, and (c) the right cap.

𝜙𝑒 = ∮ 𝐸⃗ . 𝑑𝐴 = ∫ 𝐸⃗ . 𝑑𝐴 + ∫ 𝐸⃗ . 𝑑𝐴 + ∫ 𝐸⃗ . 𝑑𝐴
𝑎 𝑏 𝑐

For the left cap, the angle 𝜃 for all points is 180°, 𝐸⃗ has a constant value, and the vectors 𝑑𝐴 are all parallel. Thus,

∫ 𝐸⃗ . 𝑑𝐴 = ∫ 𝐸𝑑𝐴𝑐𝑜𝑠180° = −𝐸∫ 𝑑𝐴 = −𝐸𝐴

𝑎

Similarly, for the right cap,

∫ 𝐸⃗ . 𝑑𝐴 = ∫ 𝐸𝑑𝐴𝑐𝑜𝑠0° = 𝐸∫ 𝑑𝐴 = 𝐸𝐴
𝑐

The angle 𝜃 for all points being 0° here.

Finally, for the cylinder wall,

∫𝐸⃗ . 𝑑𝐴 = 0
𝑏

Because, 𝜃 = 90°; hence, 𝐸⃗ . 𝑑𝐴 = 0 for all points on the cylindrical surface. Thus the total flux is,
𝜙𝑒 = −𝐸𝐴 + 0 + 𝐸𝐴 = 0
This result is expected, because the filed lines that enter at the left goes out from the right end.

Gauss’s Law

Statement:

1
The total electric flux through any close surface is times the total charge enclosed by the surface.
𝜖0

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Explanation
The Gauss’s law gives the relation between total flux and total charge enclosed by the surface.
Consider a collection of positive and negative charges in a certain region of space. According to Gauss’s law:

𝑞
𝜙𝑒 = − − − −(1)
𝜖0
Where 𝑞 is the net charge enclosed by the surface. Also,

𝜙𝑒 = ∮ 𝐸⃗ . 𝑑𝐴 − − − (2)

Comparing (1) and (2) we have,

𝑞
∮ 𝐸⃗ . 𝑑𝐴 =
𝜖0

Thus we can describe the Gauss’s law as

1
The surface normal integral of electric field intensity is equal to 𝜖 times the total charge
0

enclosed by the surface.

Electric Field due to Spherical Shell of Charge

Question: Show that the uniform spherical shell of charge behaves, for all external points, as if all its

charges were concentrated at its center.

Proof: Consider a thin spherical shell of radius ‘𝑅’ which have the charge
‘𝑞’ with constant surface charge density ‘𝜎’.

Consider a point ‘𝑃’ outside the shell. We want to find out electric

field intensity due to this charge distribution. For this we consider a

spherical Gaussian surface of radius 𝑟 > 𝑅 which passes through point

′𝑃′ as shown in the figure below.

According to Gauss’s law,

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𝑞
∮ 𝐸⃗ . 𝑑𝐴 =
𝜖0

𝑞
∮ 𝐸𝑑𝐴𝑐𝑜𝑠0° = [𝐸⃗ ∥ 𝑑𝐴]
𝜖0

𝑞
𝐸 ∮ 𝑑𝐴 = [ 𝐸 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡]
𝜖0

𝑞
𝐸(4𝜋𝑟 2 ) =
𝜖0

1 𝑞
𝐸=
4𝜋𝜖0 𝑟 2

Thus the uniform spherical shell of charge behaves like a point charge for all
the points outside the shell.

Question: Show that the uniform spherical shell of charge exerts no

electrostatic force on a charged particle placed inside the shell.

Consider a point ‘𝑃’ inside the shell. We want to fine out electric field
intensity ‘𝐸’ at point ‘𝑃’ due to this symmetrical charge distribution. For this we consider a spherical
Gaussian surface of radius 𝑟 < 𝑅 which passes through point ‘𝑃’ as shown in the figure below.

According to Gauss’s law,

𝑞
∮ 𝐸⃗ . 𝑑𝐴 =
𝜖0
Because the Gaussian surface enclose no charge, therefore ‘q = 0’,

∮ 𝐸⃗ . 𝑑𝐴 = 0

∮ 𝐸𝑑𝐴𝑐𝑜𝑠0° = 0

𝐸 ∮ 𝑑𝐴 = 0

As 𝑑𝐴 ≠ 0, therefore
𝐸=0

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So the electric field does not exist inside a uniform shell of charge. So the test charge placed inside the
charged shell would experience no force.

Deduction of Coulomb’s Law from Gauss’s Law

Coulomb’s law can be deduced from Gauss’s law under certain symmetry consideration. Consider positive point
charge ‘𝑞’. In order to apply the Gauss’s law, we assume a spherical Gaussian surface as shown in the figure below.

Considering the integral form of Gauss’s law,

𝑞
∮ 𝐸⃗ . 𝑑𝐴 =
𝜖0

Because the both vectors 𝐸⃗ and 𝑑𝐴 are directed radially outward, so

𝑞
∮ 𝐸𝑑𝐴𝑐𝑜𝑠0° =
𝜖0

As E is constant for all the points on the spherical Gaussian surface,

𝑞
𝐸 ∮ 𝑑𝐴 =
𝜖0

𝑞
𝐸(4𝜋𝑟 2 ) =
𝜖0

1 𝑞
𝐸=
4𝜋𝜖0 𝑟 2

This equation gives the magnitude of electric field intensity 𝐸 at any point which is at the distance ‘𝑟’
from an isolated point charge ‘𝑞’.
From the definition of electric field intensity, we know that

𝐹 = 𝑞0 𝐸

Where 𝑞0 is the point charge placed at a point at which the value of electric field intensity has to be
determined Therefore

1 𝑞𝑞0
𝐹=
4𝜋𝜖0 𝑟 2

This is the mathematical form of Coulomb’s law.

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ELECTRIC POTENTIAL

The energy approach in the study of dynamics of the particles can yield not only the simplification but also new
insights. One advantage of energy method is that, although force is a vector, energy is a scalar. In problems
involving vector forces and fields, calculations involving sums and integrals are often complicated
In this chapter, we introduce the energy method to the study of
electrostatics.
Potential Difference

Potential difference ′Δ𝑉′ between two point is defined as “the amount of work done ′Δ𝑊′ per unit
charge ′𝑞0 ′ in moving it from one point to the other against the electric field and by keeping the system in
equilibrium”. Mathematically

Δ𝑊
Δ𝑉 =
𝑞0
Suppose a unit positive test charge ′𝑞0 ′ is moved from one point ′𝑎′ to the point ′𝑏′ in the electric

field 𝐸⃗ of a large positive charge ′𝑞′ as shown in figure below:

The work done in moving ′𝑞0 ′ from point ′𝑎′ to the point ′𝑏′ against the electric field 𝐸⃗ is
𝑏

𝑊𝑎→𝑏 = ∫ 𝐹 . 𝑑𝑟
𝑎

The electrical force of magnitude 𝐹 = −𝑞0 𝐸⃗ must have to supplied in order to move ′𝑞0 ′ against the electric
field. Therefore
𝑏

𝑊𝑎→𝑏 = ∫൫−𝑞0 𝐸⃗ ൯. 𝑑𝑟
𝑎
𝑏

𝑊𝑎→𝑏 = −𝑞0 ∫ 𝐸⃗ . 𝑑𝑟
𝑎 𝑏
𝑊𝑎→𝑏
= − ∫ 𝐸⃗ . 𝑑𝑟
𝑞0
𝑎

𝑊𝑎→𝑏
as, 𝑞0
= Δ𝑉 = 𝑉𝑏 − 𝑉𝑎

Therefore, the electrical potential difference between two points in an electrical field will be,
𝑏

𝑉𝑏 − 𝑉𝑎 = − ∫ 𝐸⃗ . 𝑑𝑟
𝑎

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Absolute Electrical Potential at a Point

Absolute electric potential at a point is defined as “the amount of work done per unit charge in

moving it from infinity to a specific field point against the electric field and by keeping the system in
equilibrium”.
To find the absolute potential, the reference point is selected at which potential is zero. This point is
situated at infinity i.e., out of the electric field. Thus, in equation (1)
𝑉𝑎 = 𝑉(∞) = 0
Thus,
𝑏

𝑉𝑏 − 0 = − ∫ 𝐸⃗ . 𝑑𝑟
∞

If the distance from the point ′𝑏′ to the charge ′𝑞′ is ′𝑟′, then in general
𝑟

𝑉(𝑟) = − ∫ 𝐸⃗ . 𝑑𝑟
∞

Expression for the Electric Potential Difference due to a Point Charge

The potential difference between two points is the amount of work done per unit charge ′𝑞0 ′ in

moving it from one point to the other against the electric field ′𝐸⃗ ′. Mathematically, it is described as:

𝑟𝑏

𝑉𝑏 − 𝑉𝑎 = − ∫ 𝐸⃗ . 𝑑𝑟
𝑟𝑎

1 𝑞
But the electric field intensity due to point charge: 𝐸⃗ = 𝑟̂
4𝜋𝜖0 𝑟 2

So,
𝑟𝑏
1 𝑞
𝑉𝑏 − 𝑉𝑎 = − ∫ 𝑟̂ . 𝑑𝑟
4𝜋𝜖0 𝑟 2
𝑟𝑎

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𝑟𝑏
𝑞 1
𝑉𝑏 − 𝑉𝑎 = − ∫ 2 𝑟̂ . 𝑑𝑟
4𝜋𝜖0 𝑟
𝑟𝑎

As, 𝐸⃗ is directed radially outward, therefore 𝑟̂ ∥ 𝑑𝑟

hence, 𝑟̂ . 𝑑𝑟 = |𝑟̂ ||𝑑𝑟|𝑐𝑜𝑠0° = (1)(𝑑𝑟)(1) = 𝑑𝑟

𝑞 𝑟 𝑑𝑟
So, 𝑉𝑏 − 𝑉𝑎 = − 4𝜋𝜖 ∫𝑟 𝑏 2
𝑟 0 𝑎

𝑞 1 𝑟𝑏 𝑞 1 𝑟𝑏
𝑉𝑏 − 𝑉𝑎 = − |− | = | |
4𝜋𝜖0 𝑟 𝑟𝑎 4𝜋𝜖0 𝑟 𝑟𝑎

𝑞 1 1
𝑉𝑏 − 𝑉𝑎 = [ − ]
4𝜋𝜖0 𝑟𝑏 𝑟𝑎

This is the expression for the potential difference between two points ′𝑎′ 𝑎𝑛𝑑 ′𝑏′.

Expression for the Absolute Electric Potential due to a Point Charge

The electric potential at any point is the amount of work done per unit charge in moving a unit

positive charge (test charge) from infinity to that point, against the electric field.
If the point ′𝑎′ is at infinity then

𝑉𝑎 = 𝑉(∞) = 0, 𝑟𝑎 = ∞

Putting this value in equation in the expression of electric potential difference due to point charge, we get:

𝑞 1 1
𝑉𝑏 − 0 = [ − ]
4𝜋𝜖0 𝑟𝑏 ∞

𝑞 1
𝑉𝑏 = [ ]
4𝜋𝜖0 𝑟𝑏

In general, the electric potential at point due to a point charge ′𝑞′ is

1 𝑞
𝑉=
4𝜋𝜖0 𝑟

Example-. Two protons in the nucleus of 𝑼𝟐𝟑𝟖 are 𝟔 𝒇𝒎 apart. What potential energy associated
with the electric force that acts between them?
Solution:

𝑟 = 6 𝑓𝑚 = 6 × 10−15 𝑚

Δ𝑈 = ?

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−19
𝑞1 = 𝑞2 = 1𝑒 = 1.6 × 10 𝐶

As,

q 1.6 × 10−19
ΔV = k = 9 × 109 × = 2.34 × 105 V
r 6 × 10−15

Now, Δ𝑈 = 𝑞1 . Δ𝑉 = 1.6 × 10−19 × 2.34 × 105 = 3.744 × 10−14 𝐽

3.744 × 10−14
Δ𝑈 = = 2.34 × 105 𝑒𝑉
1.6 × 10−19

Example-. What is the electric potential at the surface of the gold nucleus. The radius of the gold
nucleus is 𝟕 × 𝟏𝟎−𝟏𝟓 𝒎 and atomic number is 79.

Solution: 𝑟 = 7 × 10−15 𝑚

𝑍 = 79

𝑞 = 79𝑒 = 79 × 1.6 × 10−19 𝐶

𝑉 =?

𝑞 79×1.6×10−19
As, 𝑉 = 𝑘 𝑟 = 9 × 109 × = 1.6 × 107 𝑉
7×10−15

CAPACITORS AND DIELECTRICS

A capacitor is a device that stores energy in an electrostatic field. A flash bulb, for example, requires a short
burst of electric energy that exceeds what a battery can generally provide. We can draw energy relatively
slowly (over several seconds) from battery into the capacitor which releases the energy rapidly (within
milliseconds) through the bulb. Much larger capacitors are used to provide intense laser pulses in attempts to
induce thermonuclear fusion in tiny pellets of hydrogen. In this case the power level is about 104 but it last
for only about 10−9 𝑠

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Capacitors can also be used to produce electric fields, such as the parallel plate device that deflects beam of
charged particles.

In this chapter, we consider the electrostatic field and stored energy of capacitor.

Capacitors have other important functions in electronic circuits, especially for time varying voltages and
currents. For transmitting and receiving radio and TV signals, capacitors are fundamental components of
electromagnetic oscillators.
CAPACITOR

Capacitor is a device winch is used to store charge. A simple capacitor consists of two conductors

which are separated a small distance. There may be vacuum or some dielectric medium between the conductors
of a capacitor.
When the plates of a capacitor are connected with the terminals of the battery of emf V, then the charge
q is stored in the capacitor. This charge stored is directly proportional to the potential difference applied between
the plates.
𝑞∝𝑉
𝑞 = 𝐶𝑉

Here C is constant of proportionality, called the capacitance of a capacitor. The capacitance of a

capacitor is its ability to store electrical charge. The SI unit of capacitance is farad which can be defined as “If
one coulomb of charge given to the plates to produce a potential difference of one volt, then capacitance of the
capacitor is one farad”.

Example- A storage capacitor on a random access memory (VRAM) chip has a capacitance of 55 fF. If it
is charged to 5.3 V, how many excess electrons are there on its negative plate?

Solution:

𝐶 = 55𝑓𝐹 = 55 × 10−15 𝐹,

𝑉 = 5.3 𝑉

𝑞 =?

For case of a capacitor,

𝑞 = 𝐶𝑉 = 55 × 10−15 × 5.3 = 291 × 10−15 𝐶

Number of electrons 𝑛 =?

As 𝑞 = 𝑛𝑒

𝑞 291 × 10−15 𝐶
𝑛= = = 181.8 × 104 = 1.818 × 106 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
𝑒 1.6 × 10−19 𝐶

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Capacitance of Parallel Plate Capacitor

Let the two plates of a parallel plate capacitor has area 𝐴, and charge 𝑞 on them. The plates are separated by a
distance 𝑑. Suppose that the length of plate is very large as compared to the distance between the plates. So, 𝐸
inside the plates of the capacitor is uniform. We want to find out the expression of electric field between the
plates of capacitor.

For this we consider a box shaped Gaussian surface which encloses positive charge +𝑞 on the top plate. We can
write

𝜀𝑜 ∮𝐸⃗ . 𝑑𝐴 = 𝑞

∴ 𝜀𝑜 ∮𝐸 𝑑𝐴 𝐶𝑜𝑠𝜃 = 𝑞

∴ 𝜀𝑜 ∮𝐸 𝑑𝐴 𝐶𝑜𝑠0° = 𝑞

𝑆𝑖𝑛𝑐𝑒 𝐸 𝑎𝑛𝑑 𝑑𝐴 𝑎𝑟𝑒 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙, 𝑖. 𝑒. 𝜃 = 0°

∴ 𝜀𝑜 𝐸 ∮ 𝑑𝐴 = 𝑞

∴ 𝜀𝑜 𝐸 𝐴 = 𝑞

𝑞
∴𝐸 =
𝜀𝑜 𝐴

The equation is expression of electric field intensity inside the plates of capacitor.

Now potential difference between the plates

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𝑞𝑑
𝑉 = 𝐸𝑑 =
𝜀𝑜 𝐴

Hence the capacitance between the plates,

𝑞 𝑞 𝑞𝜀𝑜 𝐴 𝜀𝑜 𝐴
𝐶= = = =
𝑉 𝑞𝑑 𝑞𝑑 𝑑
𝜀𝑜 𝐴

This is the expression for capacitance of a parallel plate capacitor with the free space between the plates.

Capacitance with Dielectrics

Consider a parallel plate capacitor which is connected with a battery of emf ′𝑉′. Let 𝐴 is the area

of each plate and ′𝑑′ is the separation between the plates.

If 𝑞 charge is stored in the capacitor when there is vacuum or air as medium between the plates, then

𝑞 = 𝐶𝑉

Where 𝐶 is the capacitance of the capacitor, which, for the case of parallel plate capacitor is expressed as

𝜖0 𝐴
𝐶=
𝑑
Micheal Faraday, in 1937, investigated that if the space between the plates of a parallel plate capacitor
is filled with some dielectric medium, then the charge stored in the capacitor increased to 𝑞′. And, hence, the
capacitance of the capacitor also increases to 𝐶′.
Therefore, the relation between the stored charges in the capacitor to the capacitance will become

𝑞′ = 𝐶 ′𝑉

The factor by which the capacitance of capacitor increases as compared to the capacitance with air as the
medium is called the dielectric constant 𝜅𝑒 . The dielectric constant 𝜅𝑒 is a dimensionless quantity. It is also
called relative permittivity of the medium. The dielectric constant is described mathematically as:

𝐶′
𝜅𝑒 =
𝐶

𝐶 ′ = 𝜅𝑒 𝐶

𝜖0 𝐴
𝐶 ′ = 𝜅𝑒 ( )
𝑑

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′
So, 𝐶 > 𝐶 by a factor of ′𝜅𝑒 ′

Effects of Dielectric Medium

 When a dielectric medium, having dielectric constant ′𝜅𝑒 ′, is placed between the point charges, then

the electrical force between two point charges decreases by a factor of ′𝜅𝑒 ′. The expression of electrical
force 𝐹𝑒 between two point charges 𝑞1 and 𝑞2 , when the dielectric medium is place between them, is
1 𝑞1 𝑞2
𝐹𝑒 =
4𝜋𝜖0 𝜅𝑒 𝑟 2

 The electrical field intensity 𝐸 due to a point charge between two point charges decreases, in the
presence of a dielectric medium. If ′𝜅𝑒 ′, is the dielectric constant of the corresponding dielectric
medium, then the electric field intensity at any point due to point charge will be
1 𝑞
𝐸=
4𝜋𝜖0 𝜅𝑒 𝑟2

 The electric field near the surface of charged conductor, which is immersed in a dielectric medium of
dielectric constant 𝜅𝑒 is
𝜎
𝐸=𝜖
0 𝜅𝑒

where 𝜎 is the uniform charge density of conductor.

Dielectrics: An Atomic View

Dielectrics are the insulating materials through which the electric current cannot pass easily, because

these materials have very high value of electrical resistance. For example, paper, pyrex, polystyrence,
transformer oil, pure water, silicon etc.
There are two types is dielectrics

(i) Polar Dielectrics

(ii) Non-Polar Dielectrics

Polar Dielectrics

The dielectric materials which have permanent electric dipoles moment are called polar dielectrics.

These materials consist of molecules which are permanent dipoles.

In the absence of external electric field, the polar molecules are randomly oriented. As the result, these
materials have no net dipole moment.

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When the external electric field is applied, then all dipoles tend to align themselves with external electric
field. But the thermal agitation tends to keep the dipoles randomly oriented. Hence the partial alignment of
electric dipoles is produced in a polar dielectric medium for specific electric field strength. However, the
alignment of dipoles can be increased by increasing the external electric field and decreasing the temperature.

Non-Polar Dielectrics

The dielectric materials which don't have permanent dipole moments are called non-polar

dielectrics. In the absence of external electric field, the centers of positive and negative charges coincide. When
the electric field of strength ′𝐸0 ′ is applied, then it tends to separate to positive and negative charges on the
atoms of molecules. As the result, the atoms and molecules of dielectric become dipoles, called induced dipoles
and this process is called electric polarization.

The gradient vector: grad

 A vector field, called the gradient, written: grad f or ∇f

can be associated with a scalar field f.

 At every point the direction of the vector field (∇f) is

 orthogonal to the scalar field contour and

 in the direction of the maximum rate of change of f.

The vector differential operator “∇”, called grade or nabla, is defined in three dimensions to be:

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𝜕 𝜕 𝜕
∇=i 𝜕𝑥
+j 𝜕𝑦
+k 𝜕𝑧

The gradient of a function, f (x, y), in two dimensions is defined as:

𝜕𝑓 𝜕𝑓
∇f = i 𝜕𝑥 + j 𝜕𝑦

The gradient of a function, f (x, y, z), in three dimensions is defined as:

𝜕𝑓 𝜕𝑓 𝜕𝑓
∇f = i 𝜕𝑥 + j 𝜕𝑦 + k 𝜕𝑧

Example- Calculate the the gradient of the function f (x, y) = x + y2 +z4.

Poission’s and Laplace’s Equations

Poission’s equation is an elliptic partial differential equation of broad utility in theoretical physics. For example, the
solution of poission’s equation is the potential field caused by a given electric charge or mass density distribution; with
the potential field known, one can calculate electrostatic or gravitational force. It is a generalization of Laplace’s
equation.

Obtaining Poission’s and Laplace’s equation is simple. From the point form of Gauss’s law,

∇. D=ρ …. …. …. (1)

The definition of D,

D = ϵE …. ….. …. (2)

And the gradient relationship,

E = - ∇V …. …. … (3)

By substitution we have,

∇ . D = ∇ . (ϵE )= - ∇ . (ϵ∇V) = ρ

ρ
Or, ∇ . ∇V = - ….. ….. ….. (4)
ϵ

For a homogeneous region for which ϵ is constant.

Equation (4) is Poission’s equation. In Cartesian coordinates,

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𝜕𝐴 𝜕𝐴 𝜕𝐴
∇. A= 𝜕𝑥
+ 𝜕𝑦
+ 𝜕𝑧

𝜕𝑉 𝜕𝑉 𝜕𝑉
∇V = i +j +k
𝜕𝑥 𝜕𝑦 𝜕𝑧

And therefore,

𝜕 𝜕𝑉 𝜕 𝜕𝑉 𝜕 𝜕𝑉
∇ . ∇V = ( )
𝜕𝑥 𝜕𝑥
+ ( )
𝜕𝑦 𝜕𝑦
+ ( )
𝜕𝑧 𝜕𝑧

𝜕 𝜕𝑉 𝜕 𝜕𝑉 𝜕 𝜕𝑉
∇ 2V = ( ) + ( ) + ( ) .... …. …. (5)
𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑧 𝜕𝑧

So the Poission’s equation is,

𝜕 𝜕𝑉 𝜕 𝜕𝑉 𝜕 𝜕𝑉 ρ
∇ 2V = ( )
𝜕𝑥 𝜕𝑥
+ ( ) + ( ) =-
𝜕𝑦 𝜕𝑦 𝜕𝑧 𝜕𝑧 ϵ

If ρ= 0, indicating zero volume charge density, but allowing point charges, line charge and surface charge density to
exist at singular location as sources of the field, then

∇ 2V = 0

Which is Laplace’s equation. The ∇ 2 operation is called the Laplacian of V. In Cartesian coordinates Laplace’s equation
is,

𝜕 𝜕𝑉 𝜕 𝜕𝑉 𝜕 𝜕𝑉
∇ 2V = ( )
𝜕𝑥 𝜕𝑥
+ ( ) + ( ) = 0.
𝜕𝑦 𝜕𝑦 𝜕𝑧 𝜕𝑧

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CURRENT AND RESISTANCE
In this chapter we shall study electric currents, i.e., of charges in motion. Example of electric currents
abound, ranging from large currents that constitute lightning strokes to the tiny nerve current that regulate
our muscular activity. We are familiar with currents resulting from charges flowing through solids conductors
(household wiring, light bulbs), semiconductors (integrated circuits), gases (fluorescent lamps), liquids
(automobile batteries), and even evacuated spaces (TV picture tubes)

ELECTRIC CURRENT

The time rate of flow of charge through a conductor is called current. If a charge 𝑑𝑞 flows through
any cross-section of a conductor in time 𝑑𝑡, then the current I is given by

𝑑𝑞
𝐼=
𝑑𝑡

The SI unit of current is Ampere, which can be defined as, “when one coulomb charge flows through
a cross-section in one second, then the current flowing is one ampere”.

Example- A current of 4.82A exist in a resistor for 4.6 minutes. (a) Find out charge, (b) How many electrons
pass through resistor in this time?

Solution. 𝐼 = 4.82𝐴

𝑡 = 4.6 𝑚𝑖𝑛 = 4.6 × 60 𝑠

𝑞 =?
𝑛 =?

𝑞
(a) As, 𝐼 = 𝑡

𝑞 = 𝐼𝑡 = 4.82 × 4.6 × 60 = 1.33 × 103 𝐶

(b) As, 𝑞 = 𝑛𝑒

𝑞 1.33 × 103
𝑛= = = 8.3 × 1021 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
𝑒 1.6 × 10−19

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DIRECTION OF CURRENT

In metals, the charge carriers are electrons. But in electrolytes, the current flow due to motion of

negative and positive ions. A positive charge moving in one direction is equivalent in all external effects to a
negative charge moving in the opposite the opposite direction. Hence for simplicity and algebraic consistency,
we adopt the following convention:
The direction of current is the direction that positive charges would move, even if the actual charge
carriers are negative. Thus, the direction of current is taken from the point of higher potential to the point of
lower potential.
Even though we assign a direction, current is a scalar quantity, not a vector. The arrow that we draw to
indicate the direction of current merely show the sense of charge flow through the wire and is not be taken as a
vector. Current does not obey the law of vector addition. Changing the direction of wires does not change the
way the currents are added.

OHM’S LAW

It states that

“The current flowing through a conductor is directly proportional to the applied potential difference
if all physical states remain same.”

If 𝑉 is the potential difference between the ends of conductor and 𝐼 is the current flowing through it, then the
Ohm’s law is described mathematically as:
𝐼∝𝑉
𝐼 = 𝐺𝑉
Where, G is the constant of proportionality, sometimes called ‘conductance’. It is related to the resistance by,
1
𝐺=
𝑅
So,
1
𝐼= 𝑉
𝑅
𝑜𝑟, 𝑉 = 𝐼𝑅

Where, 𝑅 is the resistance of the conductor. It is described as the opposition offered by conductor to the flow of
current. In system international, its unit is ohm. It is a macroscopic quantity. Its corresponding microscopic
quantity is resistivity.

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Kirchhoff’s Law

Kirchhoff’s First Law/Current law:

Kirchhoff’s current law states that:

‘The algebraic sum of the currents meeting at any junction in a circuit is zero’.

The convention is that, the current flowing towards a junction is positive and the current flowing away from
the junction is negative.

Let 𝐼1 , 𝐼2 , 𝐼3 , 𝐼4 𝑎𝑛𝑑 𝐼5 be the currents passing through the conductors

respectively. According to Kirchhoff’s first law.

𝐼1 + 𝐼2 + (−𝐼3 ) + (− 𝐼4 ) + (−𝐼5 ) = 0

𝑜𝑟 𝐼1 + 𝐼2 = 𝐼3 + 𝐼4 + 𝐼5

i.e., the sum of the currents entering the junction is equal to the sum of the currents leaving the junction.

Kirchhoff’s Second Law/Voltage law:

Kirchhoff’s Voltage rule is a particular way of stating the law of conservation of energy for a charge

carrier travelling in a closed circuit. The Kirchhoff’s second rule is described as:

“The algebraic sum of all differences in potential around a complete circuit loop is zero”.

Mathematically, ∑𝑉 = 0

[Our sign convention for applying signs to the voltage polarities in our KVL equations will be as follows: when traversing
the loop, if the negative terminal of a voltage difference is encountered before the positive terminal, the voltage difference
will be interpreted as positive in the KVL equation. If the positive terminal is encountered first, the voltage difference will
be interpreted as negative in the KVL equation. We use this sign convention for convenience; it is not required for proper
application of KVL, as long as the signs on the voltage differences are treated consistently.]

Let’s apply the Kirchhoff’s voltage law for the single circuit loop shown in
figure below:

Since the two resistors, R1 and R2 are wired together in a series connection,
they are both part of the same loop so the same current must flow through
each resistor.

Thus, the voltage drop across resistor, 𝑅1 = 𝐼𝑅1 and the voltage drop
across resistor, 𝑅2 = 𝐼𝑅2 . Now, by KVL:

𝑉𝑠 + (−𝐼𝑅1 ) + (−𝐼𝑅2 ) = 0

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∴ 𝑉𝑠 = 𝐼𝑅1 + 𝐼𝑅2

∴ 𝑉𝑠 = 𝐼(𝑅1 + 𝑅2 )

𝑉𝑠
∴𝐼=
𝑅1 + 𝑅2

This expression gives the current flow through a circuit containing single loop.

Example- Find the current through each battery in the circuit shown below.

Solution:

Applying Kirchhoff’s 1st law on junction b;

𝐼1 + 𝐼2 + 𝐼3 = 0

∴ 𝐼3 = −(𝐼1 + 𝐼2 )

Applying Kirchhoff’s 2nd law across the loop abcda;

9 − 18 𝐼1 + 3 𝐼3 = 0

. ⇒ 9 − 18 𝐼1 + 3 (−𝐼1 − 𝐼2 ) = 0

. ⇒ 21 𝐼1 + 3𝐼2 = 9 − − − −(𝑖)

Applying Kirchhoff’s 2nd law across the loop dbcd;

3 − 6 𝐼2 + 3 𝐼3 = 0

⇒ 3 − 6 𝐼2 + 3 (−𝐼1 − 𝐼2 ) = 0

⇒ 3 𝐼1 + 9 𝐼2 = 3 − − − (𝑖𝑖)

Now,

7 × (𝑖𝑖) − (𝑖) ⇒ 60 𝐼2 = 12

⇒ 𝐼2 = 0.2 𝐴

And,

3 × (𝑖) − (𝑖𝑖) ⇒ 60 𝐼1 = 24

⇒ 𝐼1 = 0.4 𝐴

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RC CIRCUIT (CHARGING OF A CAPACITOR)

A circuit containing a series combination of a resistor and a capacitor is called an 𝑅𝐶 circuit. Consider an 𝑅𝐶
circuit in series with the battery of emf 𝜀 as shown in figure below:

When the switch 𝑆 is closed , the capacitor starts charging. By using Kirchhoff’s 2nd rule, we get:

ℰ − 𝑉𝑅 − 𝑉𝐶 = 0

Where, 𝑉𝑅 → 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑡ℎ𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟 𝑎𝑛𝑑 𝑉𝐶 → 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑡ℎ𝑒 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟.
𝑞
⇒ ℰ = 𝑖𝑅 + − − − − − (1)
𝐶
𝑑𝑞
Now, as 𝑖 = 𝑑𝑡
, therefore, equation (1) becomes,

𝑑𝑞 𝑞
ℰ= 𝑅+
𝑑𝑡 𝑐

𝑑𝑞
⇒ ℰ𝐶 = 𝑅𝐶 + 𝑞
𝑑𝑡
𝑑𝑞
⇒ 𝑅𝐶 = ℰ𝐶 − 𝑞
𝑑𝑡

𝑑𝑞 1
⇒ = 𝑑𝑡
ℰ𝐶 − 𝑞 𝑅𝐶

Integrating, we get;

𝑑𝑞 𝑑𝑡
∫ =∫
ℰ𝐶 − 𝑞 𝑅𝐶

𝑡
⇒ − ln(ℰ𝐶 − 𝑞) = + 𝐴 − − − (2)
𝑅𝐶

Where 𝐴 is the constant of integration. To find 𝐴, we make use of initial conditions.

At 𝑡 = 0; 𝑞 = 0, we have from (2),

− ln(ℰ𝐶) = 𝐴

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Now the equation (2) will become,

𝑡
− ln(ℰ𝐶 − 𝑞) = − ln(ℰ𝐶)
𝑅𝐶
𝑡
⇒ ln(ℰ𝐶 − 𝑞) − ln(ℰ𝐶) = −
𝑅𝐶
ℰ𝐶 − 𝑞 𝑡
⇒ ln ( )=−
ℰ𝐶 𝑅𝐶
ℰ𝐶 − 𝑞 𝑡
⇒ = 𝑒 −𝑅𝐶
ℰ𝐶
𝑡
⇒ ℰ𝐶 − 𝑞 = ℰ𝐶𝑒 −𝑅𝐶
𝑡
⇒ 𝑞 = ℰ𝐶 − ℰ𝐶𝑒 −𝑅𝐶
𝑡
⇒ 𝑞 = ℰ𝐶 (1 − 𝑒 −𝑅𝐶 )

This equation shows that at 𝑡 = ∞; 𝑞 = ℰ𝐶 = 𝑞0 , where 𝑞0 is the

maximum value of charge on the capacitor. Therefore,
𝑡
𝑞 = 𝑞0 (1 − 𝑒 −𝑅𝐶 ) − − − −(3)

This equation gives the growth of charge in an 𝑅𝐶 circuit. The

equation shows that the charge 𝑞 goes on increasing and ultimately
attains the maximum value 𝑞0 after a long time.

Capacitive Time Constant:

In the equation (3), the factor 𝑅𝐶 has the dimensions of time and is called capacitive time constant. It is denoted
by 𝜏𝑐 .

Hence, the equation (3) is written as:

𝑡
−
𝑞 = 𝑞0 (1 − 𝑒 𝜏𝑐 )

Special Cases:

 At 𝑡 = 0; 𝑞=0

 At 𝑡 = ∞; 𝑞 = 𝑞0 = ℰ𝐶
𝜏
− 𝑐
 At 𝑡 = 𝜏𝑐 ; 𝑞 = 𝑞0 (1 − 𝑒 𝜏𝑐 )

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−1 ) 1
. ⇒ 𝑞 = 𝑞0 (1 − 𝑒 = 𝑞0 (1 − 𝑒
) = 𝑞0 (1 − 0.37)

. ⇒ 𝑞 = 0.63 𝑞0

So, the capacitive time constant is the time after which the charge on the capacitor grows to 63% of its maximum
value.
Example- A 𝟏𝟓𝟎𝑭 capacitor is connected through a 𝟓𝟎𝟎 resistor to 𝟒𝟎𝑽 battery. (a) What is the time
constant of the circuit? (b) What is the final charge 𝒒𝟎 on a capacitor plate? (c) How long does it take for
the charge on a capacitor plate to reach 𝟎. 𝟖𝒒𝟎 ?

Solution:
(a) 𝜏𝑐 = 𝑅𝐶 = 500 × 150 × 10−6 = 75 × 10−3 𝑠𝑒𝑐 = 75 𝑚𝑠
(b) 𝑞0 = ℰ𝐶 = 40 × 150 × 10−6 = 6 × 10−3 = 6 𝑚𝐶
(c) 𝑞 = 𝑞0 (1 − 𝑒 −𝑡/𝑅𝐶 )
𝑡
or, 0.8 𝑞0 = 𝑞0 (1 − 𝑒 −𝑅𝐶 )

𝑡
or, 0.8 = 1 − 𝑒 −𝑅𝐶
𝑡
or, 𝑒 −𝑅𝐶 = 0.2
𝑡
or, − = ln(0.2)
𝑅𝐶

or, −𝑡 = −1.6 × 𝑅𝐶

or, 𝑡 = 1.6 × 75 × 10−3

∴ 𝑡 = 120 × 10−3 sec = 120 𝑚𝑠

Example- A resistor 𝑹 = 𝟔. 𝟐𝑴𝛀 and a capacitor 𝑪 = 𝟐. 𝟒𝝁𝑭 are connected in series and a 12V battery of
negligible internal resistance is connected across their combination. (a) What is capacitive time constant of
this circuit. (b) At what time after the battery is connected does the potential difference across the capacitor
equal to 5.6 V.
Solution:

𝑅 = 6.2𝑀Ω = 6.2 × 106 Ω

𝐶 = 2.4 𝜇𝐹 = 2.4 × 10−6 𝐹

ℰ = 12 𝑉

(a) 𝜏𝑐 = 𝑅𝐶 = 6.2 × 106 × 2.4 × 10−6 = 15 𝑠𝑒𝑐

𝑞
(b) Potential across capacitor, 𝑉𝑐 = 𝐶 = 5.6 𝑉

As,

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𝑡
𝑞 = ℰ𝐶 (1 − 𝑒 −𝑅𝐶 )

𝑞 𝑡
−
⇒ = ℰ (1 − 𝑒 𝜏𝑐 )
𝐶

𝑡
−
⇒ 𝑉𝑐 = ℰ (1 − 𝑒 𝜏𝑐 )

𝑉𝑐 𝑡
−
⇒ = 1 − 𝑒 𝜏𝑐
ℰ
𝑡 𝑉𝑐
−
⇒𝑒 𝜏𝑐 =1−
ℰ
𝑡 𝑉𝑐
⇒− = ln (1 − )
𝜏𝑐 ℰ

𝑉𝑐
⇒ 𝑡 = −𝜏𝑐 ln (1 − )
ℰ

5.6
∴ 𝑡 = −15 ln (1 − ) = 9.4 𝑠𝑒𝑐
12

DECAY OF CHARGE IN AN RC CIRCUIT (DISCHARGING)

Consider the circuit which consists of a capacitor carrying an initial charge 𝑞, a resistor 𝑅, and a switch 𝑆 as
shown in figure below:

𝑞
When the switch is open, a potential difference 𝑉𝑐 = exists across the capacitor and 𝑉𝑅 = 0 because 𝑖 = 0. If
𝐶

the switch is closed at 𝑡 = 0, the capacitor begins to discharge through the resistor. Applying Kirchhoff’s rule:

−𝑉𝑅 − 𝑉𝐶 = 0
𝑞
⇒ 𝑖𝑅 = −
𝐶
𝑑𝑞 𝑞
⇒ =−
𝑑𝑡 𝑅𝐶
𝑑𝑞 1
⇒ =− 𝑑𝑡
𝑞 𝑅𝐶

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Integrating both sides, we obtain

𝑑𝑞 𝑑𝑡
∫ = −∫
𝑞 𝑅𝐶

𝑡
⇒ ln 𝑞 = − + 𝐴 − − − −(1)
𝑅𝐶

By applying the initial conditions, 𝑎𝑡 𝑡 = 0; 𝑞 = 𝑞0 , 𝑤𝑒 𝑔𝑒𝑡 𝑓𝑟𝑜𝑚 (1)

𝐴 = ln 𝑞0

The equation (1) implies:

𝑡
ln 𝑞 = − + ln 𝑞0
𝑅𝐶
𝑞 𝑡
⇒ ln =−
𝑞0 𝑅𝐶

𝑞 𝑡
⇒ = 𝑒 −𝑅𝐶
𝑞0
𝑡
∴ 𝑞 = 𝑞0 𝑒 −𝑅𝐶 − − − (2)

This is the expression for the decay of charge of capacitor in an 𝑅𝐶 circuit.

Capacitive Time Constant:

The factor 𝑅𝐶 = 𝜏𝑐 is called the capacitive time constant. The equation (2) will become:
𝑡
−
𝑞 = 𝑞0 𝑒 𝜏𝑐

Special Cases:

 At 𝑡 = 0; 𝑞 = 𝑞0

 At 𝑡 = ∞; 𝑞=0
𝜏
− 𝑐
 At 𝑡 = 𝜏𝑐 ; 𝑞 = 𝑞0 𝑒 𝜏𝑐 = 𝑞0 𝑒 −1 = 0.37 𝑞0

So, after time 𝝉𝒄 , the charge of capacitor reduces to 𝟑𝟕% of the initial value.

Example- A capacitor 𝑪 discharge through a resistor 𝑹. (a) After how many time constant does its charge
fall to one half of its initial value?

Solution: As we know,
𝑡
𝑞 = 𝑞0 𝑒 −𝜏
𝑞0
As we want to find out the time in which 𝑞 = 2
. Therefore,

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𝑞0 𝑡
= 𝑞0 𝑒 −𝜏
2
1 𝑡
⇒ = 𝑒 −𝜏
2
𝑡 1
⇒ − = ln ( )
𝜏 2
𝑡
⇒ = ln(2)
𝜏

⇒ 𝑡 = 𝜏 ln(2) = 0.693 𝜏

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Energy Stored in an Electric Field

Consider a capacitor with the capacitance C, which is connected to the battery of emf V. If dq
charge is transferred from one plate to other, then the work done dW will be,

𝑑𝑊 = 𝑉𝑑𝑞

This work done is stored in the form of electric potential energy dU

𝑑𝑈 = 𝑉𝑑𝑞

When the capacitor is fully charged then the total energy stored is,

𝑈 = ∫ 𝑑𝑈 = ∫ 𝑉𝑑𝑞
0

𝑞
As, 𝑞 = 𝐶𝑉; ⇒ 𝑉 =
𝐶

𝑞
𝑞
𝑈 = ∫ 𝑑𝑞
𝐶
0

𝑞
1
𝑈 = ∫ 𝑞𝑑𝑞
𝐶
0

𝑞
1 𝑞2
𝑈= [ ]
𝐶 2 0

1 𝑞2
∴𝑈= ( ) − − − − − (1)
2 𝐶

Again as, 𝑞 = 𝐶𝑉, we can write,

1 𝐶2𝑉2
𝑈= ( )
2 𝐶

1
𝑈 = 𝐶𝑉 2
2

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The energy stored in the capacitor is the energy store in the electric field between its plates. So, the
energy stored can be expressed in terms of electric filed strength 𝐸.

𝜖0 𝐴
As, 𝐸 = 𝑉𝑑 and 𝐶 = 𝑑
, therefore,

1 𝜖0 𝐴
𝑈= ( ) (𝐸 2 𝑑2 )
2 𝑑

1
𝑈 = 𝜖0 𝐸 2 𝐴𝑑
2

This is the expression of energy stored in the electric filed between the plates of capacitor.

Energy Density:

The energy density ′𝑢′ is described as the energy stored ′𝑈′ per unit volume ′𝑉′.

Mathematically,

1
𝑈 𝑈 (2 𝜖0 𝐸 2 𝐴𝑑)
𝑢= = =
𝑉 𝐴𝑑 𝐴𝑑

1
∴ 𝑢 = 𝜖0 𝐸 2
2

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Magnetism
Magnet:

The material which has directional, attractive and repulsive properties and the material by which magnetic
properties can be given to other materials are called magnets. Its chemical formula is Fe3O4.

Magnetism:

The directional and attractive property of a magnet is called its magnetism.

Magnetic field:

The region around a magnet up to which it can show its magnetic properties is called magnetic field.

Curie point or temperature a magnet:

A particular temperature at which the magnetism of a magnet is destroyed completely is called the Curie
point.

Magnetic permeability of a medium:

The characteristic of a material for which magnetic field lines can pass more easily through a material
compared to another material is called magnetic permeability.

In other word, the ratio of the magnetic flux density created in a medium to the magnetic intensity is called
absolute permeability of that medium. It is denoted by μ and its unit is Henry/meter (H/m).

𝑀𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑓𝑙𝑢𝑥 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝐵

i.e. 𝑀𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑝𝑒𝑟𝑚𝑖𝑎𝑏𝑖𝑙𝑖𝑡𝑦 = 𝑀𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦
;∴ 𝜇 = 𝐻.

Magnetic induction:

The process by which a magnetic material becomes a magnet is called magnetic induction.

Magnetic materials and Types of them:

The materials which can be attracted and which can be transformed into magnets are called magnetic
materials. For example, Iron, nickel, cobalt etc.

There are three types of magnetic materials those are given below,

(i) Diamagnetic substance (μ <1); Gold, Copper etc.

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(ii) Paramagnetic substance (μ >1); Platinum, Manganese etc.

(iii) Ferromagnetic substance (μ >>1); Iron, Cobalt etc.

MAGNETIC FIELD EFFECTS

Magnetic Field

Magnetic field is the region or space around any charge within which its influence can be felt by
other magnetic substances.

The magnetic field around any magnet is considered as closely spaced magnetic field lines. The
magnetic field lines of a bar magnet can be traced with the aid of a compass as shown in the figure below:

In addition to a bar magnet, a moving charge or a current creates a magnetic field in the surrounding
space (in addition to its electric field). The magnetic field exerts a force on any other moving charge or
current that is present in the field.

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Biot-Savart’s law:

Statement: When current flows through a conductor, magnetic field is generated around it. According to
this law, the magnitude of magnetic field produced around a conductor of small length due to the flow of
current through it is –

 Directly proportional to the amount of current,

 Directly proportional to the length of the conductor,
 Directly proportional to the sine of the angle between the
lines connecting the point from the mid-point of the
conductor and the conductor itself,
 Inversely proportional to the square of thedistance
between the point and the mid-point of the conductor.

Explanation:

According to this law, the magnetic flux density 𝑑𝐵 at a point 𝑃 due to a small element 𝑑𝑙 of a conductor
carrying a current 𝐼is given by,

𝐼 𝑑𝑙 𝑠𝑖𝑛𝜃
𝑑𝐵 ∝ … … … … … (1)
𝑟2

Where, 𝑟 is the distance from the point 𝑃 to the mid-point of the element 𝑑𝑙 and 𝜃 is the angle between the
element and the line joining it to 𝑃as shown in Fig-3.

From equation (1) we can write,

𝐼 𝑑𝑙 𝑠𝑖𝑛𝜃
𝑑𝐵 = 𝐾 … … … … . (2)
𝑟2

Where, 𝐾 is a proportionality constant whose value depends on nature of the medium and the unit of the
quantities.

For air medium and in S.I. unit

𝜇0
𝐾=
4𝜋

Where, 𝜇0 = 4𝜋 × 10−7 H/m.

So, putting this value in (2) we get,

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𝜇0 𝐼. 𝑑𝑙. 𝑠𝑖𝑛𝜃
𝑑𝐵 =
4𝜋 𝑟2

For other medium we get,

𝜇 𝐼. 𝑑𝑙. 𝑠𝑖𝑛𝜃
𝑑𝐵 =
4𝜋 𝑟2

where, 𝜇is the permeability of that medium.

FARADAY’S LAW OF ELECTROMAGNETIC INDUCTION

Magnetic Flux

The number of magnetic lines of force passing normally through certain area is called magnetic
flux. It is denoted by 𝜙𝐵 . It is a scalar quantity and its SI unit is weber (Wb). It is measured by the product
of magnetic field strength and the component of vector area parallel to magnetic field.

If 𝑑𝑨 is the vector area element of the surface placed in uniform magnetic field of magnetic field
strength B as shown figure below.

The magnetic flux𝑑𝜙𝐵 through𝑑𝑨is given by:

𝑑𝜙𝐵 = 𝑩. 𝑑𝑨

𝜙𝐵 = ∫ 𝑩. 𝑑𝑨

𝜙𝐵 = ∫ 𝐵𝑑𝐴 cos𝜃

Where 𝜃is the angle between magnetic field strength and vector the area element.

9.2 Faraday’s Law of ElectromagneticInduction

Statement:

The induced 𝑒𝑚𝑓 in a circuit is equal to the negative of rate at which the magnetic flux through
the circuit is changing with time.

Or

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The magnitude of induced 𝑒𝑚𝑓 in a circuit is directlyproportional to the rate of change of magnetic
flux.

*Explanation

When a magnet is moved toward the loop, the

ammeter needle deflects in one direction, as shown in the
figure (a). When the magnet is brought to rest and held
stationary relative to the loop figure (b), no deflection is
observed. When the magnet is moved away from the
loop, the needle deflects in the opposite direction, as
shown in figure

(c). Finally, if the magnet is held stationary and

the loop is moved either toward or away from it, the
needle deflects.

From these observations, we conclude that the loop

detects that the magnet is moving relative to it and we
relate this detection to a change in magnetic field. Thus,
it seems that a relationship exists between current and
changing magnetic field.

These results are quite remarkable in view of the fact that a current is set up even though no batteries are
present in the circuit. We call such a current the induced current which is produced by an induced emf. This
phenomenon is called electromagnetic induction.

Mathematical form of Faraday’s Law

If ℰ is the induced emf due to change in flux 𝑑𝜙𝐵 in time 𝑑𝑡, then we can describe the Faraday’s law of
electromagnetic induction as:

𝑑𝜙𝐵
ℰ= −
𝑑𝑡

If there are 𝑁 number of turns in a coil, then,

𝑑(𝑁𝜙𝐵 )
ℰ=−
𝑑𝑡

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𝑑𝜙𝐵
ℰ = −𝑁
𝑑𝑡

9.3 Lenz law

The direction of induced current is such that it opposes its own cause.

Explanation:

Consider a bar magnet moves toward a stationary metal loop, as in figure (a). As the magnet moves
to the right toward the loop, the external magnetic flux through the loop increases with time. As the result,
the induced current set up in the loop which produces magnetic
field, as illustrated in figure (b). Knowing that like magnetic poles
repel each other, we conclude that the left face of the current loop
acts like a north pole and that the right face acts like a south pole.

If the magnet moves to the left, as in figure (c), its

flux through the area enclosed by the loop decreases in time. Now
the induced current in the loop produces the magnetic field as
shown in figure (d). In this case, the left face of the loop is a south
pole and the right face is a north pole.

9.3.1 Lenz Law and Conservation of Energy

“Lenz law is the statement of law of conservation of energy

for the circuit involving induced current”

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To understand this statement, consider a conducting bar moving to the right on two parallel rails
in the presence of a uniform magnetic field as shown in the figure below. As the bar moves to the right,
the magnetic flux through the area enclosed by the circuit increases with time because the area increases.
As the result, the induced current must be directed counterclockwise when the bar moves to the right.
Since the current carrying bar is moving in the magnetic field, it will experience a magnetic force. By
using right hand rule, the direction of is opposite to that of, that tends to stop the rod. An external
dragging force must be applied to keep the rod moving in the magnetic field.

This dragging force provides the energy for the induced currents to flow. This energy is the
source of induced current. Thus, the electromagnetic induction is exactly according to the law of
conservation of energy.

If the bar is moving to the left, as in figure (b), the external magnetic flux through the area
enclosed by the loop decreases with time. Because the field is directed into the page, the direction of the
induced current must be clockwise.

9.4 Self-Induction

The property of a coil which enables to produce an opposing induced 𝑒𝑚𝑓 in it when the current in the
coil changes is called self-induction.

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Coefficient of self induction

When a current I flows through a coil, the magnetic flux (φ) linked with the coil is proportional to the
current.
𝜙∝𝐼
𝜙 = 𝐿𝐼
Where L is a constant of proportionality and is called coefficient of self induction or self inductance.
From Faraday’s law of induction.
𝑑𝜙
ℰ=−
𝑑𝑡
So,
𝑑(𝐿𝐼)
ℰ=−
𝑑𝑡
𝑑𝐼
ℰ = −𝐿
𝑑𝑡
If there are N no. of turns in the coil then,
𝑑𝐼
ℰ = −𝑁𝐿
𝑑𝑡
The unit of self inductance is henry.

9.5 Mutual induction

The phenomenon of producing an induced emf in a coil due to the change in current in the other coil is
known as mutual induction.

Coefficient of mutual induction

When a current I flows through a coil, the magnetic flux (φ) linked with the other coil is proportional to the
current.

Figure: Mutual Induction

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𝜙∝𝐼
𝜙 = 𝑀𝐼
Where M is a constant of proportionality and is called coefficient of mutual induction or mutual inductance.
From Faraday’s law of induction.
𝑑𝜙
ℰ=−
𝑑𝑡
So,
𝑑(𝑀𝐼)
ℰ=−
𝑑𝑡
𝑑𝐼
ℰ = −𝑀
𝑑𝑡
If there are N no. of turns in the coil then,
𝑑𝐼
ℰ = −𝑁𝑀
𝑑𝑡
The unit of mutual inductance is henry.
10.1 Energy Stored in the Magnetic Field

Consider a resistor and an inductor is connected in series with a battery of emfℰ. According to the
Kirchhoff‟s 2nd rule;

ℰ − 𝑉𝑅 − 𝑉𝐿 = 0

ℰ = 𝑉𝑅 + 𝑉𝐿

𝑑𝐼
ℰ = 𝐼𝑅 + 𝐿
𝑑𝑡

Multiplying both sides by 𝐼, we have:

𝑑𝐼
ℰ𝐼 = 𝐼 2 𝑅 + 𝐿𝐼
𝑑𝑡
𝑑𝐼
Hereℰ𝐼is the power supplied by the source,𝐼 2 𝑅 is the power dissipated in resistance and𝐿𝐼 𝑑𝑡is the energy
supplied per time in the inductance coil, where the magnetic field also exists.

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If 𝑈𝐵 is the total energy stored in the magnetic field of inductance coil, then the energy stored per
unit time is:

𝑑𝑈𝐵 𝑑𝐼
= 𝐿𝐼
𝑑𝑡 𝑑𝑡

𝑑𝑈𝐵 = 𝐿𝐼 𝑑𝐼

Integrating both sides, we get:

𝑈𝐵 𝐼

∫ 𝑑𝑈𝐵 = ∫ 𝐿𝐼 𝑑𝐼
0 0

𝐼
𝐼2
𝑈𝐵 = 𝐿 [ ]
2 0

1
𝑈𝐵 = 𝐿𝐼 2
2

This is the expression of energy stored in the magnetic field of a current carrying inductor.

Maxwell’s Equations

Maxwell’s equations describe all (classical) electromagnetic phenomena:

𝜕𝑩
∇×E=
𝜕𝑡

𝜕𝑫
∇×H=J+
𝜕𝑡

∇. D=ρ

∇. B=0

The first is Faraday’s law of induction, the second is Amp`ere’s law as amended by Maxwell to include the
displacement current ∂D/∂t, the third and fourth are Gauss’ laws for the electric and magnetic fields.

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The displacement current term ∂D/∂t in Amp`ere’s law is essential in predicting the existence of
propagating electromagnetic waves. Its role in establishing charge conservation is discussed in Sec. 1.7.
Eqs. (1.1.1) are in SI units. The quantities E and H are the electric and magnetic field intensities and are
measured in units of [volt/m] and [ampere/m], respectively. The quantities D and B are the electric and
magnetic flux densities and are in units of [coulomb/m2] and [weber/m2], or [tesla]. D is also called the
electric displacement, and B, the magnetic induction. The quantities ρ and J are the volume charge density
and electric current density (charge flux) of any external charges (that is, not including any induced
polarization charges and currents.) They are measured in units of [coulomb/m3] and [ampere/m2]. The
right-hand side of the fourth equation is zero because there are no magnetic monopole charges. Eqs.
(1.3.17)–(1.3.19) display the induced polarization terms explicitly. The charge and current densities ρ, J
may be thought of as the sources of the electromagnetic fields. For wave propagation problems, these
densities are localized in space; for example, they are restricted to flow on an antenna. The generated
electric and magnetic fields are radiated away from these sources and can propagate to large distances to
the receiving antennas. Away from the sources, that is, in source-free regions of space, Maxwell’s equations
take the simpler form:

𝜕𝑩
∇×E=
𝜕𝑡

𝜕𝑫
∇×H=
𝜕𝑡

∇. D=0

∇. B=0

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