Two charges $5 \times 10^{-8} \; C$ and $-3 \times 10^{-8} \; C$ are located $16 \; cm$ apart. At what point$(s)$ on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

(N/A) Given charges are $q_{1} = 5 \times 10^{-8} \; C$ and $q_{2} = -3 \times 10^{-8} \; C$. The distance between them is $d = 16 \; cm = 0.16 \; m$.
Case $1$: Point $P$ is between the charges at a distance $r$ from $q_{1}$.
The potential $V$ at $P$ is given by $V = \frac{1}{4 \pi \varepsilon_{0}} \left( \frac{q_{1}}{r} + \frac{q_{2}}{d-r} \right)$.
Setting $V = 0$,we get $\frac{q_{1}}{r} = -\frac{q_{2}}{d-r}$.
Substituting values: $\frac{5 \times 10^{-8}}{r} = -\frac{-3 \times 10^{-8}}{0.16-r}$.
$5(0.16 - r) = 3r \Rightarrow 0.8 - 5r = 3r \Rightarrow 8r = 0.8 \Rightarrow r = 0.1 \; m = 10 \; cm$.
Case $2$: Point $P$ is outside the charges at a distance $s$ from $q_{1}$.
The potential $V$ at $P$ is given by $V = \frac{1}{4 \pi \varepsilon_{0}} \left( \frac{q_{1}}{s} + \frac{q_{2}}{s-d} \right)$.
Setting $V = 0$,we get $\frac{q_{1}}{s} = -\frac{q_{2}}{s-d}$.
Substituting values: $\frac{5 \times 10^{-8}}{s} = -\frac{-3 \times 10^{-8}}{s-0.16}$.
$5(s - 0.16) = 3s \Rightarrow 5s - 0.8 = 3s \Rightarrow 2s = 0.8 \Rightarrow s = 0.4 \; m = 40 \; cm$.
Thus,the potential is zero at $10 \; cm$ from the positive charge between the charges and at $40 \; cm$ from the positive charge outside the system.