JEE Main 2025 — Electrostatics Question with Solution

The work done on an electric dipole in a uniform electric field when it is rotated from an angle ${\theta }_1$ to an angle ${\theta }_2$ is given by:

$$W=-pE(\cos {\theta }_2-\cos {\theta }_1)$$

where:

$p=q\cdot d$ is the dipole moment,

$E$ is the electric field strength,

${\theta }_1$ and ${\theta }_2$ are the initial and final angles between the dipole moment and the electric field.

Given:

Charge $q=4\mu \mathrm{C}=4\times 10^{-6}\mathrm{C}$

Distance $d=18\mathrm{cm}=0.18\mathrm{m}$

Electric field $E=10^{4}N/C$

Initial angle ${\theta }_1=0^{\circ }$ (aligned with the field, equilibrium position)

Final angle ${\theta }_2=180^{\circ }$

First, calculate the dipole moment $p$:

$$p=q\cdot d=4\times 10^{-6}\mathrm{C}\cdot 0.18\mathrm{m}=7.2\times 10^{-7}\mathrm{Cm}$$

Calculate the work done $W$:

$$W=-pE(\cos 180^{\circ }-\cos 0^{\circ })$$

$$W=-7.2\times 10^{-7}\mathrm{Cm}\cdot 10^{4}N/C\cdot (-1-1)$$

$$W=7.2\times 10^{-7}\mathrm{Cm}\cdot 10^{4}N/C\cdot 2$$

$$W=1.44\times 10^{-2}\mathrm{J}=14.4\mathrm{mJ}$$

Therefore, the work done on the dipole in rotating it from the equilibrium position through $180^{\circ }$ is 14.4 mJ (Option D).