JEE Main 2025 — Electrostatics Question with Solution

We know, electric field due to uniformly charged infinite plane sheet,

$$E=\frac{{\sigma }_0}{2{\epsilon }_0}$$

perpendicularly directed away from the sheet for positive charge

So, $$\overrightarrow{E}=\frac{{\sigma }_0}{2{\epsilon }_0}\uparrow \theta$$

note : E is constant

Let length of dipole = d

so dipole moment, $$\overrightarrow{p}=qd\uparrow$$

We know,

Torque, $$\overrightarrow{\tau }=\overrightarrow{p}\times \overrightarrow{E}$$

$$\overrightarrow{\tau }=qd\hat{i}\times \frac{{\sigma }_0}{2{\epsilon }_0}\hat{i}$$

$$\hat{\tau }=O$$ (as $$\hat{i}\times \hat{i}=0$$)

force, on charge $$-q,\overrightarrow{F_1}=-qE\hat{i}$$

on charge $$+q,\overrightarrow{F_2}=qE\hat{i}$$

So net force on dipole, $$\overrightarrow{F}=\overrightarrow{F_1}+\overrightarrow{F_2}$$

$$=-qE\hat{i}+qE\hat{i}$$

$$\overrightarrow{F}=O$$

We know, potential energy of dipole

$$u(O)=-pE\cos \theta$$

here, $$O=0^{\circ }$$ (given)

So, $$u_{\min }=-pE$$

Hence, option B is correct.