Two point charges of magnitude $+q$ and $-q$ are placed at $\left( -\frac{d}{2}, 0, 0 \right)$ and $\left( \frac{d}{2}, 0, 0 \right)$,respectively. Find the equation of the equipotential surface where the potential is zero.

(N/A) Let the required plane lie at a distance $x$ from the origin as shown in the figure.
The potential at a general point $P$ on the surface is given by the sum of potentials due to both charges:
$V = \frac{kq}{\sqrt{(x + d/2)^2 + h^2}} - \frac{kq}{\sqrt{(x - d/2)^2 + h^2}} = 0$
$\therefore \frac{1}{\sqrt{(x + d/2)^2 + h^2}} = \frac{1}{\sqrt{(x - d/2)^2 + h^2}}$
Squaring both sides:
$(x - d/2)^2 + h^2 = (x + d/2)^2 + h^2$
Expanding the squares:
$x^2 - xd + \frac{d^2}{4} + h^2 = x^2 + xd + \frac{d^2}{4} + h^2$
Simplifying the equation:
$-xd = xd$
$2xd = 0$
Since $d \neq 0$,we must have $x = 0$.
The equation of the required equipotential surface is $x = 0$,which represents the $yz$-plane.