Two charges ${q_1}$ and ${q_2}$ are placed at $(0, 0, d)$ and$(0, 0, - d)$ respectively. Find locus of points where the potential is zero.

Let us take a point of the required plane as $(x, y, z)$. The two charges lies on $z$-axis at a separation of $2 d$.

The potential at the point $P$ due to two charges is given by $V=V_{1}+V_{2}$

$\therefore0=\frac{k q_{1}}{\sqrt{x^{2}+y^{2}+(z-d)^{2}}}+\frac{k q_{2}}{\sqrt{x^{2}+y^{2}+(z+d)^{2}}}$

$\therefore\frac{q_{1}}{\sqrt{x^{2}+y^{2}+(z-d)^{2}}}=-\frac{q_{2}}{\sqrt{x^{2}+y^{2}+(z+d)^{2}}}$

$\frac{q_{1}}{q_{2}}=-\sqrt{\frac{x^{2}+y^{2}+(z-d)^{2}}{x^{2}+y^{2}+(z+d)^{2}}}$

Compendo and dividendo,

$\frac{\sqrt{x^{2}+y^{2}+(z-d)^{2}}}{q_{1}+q_{2}}=-\frac{+\sqrt{x^{2}+y^{2}+(z+d)^{2}}}{\sqrt{x^{2}+y^{2}+(z-d)^{2}}}$

$-\sqrt{x^{2}+y^{2}+(z+d)^{2}}$

Squaring both side,

$\frac{\left(q_{1}+q_{2}\right)^{2}}{\left(q_{1}-q_{2}\right)^{2}}=- \frac{+\left(x^{2}+y^{2}+z^{2}+2 z d+d^{2}\right)}{\left(x^{2}+y^{2}+z^{2}-2 z d+d^{2}\right)}$

$=\frac{2\left(x^{2}+y^{2}+z^{2}+2 z d+d^{2}\right)}{2(2 z d)}$

$\frac{\left(q_{1}+q_{2}\right)^{2}}{\left(q_{1}-q_{2}\right)^{2}}=-\frac{x^{2}+y^{2}+z^{2}+d^{2}}{2 z d}$

$\therefore x^{2}+y^{2}+z^{2}+2 z d \frac{\left(q_{1}+q_{2}\right)^{2}}{\left(q_{1}-q_{2}\right)^{2}}+d^{2}=0$

This is the equation of sphere with centre,

$\left(0,0,-2 d\left[\frac{q_{1}^{2}+q_{2}^{2}}{q_{1}^{2}-q_{2}^{2}}\right]\right)$