Solution:

Ans. (i)

(ii) (a) For region II,

$$ E_{I I}=\frac{1}{2 \varepsilon_{0}}\left(\sigma_{1}-\sigma_{2}\right) $$

towards right side from sheet $\mathrm{A}$ to sheet $\mathrm{B}$. $1 / 2$

(b) For region III,

$$ E_{I I I}=\frac{1}{2 \varepsilon_{0}}\left(\sigma_{1}+\sigma_{2}\right) \quad 1 / 2 $$

towards right side away from the two sheets. $1 / 2$

[CBSE Marking Scheme, 2014]

Long Answer Type Questions

(5 marks each)

[AI Q. 1. (i) Derive an expression for electric field $E$ due to a dipole of length ’ $2 a$ ’ at a distant point ’ $r$ ’ from the centre of the dipole on an axial line.

(ii) Draw a graph of $E$ versus $r$ for $r»a$

(iii) If this dipole were kept in a uniform external electric field $E_{0}$, diagrammatically represent the position of the dipole in stable and unstable equilibrium and write the expressions for the torque acting on the dipole in both the cases.

U [O.D. I, II, III 2017] Ans. (i) Derivation of $E$ along the axial line of dipole 2

(ii) Graph of $E$ versus $r$

(iii) (a) Diagrams for stable and unstable equilibrium of dipole

$1 / 2+1 / 2$

(b) Torque on dipole in both the cases $1 / 2+1 / 2$

[CBSE Marking Scheme, 2017]

Detailed Answer:

(i) Derivation of $E$ along the axial line of dipole Try yourself, Similar to Question 6, Short Answer Type Questions-I

(ii) Graph of $E$ versus $r$

(iii) (a) Diagrams for stable and unstable equilibrium of dipole.

Try yourself Similar to Question 1 (ii), Short Answer Type Questions-II

(b) Torque on dipole in both the cases.

Torque on dipole $=p E \sin \theta$

for stable equilibrium

$\theta=0^{\circ}$

$\therefore$ Torque $=0$

For unstable equilibrium

$\theta=180^{\circ}$

Then new Torque $=0$