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 $ \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

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. Co =  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 $

CK = 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} $

CS = 3 × 10-12 F

CS = 3 pF

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

C1V1 = C2V2 = C3V3

As C1 = C2= C3 ,

We have, V1 = V2 = V3 = Vc (say)

Since V1 + V2 + V3 = V

3Vc = 120, Vc = 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

q1 = C1 V = 2 × 100 = 200pC

q2 = C2 V = 3 × 100 = 300pC

q3 = C3 V = 4 × 100 = 400pC

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