**COULOMB’S LAW:**

Two point electric charges q_{1} and q_{2} at rest, separated by a distance r exert a force on each other whose magnitude is given by

If between the two charges there is free space then

Where ε_{0} is the absolute electric permittivity of the free space and ε_{0} = 8.85 x 10^{-12} C^{2} / N-m^{2}

**Illustration 1. **A polythene piece rubbed with wool is found to have negative charge of 3.2 x 10^{-7 }C.

(a) Estimate the number of electrons transferred from wool to polythene.

(b) Is there a transfer of mass from wool to polythene? If yes, how much?

**Solution:** (a) Let n be the number of electrons getting transferred.

⇒ n x e = 3.2 x 10^{-7}

⇒ 1.6 x 10^{-19} n = 3.2 x 10^{-7}

⇒ n = 2 x 10^{12} ⇒ 2 x 10^{12} electrons will get transferred.

(b) Mass transferred will be product of number of electrons and the mass of electron.

If m = mass getting transferred.

⇒ m = n x (9.1 x 10^{-31}) kg

m = 2 x 10^{12} x 9.1 x 10^{-31}

m = 18.2 x 10^{-31} kg

**Illustration 2.** Calculate force between two charges of 2 C each separated by 2 m in vacuum.

**Solution:** F = kQ_{1}Q_{2}/R^{2} = 9 x 10^{9} x 2 x 2 ⇒ F = 9 x 10^{9} N

**Illustration 3. **Two particles A and B having charges 8 x 10^{-6} C and –2 x10^{-6 }C respectively are held fixed with a separation of 20 cm. Where should a third charged particle be placed so that it does not experience a net electric force?

Solution: As the net electric force on C should be equal to zero, the force due to A and B must be opposite in direction. Hence, the particle should be placed on the line AB. As A and B have charges of opposite signs, C cannot be between A and B. Also A has larger magnitude of charge than B. Hence, C should be placed closer to B than A. The situation is shown in figure. Suppose BC=x and the charge on C is Q

**Illustration 4. **A charge of -2 μC is placed at the perpendicular bisector of the line joining two point charges of 10 μC as shown in figure. What is the net force acting on the -2 μC charge ?

Components of forces parallel to AB will cancel out.

F = F_{1} cos θ + F_{2} cos θ

**Questions For You:**

**(i) **A negatively charged particle is placed exactly midway between two fixed particles having equal positive charge. What will happen to the charge ?

(a) if it is displaced at right angle to the line joining the positive charges?

(b) if it is displaced along the line joining the positive charges?

**(ii) **Is the Coulomb force between two given charges affected in anyway, if other charges are brought in the neighbourhood ?

**Coulombs law in Vector Relations**

**PRINCIPLE OF SUPERPOSITION**

This principle tells us that if charge Q is placed in the vicinity of several charges q_{1}, q_{2} ….. q_{n}, then the force on Q can be found out by calculating separately the forces F_{1} ,F_{2} …F_{n} , exerted by q_{1}, q_{2}, ….. q_{n} respectively on Q and then adding these forces vectorially. Their resultant F is the total force on Q due to all of charges.

**Illustration 5. **It is required to hold equal charges q each in equilibrium at the corners of a square of

side a. What charge when placed at the centre of the square will do this?

**Solution:**

Let the charge be Q

As ABCD is a square of side a, r=a/√2

Net force on the charge at B is

For charge q to be in equilibrium at B, the net force on it must be zero. Taking x-component