Two charges $3 \times 10^{-8} \; C$ and $-2 \times 10^{-8} \; C$ are located $15 \; cm$ apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

(N/A) Let us take the origin $O$ at the location of the positive charge. The line joining the two charges is taken to be the $x$-axis; the negative charge is taken to be on the right side of the origin.
Let $P$ be the required point on the $x$-axis where the potential is zero.
If $x$ is the $x$-coordinate of $P$,obviously $x$ must be positive. If $x$ lies between $O$ and $A$,we have:
$\frac{1}{4 \pi \varepsilon_{0}} \left[ \frac{3 \times 10^{-8}}{x \times 10^{-2}} - \frac{2 \times 10^{-8}}{(15-x) \times 10^{-2}} \right] = 0$
where $x$ is in $cm$. That is,
$\frac{3}{x} - \frac{2}{15-x} = 0$
which gives $x = 9 \; cm$.
If $x$ lies on the extended line $OA$,the required condition is:
$\frac{3}{x} - \frac{2}{x-15} = 0$
which gives $x = 45 \; cm$.
Thus,the electric potential is zero at $9 \; cm$ and $45 \; cm$ away from the positive charge on the side of the negative charge.