**COULOMB’S LAW **is also the related topic to the electrostatics.

We know that a force of attraction or repulsion acts between

two charged bodies. How is this force affected when the

quantity of the charge on the two bodies or the distance

between them is changed? In order to find the answers of

these questions, a French scientist Charles Coulomb

(1736–1806) in 1785 experimentally established the

fundamental law of electric force between two stationary

charged particles.

**Coulomb’s Law:** The force of attraction or repulsion between

two point charges is directly proportional to the product of

the quantity of charges and inversely proportional to the

square of the distance between them. Therefore

F∝q_{1}q_{2}…….. (13.1)

F=1/r^{2}…….. (13.2)

Combining Eqs. (13.1) and (13.2), we get

F=K(q_{1}q_{2}/r_{2})…….. (13.3)

Eqution (13.3) is known as Coulomb’s law.

where F is the force between the two charges and is called

the Coulomb force, q_{1} and q_{2} are the quantities of two

charges and r is the distance between the centre of two

charges (Fig. 13.12-a,b). K is the constant of proportionality

given by

k =1⁄ 4π ε_{0}

The value of k depends upon the medium between the two

charges and the system of units in which F, q, and r are

measured, ε_{0}is the permittivity of free space.

Now if the medium between the two charges is air then the

value of K in SI units will be 9 ×10^{9}Nm^{2}C^{-2}.

Coulomb’s law is true only for point charges whose sizes are

very small as compared to the distance between them. Like

other forces, electric forces also obey Newton’s third law.

**Example 13.1:**

Two bodies are oppositely charged with

500 µC and 100 µC. Find the force between the two charges if

the distance between them in air is 0.5m.

** Solution**:Given, q_{1} = 500 µC = 500 × 10^{-6 }C , q_{2} = 100 µC = 100 ×

10^{-6}C, Distance between charges r = 0.5m

Substituting these values in equation of Coulomb’s law, we

have

F=q_{1}q_{2}/4Πε_{o}r^{2}=9× 10^{6}Nm^{2}C^{-2}×500×10^{-6}C×100×10^{-6}C/(0.5m)^{2}

F=1800N