JEE Main 2026 — Electrostatics Question with Solution

Let the distance of point $x$ from the origin $O$ be $r$.

Point $x$ lies on the axial line of dipole A. The electric field at $x$ due to dipole A is directed along the horizontal axis and its magnitude is given by:
$E_A=\frac{1}{4\pi {ϵ}_0}\frac{2p_1}{r^3}$

Point $x$ lies on the equatorial line of dipole B. The electric field at $x$ due to dipole B is directed perpendicular to the horizontal axis and its magnitude is given by:
$E_B=\frac{1}{4\pi {ϵ}_0}\frac{p_2}{r^3}$

The resultant electric field makes an angle of $60^{\circ }$ with the line $Ox$ (the horizontal axis). Therefore, the tangent of this angle is the ratio of the perpendicular component of the electric field to the parallel component:
$\tan 60^{\circ }=\frac{E_B}{E_A}$

Substituting the expressions for $E_A$ and $E_B$:
$\sqrt{3}=\frac{\frac{1}{4\pi {ϵ}_0}\frac{p_2}{r^3}}{\frac{1}{4\pi {ϵ}_0}\frac{2p_1}{r^3}}$

$\sqrt{3}=\frac{p_2}{2p_1}$

$\frac{p_2}{p_1}=2\sqrt{3}$

Answer: $2\sqrt{3}$