Lecture Sheet-1 Electricity & Magnetism

Electricity & Magnetism

We are surrounded by devices that depend on the physics of electromagnetism, which is the
combination of electric and magnetic phenomena. When an ebonite rod is rubbed with fur
(cat’s hair) or cat skin, it gains a property due to which it attracts some other bodies like piece
of paper. We know that the attraction between ebonite and paper is due to an electric force. If
a certain type of stone (a naturally occurring magnet) is brought near bits of iron, the iron
will jump to the stone. We know that the attraction between magnet and iron is due to a
magnetic force.

Electrostatics
Electrostatics is the physics of stationary electric charge (i.e. charge at rest).

Concept of electric charge

Electric charge is a property that comes with particles wherever they exist. The vast amount
of charge in an everyday object is usually hidden because the object contains equal amounts
of the two kinds of charge: positive charge and negative charge. With an equality or balance
of charge, the object is said to be electrically neutral; that is, it contains no net charge.

We say that an object is charged to indicate that it has a charge imbalance, or net charge. The
imbalance is always much smaller than the total amounts of positive charge and negative
charge contained in the object. Charges with the same electrical sign repel each other, and
charges with opposite electrical signs attract each other.

Quantization of electric charge

Molecules, in general, are larger in size than atoms. A molecule usually contains a few atoms
of either the same element or of different elements. The fundamental charge on the electron
(or proton) is 1.6 × 10−19 𝑐𝑜𝑢𝑙𝑜𝑚𝑏 denoted by e. Thus the charge on the proton is +e and on
the electron is –e.

Since all matter in nature consists of an integral number of fundamental particles (atoms or
molecules), the total charge q in any matter is integral multiple of e, i.e.,𝑞 = 𝑛𝑒. In other
words, we can only have charges ±𝑒, ±2𝑒, ±3𝑒, ⋯ ⋯ , ±𝑛𝑒. This is known as principle of
quantization of charge.
Coulomb’s law
Statement: The magnitude of the force between two point charges is directly proportional to
the product of the magnitude of the charges and is inversely proportional to the square of the
distance between them and acts on the line joining between the charges.
Explanation: If 𝑞1 and 𝑞2 are the magnitude of two point charges and r is the distance
between them, then the force
𝑞1 𝑞2
𝐹∝ 2
𝑟
𝑞1 𝑞2 Fig. 1.
∴ 𝐹=𝑘 2 (1)
𝑟

Md. Matiur Rahman, Sr. Lecturer, Dept. of Medical Physics, KYAU Page: 1
Lecture Sheet-1 Electricity & Magnetism

The value of constant k depends on the choice of units of 𝐹, 𝑟 and 𝑞 and the nature of the
medium in which the charges are situated.
When expressed in M.K.S. system, the Coulomb’s law takes the form
1 𝑞1 𝑞2
𝐹= (2)
4𝜋𝜀0 𝑟 2
Where, 𝜀0 is the permittivity of the free space.

In vector forms
1 𝑞1 𝑞2 Fig. 2.
𝐹⃗ = 𝑟⃗⃗⃗ (3)
4𝜋𝜀0 𝑟 3
For radius vectors of charges q1 and q2 with respect to a fixed origin O

𝑞1 𝑞2 ⃗⃗⃗⃗
𝑟2 − ⃗⃗⃗⃗
𝑟1
𝐹⃗ = (4)
𝑟1 3
4𝜋𝜀0 |𝑟⃗⃗⃗⃗2 − ⃗⃗⃗⃗|

Limitations of Coulomb’s law: Coulomb’s law is strictly applicable to charges at rest. When
a charge is in motion forces other than electrostatic forces come into play and Coulomb’s law
cannot be used to measure the charges in motion.

Unit of charge: The S.I. unit of charge is Coulomb. It is defined from the basic unit of
current in S.I. i.e., an ampere.
A Coulomb is defined as the amount of charge that flows through a given cross-section of a
wire in one second if there is a steady current of one ampere in the wire, i.e.,

𝑞 = 𝐼𝑡

If 𝐼 = 1 𝑎𝑚𝑝 𝑎𝑛𝑑 𝑡 = 1 𝑠𝑒𝑐 then 𝑞 = 1 𝐶𝑜𝑢𝑙

Relative permittivity: Relative permittivity of a medium is denoted by 𝜀𝑟 and is defined as,

‘the ratio of the permittivity of the medium to the permittivity of the free space’.
Mathematically
𝜀
𝜀𝑟 =
𝜀0
Where, 𝜀 is the permittivity of the medium
𝜀0 is the permittivity of the free space.

Md. Matiur Rahman, Sr. Lecturer, Dept. of Medical Physics, KYAU Page: 2
Lecture Sheet-1 Electricity & Magnetism

Mathematical problems
Problem-1: The distance r between the electron and the proton in the hydrogen atom is
about 5.3 × 10−11 𝑚𝑒𝑡𝑒𝑟? What are the magnitudes of
(i) The electric force
(ii) The gravitational force between these two particles.
Solution: From Coulomb’s law,
Here,
1 𝑞1 𝑞2
𝐹𝑒 = 𝑞1 = 𝑞2 = 1.6 × 10−19 𝐶
4𝜋𝜀0 𝑟 2
𝑟 = 5.3 × 10−11 𝑚
9 × 109 × (1.6 × 10−19 )2
𝑂𝑟 𝐹𝑒 = 1
(5.3 × 10−11 )2 = 9 × 109 𝑁 − 𝑚2 /𝐶 2
4𝜋𝜀0
∴ 𝐹𝑒 = 8.1 × 10−8 𝑁 (𝑨𝒏𝒔)
𝐺 = 6.67 × 10−11 𝑁𝑚2 𝑘𝑔−2
The gravitational force is given by
𝑚1 = 9.1 × 10−31 𝑘𝑔
𝑚1 𝑚2
𝐹𝑔 = 𝐺 𝑚2 = 1.7 × 10−27 𝑘𝑔
𝑟2
𝐹𝑒 = ?
6.67 × 10−11 × 9.1 × 10−31 × 1.7 × 10−27
𝑂𝑟 𝐹𝑔 = 𝐹𝑔 = ?
(5.3 × 10−11 )2

∴ 𝐹𝑔 = 3.68 × 10−47 𝑁 (𝑨𝒏𝒔)

Problem-2: Calculate the repulsive Coulomb force exist between two protons in a nucleus of
iron. Assume a separation of 4 × 10−15 𝑚.
Solution: From Coulomb’s law, Here,
1 𝑞1 𝑞2
𝐹= 𝑞1 = 𝑞2 = 1.6 × 10−19 𝐶
4𝜋𝜀0 𝑟 2
𝑟 = 4.0 × 10−15 𝑚
9 × 109 × (1.6 × 10−19 )2
𝑂𝑟 𝐹= 1
(4.0 × 10−15 )2 = 9 × 109 𝑁 − 𝑚2 /𝐶 2
4𝜋𝜀0
∴ 𝐹 = 14 𝑁 (𝑨𝒏𝒔)
𝐹 =?
Problem-3: A negative charge of 10-6 is situated in air at the origin of a rectangular co-
ordinate system. A second negative point charge of 10-4 is situated on the positive X-axis at a
distance of 50 cm from the origin. What is the force on the second charge?
Solution: From Coulomb’s law, Here,
1 𝑞1 𝑞2
𝐹= 𝑞1 = −1 × 10−6 𝐶
4𝜋𝜀0 𝑟 2
9 −6 ) −4 )
𝑞2 = −1 × 10−4 𝐶
9 × 10 × (−1 × 10 × (−1 × 10
𝑂𝑟 𝐹= 2 𝑟 = 50 𝑐𝑚 = 0.5 𝑚
(0.5 )
1
∴ 𝐹 = 3.6 𝑁 (𝑨𝒏𝒔) = 9 × 109 𝑁 − 𝑚2 /𝐶 2
4𝜋𝜀0
Thus there is a force of 3.6 N in the positive X-direction 𝐹 =?
on the second charge. (Ans)

Md. Matiur Rahman, Sr. Lecturer, Dept. of Medical Physics, KYAU Page: 3