**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 **

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

**q _{1} = 5 × 10^{–8} C ; q_{2} = –3 × 10^{–8} C**

**⇒ V _{1} + V_{2} = o**

**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**

**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 having 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: The system represents an electric dipole of charge strength 2μC, electric dipole length 0.06 m and electric dipole moment = 0.12 × 10 ^{-6} C-m.**

**By Formula, **

**(i) For points on axial line, θ = 0°, cosθ = 1**

**The equipotential surfaces are spherical with centre at the centre of the dipole.**

**(ii) For points on equatorial line,**

**θ = 90°, cosθ = zero.**

**The equatorial plane in plane of zero potential.**

**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 ,**

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

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

**Electric Field **

**On putting the values we get ,**

**E’ = 4.4 × 10 ^{4} N/C**