Q.1. Two charges 5 × 10^{–8} C and –3 × 10^{–8} C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

**Sol.** Let q_{1} = 5 × 10^{–8} C and q_{2} = –3 × 10^{–8} C

Use the formula , Potential V = q/4πε_{o}r

Suppose Potential is zero at a distance x cm from q_{1} hence from charge q_{2} it is (16-x) cm

⇒ V = o

⇒ [q_{1}/4πε_{o}x × 10^{–2}] + [q_{2}/4πε_{o}(16-x)× 10^{–2}] = 0

On solving we get , x = 0.1 m = 10 cm

Q.2. A regular hexagon of side 10 cm has a charge 5 μC at each of its

vertices. Calculate the potential at the centre of the hexagon.

**Sol.** charge q = 5 × 10^{–6} C

The distance of each charges from the centre of hexagon is r = 10 cm = 0.1 m

Hence Net potential at the centre is V = 6 ×(q/4πε_{o}r)

On putting the values we get , V = 2.7 × 10^{–6} volt

Q.3. Two charges 2 μC and –2 μC are placed at points A and B 6 cm

apart.

(a) Identify an equipotential surface of the system.

(b) What is the direction of the electric field at every point on this

surface ?

**Sol.** (a) The plane normal to AB and passing through its mid point has zero potential everywhere .

(b) Normal to the plane in direction of AB

Q.4. A spherical conductor of radius 12 cm has a charge of 1.6 × 10^{–7}C

distributed uniformly on its surface. What is the electric field

(a) inside the sphere

(b) just outside the sphere

(c) at a point 18 cm from the centre of the sphere?

Sol. Given , q = 1.6 × 10^{–7}C , r = 12 cm = 0.12m

(a) Just inside the sphere , E = 0

(b)Just outside the sphere i.e. on the surface of sphere ,

E = q/4πε_{o}r^{2}

on putting the value we get , E = 10^{5} N/C

(b) Given , q = 1.6 × 10^{–7}C , r’ = 18 cm = 0.18m

Electric Field E’ = q/4πε_{o}r’^{2}

On putting the values we get ,

E’ = 4.4 × 10^{4} N/C