JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Electrostatics

Charge $Q$ at point $P$ sits on the axis of dipole 1 (charges $\pm q$ separated by $2a$) at distance $r$, and simultaneously on the equatorial line of dipole 2 (same charges, same separation) at the same distance $r$. For $Q$ to feel zero net force, the axial field of dipole 1 at $P$ must exactly cancel the equatorial field of dipole 2 at $P$.

Field on the axis of dipole 1, from the two point charges directly at distances $r-a$ and $r+a$:

$E_{axial}=\frac{kq}{(r-a{)}^2}-\frac{kq}{(r+a{)}^2}=\frac{4kqra}{(r^2-a^2{)}^2}$

Field on the equatorial line of dipole 2, from the two charges each at distance $\sqrt{r^2+a^2}$, components along the axis adding:

$E_{equatorial}=\frac{2kqa}{(r^2+a^2{)}^{3/2}}$

Setting these equal for force balance:

$\frac{4kqra}{(r^2-a^2{)}^2}=\frac{2kqa}{(r^2+a^2{)}^{3/2}}$

$4r^2(r^2+a^2{)}^3=(r^2-a^2{)}^4$

This relation between $a$ and $r$ has no clean closed-form solution; solving it numerically gives

$\frac{a}{r}\approx 3$