**Coulomb’s Law:**

#### Statement:

- The force of attraction or repulsion between two electric point charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between the two charges.
- The force acts along the line joining the two charges.

#### Explanation:

- Let q
_{1}and q_{2}be the two point charges separated by distance ‘r’ in a medium with dielectric constant ‘k’ Then,

F ∝ q_{1} . q_{2 } ……….. (1)

F ∝ 1/r²_{ }…………(2)

From statement (1) & (2) we get

Where F = force between the two charges,

∈_{o }= electrical permittivity of free space. = 8.85 × 10^{-12} C² N^{-1}m^{-2}.

k = dielectric constant of the medium.

This expression is called as a mathematical statement of Coulomb’s Law.

The value of the constant

Hence the equation of Coulomb’s law can be written as

**Coulomb’s Law in a Vector Form:**

Where F _{21} = Force on the second charge due to the first charge.

r _{12} = Unit vector from the first charge due to the second charge

Where F _{12} = Force on the second charge due to the first charge.

r _{21} = Unit vector from the first charge due to the second charge

as r _{12} = – r _{21}

F _{21} = – F _{12}

Thus the forces are equal but oppositely directed.

#### Limitations of Coulomb’s law:

Coulomb’s law is applicable to point charges only and is not applicable to extended bodies.

It can be applied to extended bodies by assuming them to be point charges but the distance between them should be sufficiently large,

#### Note:

- The charges possessing same nature of charge are called like charges. Like charges either all positive or all negative. Like charges always repel each other.
- The charges possessing opposite nature of charge are called, unlike charges. Unlike charges always repel each other.

#### Dimensions of Electrical Permittivity:

Hence dimensions of universal electrical permittivity are [L^{-3}M^{-1}T^{4}I²]

#### Definition of Coulomb (Using Coulomb’s Law):

By Coulomb’s Law

If q_{1} = q_{2} = q, r = 1 m and F = 9 × 10^{9} N, k =1, then

q² = 1 and q = ± 1 C

- The charge of 1 coulomb is that charge which, when placed in free space at a distance of 1 metre from an equal and similar charge, repels it with a force of 9 × 10
^{9}N

#### Definition of Coulomb (Using Electric Current):

The electric current flowing through any section of a conductor is given by

I = q/t

Where q = electric charge flowing through any section of the conductor

t = time for which the charge is flowing

q = I × t

unit of charge = unit of a current × unit of time

1 coulomb = 1 ampere × 1 second

- Hence a charge of 1 coulomb is defined as the quantity of a charge which flows per second through any cross-section of a conductor when there is a steady current of 1 ampere in the conductor.

**Dielectric Constant of Medium:**

- The ratio of electrical permittivity of the medium to the electrical permittivity of free space is called the dielectric constant of the medium.

k = ∈ / ∈_{o }

- The dielectric constant of medium is denoted by letter ‘k’. As it is a pure ratio it has no unit. For air or vacuum k = 1 for any other medium k > 1.

#### Example – 1:

- The distance between electron and proton in a hydrogen atom is 5.3 × 10
^{-11}m. What is the magnitude of the electric force between them? **Solution:****Given:**Distance between proton and electron = r = 5.3 × 10^{-11 }m, The magnitude of the charge on proton = q_{1}= 1.6 × 10^{-19 }C, Magnitude of the charge on electron = q_{2}= 1.6 × 10^{-19 }C.**To find:**Force = ?

By Coulomb’s law

**Ans:** The electric force between electron and proton is 8.202 × 10^{-8} N.

#### Example – 2:

- Two charges of magnitudes 1 μC and 2 μC are separated by a medium of dielectric constant 2. What is the magnitude of the electric force between them?
**Solution:****Given:**Distance between proton and electron = r = 10 cm = 0.1 m, The magnitude of the first charge = q_{1}= 1 μC = 1 × 10^{-6 }C, Magnitude of the charge on second charge = q_{2 }= 2 μC = 2 × 10^{-6 }C, dielectric constant of medium = k = 2.**To find:**Force = ?

By Coulomb’s law

**Ans:** The electric force between the charges is 0.9 N.

#### Example – 3:

- Two small spheres are charged positively. The combined charge being 5.0 x 10-5 C. If each sphere is repelled from other by a force of 1.0 N when they are 2.0 m apart. How is total charge distributed between two spheres.
**Solution:****Given:**Distance between two charges = r = 2.0 m, Sum of the cahrges = 5.0 × 10^{-5 }C = 50 × 10^{-6 }C, Let , The magnitude of the first charge = q_{1 }= q μC = q × 10^{-6 }C, Magnitude of the other charge = q_{2 }=(50 – q) × 10^{-6 }C, dielectric constant of medium = k = 1. Force = F = 1.0 N**To find:**charges = ?

By Coulomb’s law

By Coulomb’s law

∴ q_{1}= 38.43 m C = 38.43 × 10^{-6 }C = 3.84 × 10^{-5 }C

∴ q_{2}= 11.57 m C = 11.57 × 10^{-6 }C = 1.16 × 10^{-5 }C

**Ans:** The distribution of charge is 3.84 × 10^{-5 }C and 1.16 × 10^{-5 }C

#### Principle of Superposition of Forces:

- This principle is used for calculation of net force acting on an individual charge due to other when more than two charges are interacting with each other.
- When a number of charges are interacting, the total force on a given charge is the vector sum of individual forces exerted on a given charge by all other charges.
- Let us consider a system of point charges q
_{1}, q_{2}, q_{3}, q_{4}, ……..q_{n}. Let F_{12}, F_{13}, F_{14}, F_{15}, ……..F_{1n}be the forces acting on charge q_{1 }due to charges q_{2}, q_{3}, q_{4}, ……..q_{n }respectively. - Then according to the principle of superposition of forces, the force acting on charge q
_{1}is given by

F_{1} = F_{12} + F_{13} + F_{14} + ……+ F_{1n}

By vector representation

Applying same logic to all such pairs we have

- The same procedure can be adopted for finding the force on any other charge due to remaining charges.