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.

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. (There is no possibility of potentials due to the two charges adding up to zero for $x < 0 .$ 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, electric potential is zero at $9 cm$ and $45 cm$ away from the positive charge on the side of the negative charge. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity.