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 $ \displaystyle V = \frac{q}{4\pi \epsilon_0 r} $

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

$ \displaystyle \frac{q_1}{4\pi \epsilon_0 x} + \frac{q_2}{4\pi \epsilon_0 (16-x)} = 0$

$ \displaystyle \frac{5\times 10^{-8}}{4\pi \epsilon_0 x} + \frac{-3\times 10^{-8}}{4\pi \epsilon_0 (16-x)} = 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

$ \displaystyle V = 6 (\frac{q}{4\pi \epsilon_0 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 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, $ \displaystyle V = \frac{p cos\theta}{4\pi \epsilon_0 r^2} $

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

$ \displaystyle V \propto \frac{1}{r^2} $

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 ,

$ \displaystyle E = \frac{q}{4\pi\epsilon_0 r^2} $

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 $ \displaystyle E’ = \frac{q}{4\pi\epsilon_0 r’^2} $

On putting the values we get ,

E’ = 4.4 × 10^{4} N/C

Q:5. A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10^{–12} F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6 ?

Sol. C_{o} = 8 pF = 8 × 10^{-12} F , K = 6

In first case , $ \displaystyle C_0 = \frac{\epsilon_0 A}{d} $

By Formula , $\displaystyle C_K = \frac{K \epsilon_0 A}{d} $

In 2nd case , $ \displaystyle C_K = \frac{K \epsilon_0 A}{d/2} = \frac{2 K \epsilon_0 A}{d}$

On dividing , $ \displaystyle \frac{C_K}{C_0} = 2 K $

$ \displaystyle C_K = 2K C_0 $

C_{K} = 2 × 6 × 8 × 10^{-12} = 96pF

Q:6. Three capacitors each of capacitance 9 pF are connected in series.

(a) What is the total capacitance of the combination ?

(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply ?

Sol. C = 9 pF = 9 × 10^{-12} F , n = 3 , V = 120 volt

(a) As Three capacitors are in series

By Formul $ \displaystyle C_S = \frac{C}{n} $

$ \displaystyle C_S = \frac{9\times 10^{-12}}{3} $

C_{S} = 3 × 10^{-12} F

C_{S} = 3 pF

(b) Since in series, charge q is same on each capacitor,

C_{1}V_{1} = C_{2}V_{2} = C_{3}V_{3}

As C_{1} = C_{2}= C_{3} ,

We have, V_{1} = V_{2} = V_{3} = V_{c} (say)

Since V_{1} + V_{2} + V_{3} = V

3V_{c} = 120, V_{c} = 40V

Potential difference across each capacitor is 40V.

Q:7. Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.

(a) What is the total capacitance of the combination?

(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

Sol. (a) Total capacitance , C = 2 + 3 + 4 = 9 pF

(b)As all capacitors are connected in parallel, hence potential Difference across each capacitor is same .

V = 100 volt

q_{1} = C_{1} V = 2 × 100 = 200pC

q_{2} = C_{2} V = 3 × 100 = 300pC

q_{3} = C_{3} V = 4 × 100 = 400pC