NCERT : Electrostatic Potential & Capacitance

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 q1 = 5 × 10–8 C and q2 = –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 q1 hence from charge q2 it is (16-x) cm

q1 = 5 × 10–8 C ; q2 = –3 × 10–8 C

⇒ V1 + V2 = 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 × 10–7C 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 × 10–7C , 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 = 105 N/C

(c) Given , q = 1.6 × 10–7C , 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 × 104 N/C

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