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

(N/A) Let the coordinates of a point on the locus be $(x, y, z)$. The potential $V$ at this point due to the two charges is the sum of the individual potentials: $V = V_1 + V_2 = 0$.
Using the formula for electric potential $V = \frac{kq}{r}$,we have:
$\frac{kq_1}{\sqrt{x^2 + y^2 + (z - d)^2}} + \frac{kq_2}{\sqrt{x^2 + y^2 + (z + d)^2}} = 0$
Rearranging the terms:
$\frac{q_1}{\sqrt{x^2 + y^2 + (z - d)^2}} = -\frac{q_2}{\sqrt{x^2 + y^2 + (z + d)^2}}$
Squaring both sides:
$\frac{q_1^2}{x^2 + y^2 + (z - d)^2} = \frac{q_2^2}{x^2 + y^2 + (z + d)^2}$
Let $\lambda^2 = \frac{q_1^2}{q_2^2}$. Then:
$\lambda^2 (x^2 + y^2 + z^2 + 2zd + d^2) = x^2 + y^2 + z^2 - 2zd + d^2$
$(\lambda^2 - 1)(x^2 + y^2 + z^2 + d^2) = -2zd(1 + \lambda^2)$
$x^2 + y^2 + z^2 + 2zd \left( \frac{\lambda^2 + 1}{\lambda^2 - 1} \right) + d^2 = 0$
Substituting $\lambda^2 = \frac{q_1^2}{q_2^2}$:
$x^2 + y^2 + z^2 + 2zd \left( \frac{q_1^2 + q_2^2}{q_1^2 - q_2^2} \right) + d^2 = 0$
This represents a sphere.