Class 12 Physics Chapter 2 Electrostatic Potential And Capacitance Solutions: Free PDF and Important Formulas

Ans.2.1

Let the charges be

$q_1$ = 5  $\times {10}^{-8}$ C and  $q_2$ = -3  $\times {10}^{-8}$ C

Distance between the two charges, d = 16 cm = 0.16 m

Consider a point P on the line joining the two charges

R = distance of the point P from the charge  $q_1$

Let the electrical potential (V) at point P is zero

Potential at point P caused by charges  $q_1$ and  $q_2$ respectively.

V =  $\frac{1}{4\pi {\epsilon }_0}\times \frac{q_1}{r}$  $+$  $\frac{1}{4\pi {\epsilon }_0}\times \frac{q_2}{(d-r)}$ …………………………..(1)

Where  ${\epsilon }_0$ = permittivity of free space

For V = 0, the equation (1) becomes

$\frac{1}{4\pi {\epsilon }_0}\times \frac{q_1}{r}=-\frac{1}{4\pi {\epsilon }_0}\times \frac{q_2}{(d-r)}$ or  $\frac{q_1}{r}$ =  $-\frac{q_2}{(d-r)}$ or  $\frac{5\mathrm{}\times {10}^{-8}}{r}$ =  $-\frac{-3\mathrm{}\times {10}^{-8}}{(d-r)}$

$\frac{5\mathrm{}\times {10}^{-8}}{r}$ =  $\frac{3\mathrm{}\times {10}^{-8}}{(d-r)}$ or  $\frac{r}{(d-r)}$ =  $\frac{5\mathrm{}\times {10}^{-8}}{3\mathrm{}\times {10}^{-8}}$ or  $\frac{r}{(d-r)}$ =  $\frac{5}{3}$

3r = 5d -5r or r = (5d/8) = 0.1 m = 10 cm

Therefore the potential is zero at a distance of 10 cm from the charge  $q_1$

Suppose the point P is outside the system of two charges, as shown in the following figure.

V =  $\frac{1}{4\pi {\epsilon }_0}\times \frac{q_1}{s}$  $+$  $\frac{1}{4\pi {\epsilon }_0}\times \frac{q_2}{(s-d)}$ ……………………………(2)

For V = 0, we get

$\frac{1}{4\pi {\epsilon }_0}\times \frac{q_1}{s}=-\frac{1}{4\pi {\epsilon }_0}\times \frac{q_2}{(s-d)}$ or  $\frac{q_1}{s}=-\frac{q_2}{(s-d)}$ or  $\frac{5\mathrm{}\times {10}^{-8}}{s}=-\frac{-3\mathrm{}\times {10}^{-8}}{(s-d)}$

$\frac{5}{s}$ =  $\frac{3}{s-0.16}$ or 5s – 0.8 = 3s or s =  $\frac{0.8}{2}$ = 0.4 m = 40 cm

Therefore, the potential is zero at a distance 40 cm from  $q_1$ , outside the system of charge.

Q.2.2 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.

Ans.2.2

The charge at each corner,  $q_1$ = 5  $\mu C=5\times {10}^{-6}$ C

The sides of the hexagon, AB = BC = CD = DE = EF =FA = 10 cm = 0.1 m

Let O be the centre. The electric potential at O is given by

V =  $\frac{1}{4\pi {\epsilon }_0}\times \frac{6q_1}{d}$ , where  ${\epsilon }_0$ = permittivity of free space = 8.854  $\times {10}^{-12}$  $C^2{N}^{-1}$  $m^{-2}$

Hence V =  $\frac{1}{4\times \pi \times 8.854\mathrm{}\times {10}^{-12}}\times \frac{6\times 5\times {10}^{-6}}{0.1}$ = 2.696  $\times {10}^6$ V

Therefore, the potential at the centre is 2.7  $\times {10}^6$ V

Q.2.3 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?

Ans.2.3 The arrangement is represented in the adjoining figure.

An equipotential surface is the plane on which total potential is zero everywhere. The plane is normal to line AB. The plane is located at the midpoint of the line AB because the magnitude of the charge is same.

The direction of the electric field at every point on that surface is normal to the plane in the direction of AB.

Q.2.4 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?

Ans.2.4 Radius of the spherical conductor, r = 12 cm = 0.12 m

Uniformly distributed charge, q = 1.6  $\times {10}^{-7}$ C

The electrical field inside the conductor is zero.

Electric field E just outside the conductor is given by

E =  $\frac{1}{4\pi {\epsilon }_0}\times \frac{q}{r^2}$ where  ${\epsilon }_0$ = permittivity of free space = 8.854  $\times {10}^{-12}$  $C^2{N}^{-1}$  $m^{-2}$

E =  $\frac{1}{4\times \pi \times 8.854\mathrm{}\times {10}^{-12}}\times \frac{1.6\mathrm{}\times {10}^{-7}}{{0.12\mathrm{}}^2}$ = 99863.8 N  $C^{-1}$  $\cong {10}^5$ N  $C^{-1}$

At a point 18 cm from the centre of the sphere. Hence r = 18 cm = 0.18 m

From the above equation we get

E at 18 cm from the centre =  $\frac{1}{4\times \pi \times 8.854\mathrm{}\times {10}^{-12}}\times \frac{1.6\mathrm{}\times {10}^{-7}}{{0.18\mathrm{}}^2}$ = 4.438  $\times {10}^4$ N  $C^{-1}$

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