NCERT Solution : 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 V = q/4πεor

Suppose Potential is zero at a distance x cm from q1 hence from charge q2 it is (16-x) cm

⇒ V = o

⇒ [q1/4πεox × 10–2] + [q2/4πεo(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/4πεor)

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 × 10–7C
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 ,

E = q/4πεor2

on putting the value we get , E = 105 N/C

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

Electric Field E’ = q/4πεor’2

On putting the values we get ,

E’ = 4.4 × 104 N/C

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