JEE Main 2025 — Electrostatics Question with Solution

Given values:

The dipole moment is $p=6\times 10^{-6}\mathrm{Cm}$.

The electric field is $E=10^{6}V/m$.

Formula for work done:

Work required to rotate the dipole is $W=\Delta U=-pE(\cos {\theta }_f-\cos {\theta }_i)$ where ${\theta }_i$ is the initial angle between the dipole and the electric field, and ${\theta }_f$ is the final angle.

Angles:

At the start, the dipole is parallel to the field (${\theta }_i=0^{\circ }$), so $\cos {\theta }_i=1$.

At the end, the dipole is opposite the field (${\theta }_f=180^{\circ }$), so $\cos {\theta }_f=-1$.

Calculation:

$W=-pE(\cos 180^{\circ }-\cos 0^{\circ })=-pE(-1-1)=2pE$

Plug in the values:

$W=2\times (6\times 10^{-6})\times (10^6)=2\times 6=12\mathrm{J}$