JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Electrostatics

A point on the $y$-axis lies on the equatorial plane of a dipole pointing along $x$. On the equatorial line the potential vanishes, $V=0$, because the point is equidistant from the two ends of the dipole. The equatorial field points opposite to the dipole moment and has magnitude $\frac{kp}{d^3}$, so as a vector $\vec{E}=-\frac{\vec{p}}{4\pi {\epsilon }_0{d}^3}$.