Question

A positive point charge $q$ is fixed at origin. A short dipole with a dipole moment $p$ is placed along the $x-$axis far away from the origin with $p$ pointing along the $+x-$axis. The kinetic energy of the dipole when it reaches distance $d$ from origin is

• A. $\frac{3qp}{8\pi {ϵ}_0{d}^2}$
• B. $\frac{qp}{4\pi {ϵ}_0{d}^2}$
• C. $\frac{2qp}{4\pi {ϵ}_0{d}^2}$
• D. $\frac{qp}{16\pi {ϵ}_0{d}^2}$

Answer 1

The correct option is B $\frac{qp}{4\pi {ϵ}_0{d}^2}$

Electric field due to the point charge (placed at origin) at a distance $x$ from origin:
$\overrightarrow{E}=\frac{kq}{x^2}\hat{i}$

Initially when dipole is at infinity (very far distance from $q$)

$U_i=0$

When the dipole is at a distance $x=d$ from origin.

$U_f=-\overrightarrow{p}\cdot \overrightarrow{E}$

$\Rightarrow U_f=-\left(p\hat{i}\right)\cdot \left(\frac{kq}{d^2}\hat{i}\right)$

$\Rightarrow U_f=-\frac{kpq}{d^2}$

Applying energy conservation,

$\Rightarrow$ Gain in K.E of dipole = loss in electrostatic potential energy

$\Rightarrow K.E_f-K.E_i=U_i-U_f$

$\Rightarrow K.E_f-0=0-(-\frac{kpq}{d^2})$

$\therefore K.E_f=\frac{kqp}{d^2}=\frac{qp}{4\pi {ϵ}_0{d}^2}$

Therefore, option $\left(b\right)$ is the correct choice.

Tip : Electric field due to a point charge is always along line joining charge and point. Thus, at $x-$axis, $\overrightarrow{E}=\frac{kq}{x^2}\hat{i}$