QJEEM25Phb3b2: JEE Main 2025 Physics - Class 12 JEEM25Phb3b2 Solution: An electric dipole is placed at a distance of 2 cm from an…

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

$E = {{{\sigma _0}} \over {2{\varepsilon _0}}}$

perpendicularly directed away from the sheet for positive charge

So, $\overrightarrow E = {{{\sigma _0}} \over {2{\varepsilon _0}}} \uparrow \theta$

note : E is constant

Let length of dipole = d

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

We know,

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

$\overrightarrow \tau = qd\widehat i \times {{{\sigma _0}} \over {2{\varepsilon _0}}}\widehat i$

$\widehat \tau = O$ (as $\widehat i \times \widehat i = 0$ )

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

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

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

$= - qE\widehat i + qE\widehat i$

$\overrightarrow F = O$

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.