CBSE 2025, Set 3

Question

(b) (i) An electric dipole of dipole moment $p$ consists of point charges $q$ and $-q$, separated by $2a$. Derive an expression for electric potential in terms of its dipole moment at a point at a distance $x (x >> a)$ from its centre and lying (I) along its axis, and (II) along its bisector line.
(ii) An electric dipole of dipole moment $\vec{p} = (0.8\hat{i} + 0.6\hat{j}) 	imes 10^{-29}	ext{ C m}$ is placed in an electric field $\vec{E} = 1.0 	imes 10^7 \hat{k}\ 	ext{V/m}$. Calculate the magnitude of the torque acting on it and the angle it makes with the x-axis, at this instant.

Edvaya · Accepted Answer

The correct answer is Subjective. **(i) Electric Potential due to a Dipole:**

**I. Along its axis:**
Consider a point P on the axis at a distance $x$ from the centre of the dipole.

![Axial Potential](https://www.edvaya.com/images/cbse/physics/2025/set3/q33b-axial-potential-derivation-answer.webp)

Potential at P due to $-q$: $V_- = \frac{-kq}{x + a}$ (where $k = \frac{1}{4\pi\epsilon_0}$)
Potential at P due to $+q$: $V_+ = \frac{kq}{x - a}$
Net potential $V = V_+ + V_-$
$$V = kq \left( \frac{1}{x - a} - \frac{1}{x + a} ight)$$
$$V = kq \left( \frac{x + a - (x - a)}{x^2 - a^2} ight) = \frac{kq(2a)}{x^2 - a^2}$$
Since dipole moment $p = q(2a)$:
$$V = \frac{kp}{x^2 - a^2}$$
For $x >> a$, $a^2$ can be neglected:
$$\boxed{V = \frac{kp}{x^2} = \frac{1}{4\pi\epsilon_0} \frac{p}{x^2}}$$

**II. Along its bisector (equatorial) line:**
Consider a point P on the equatorial line at a distance $x$ from the centre.

![Equatorial Potential](https://www.edvaya.com/images/cbse/physics/2025/set3/q33b-equatorial-potential-derivation-answer.webp)

The distance of P from both $+q$ and $-q$ is $\sqrt{x^2 + a^2}$.
Potential at P due to $-q$: $V_- = \frac{-kq}{\sqrt{x^2 + a^2}}$
Potential at P due to $+q$: $V_+ = \frac{kq}{\sqrt{x^2 + a^2}}$
Net potential $V = V_+ + V_- = \frac{kq}{\sqrt{x^2 + a^2}} - \frac{kq}{\sqrt{x^2 + a^2}}$
$$\boxed{V = 0}$$

**(ii) Torque on Dipole:**
Given:
$\vec{p} = (0.8\hat{i} + 0.6\hat{j}) 	imes 10^{-29}	ext{ C m}$
$\vec{E} = 1.0 	imes 10^7 \hat{k}	ext{ V/m}$

Torque $\vec{	au} = \vec{p} 	imes \vec{E}$
$\vec{	au} = [(0.8\hat{i} + 0.6\hat{j}) 	imes 10^{-29}] 	imes [1.0 	imes 10^7 \hat{k}]$
$\vec{	au} = 10^{-22} [(0.8\hat{i} 	imes \hat{k}) + (0.6\hat{j} 	imes \hat{k})]$
Using cross product rules ($\hat{i} 	imes \hat{k} = -\hat{j}$, $\hat{j} 	imes \hat{k} = \hat{i}$):
$\vec{	au} = 10^{-22} [-0.8\hat{j} + 0.6\hat{i}]$
$\vec{	au} = (0.6\hat{i} - 0.8\hat{j}) 	imes 10^{-22}	ext{ N m}$

**Magnitude of Torque:**
$|\vec{	au}| = \sqrt{(0.6)^2 + (-0.8)^2} 	imes 10^{-22} = \sqrt{0.36 + 0.64} 	imes 10^{-22} = \sqrt{1.0} 	imes 10^{-22}$
$$\boxed{|\vec{	au}| = 10^{-22}	ext{ N m}}$$

**Angle with x-axis:**
Let the torque vector make an angle $\alpha$ with the x-axis.

![Dipole Angle](https://www.edvaya.com/images/cbse/physics/2025/set3/q33b-dipole-torque-angle-answer.webp)

$	an \alpha = \frac{	au_y}{	au_x} = \frac{|-0.8|}{0.6} = \frac{4}{3}$
$$\boxed{\alpha = 	an^{-1}\left(\frac{4}{3}ight) = 53^\circ}$$

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