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.

There are two charges,

$q_{1}=5 \times 10^{-8} \,C$

$q_{2}=-3 \times 10^{-8} \,C$

Distance between the two charges, $d =16 \,cm =0.16 \,m$

Consider a point $P$ on the line joining the two charges, as shown in the given figure.

$r=$ Distance of point $P$ from charge $q_{1}$ Let the electric potential $(V)$ at point $P$ be zero. Potential at point $P$ is the sum of potentials caused by charges $q_{1}$ and $q_{2}$ respectively.

Where, $\therefore V=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1}}{r}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{2}}{(d-r)} \dots(i)$

$\varepsilon_{0}=$ Permittivity of free space For $V =0,$ equation $(i)$ reduces to $0=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1}}{r}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{2}}{(d-r)}$

$\Rightarrow \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1}}{r}=-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{2}}{(d-r)}$

$\Rightarrow \frac{q_{1}}{r}=-\frac{q_{2}}{(d-r)}$

$\Rightarrow \frac{5 \times 10^{-8}}{r}=-\frac{\left(-3 \times 10^{-8}\right)}{(0.16-r)}$

$\Rightarrow 5(0.16-r)=3 r$

$\Rightarrow 0.8=8 r \Rightarrow r=0.1 \,m =10 \,cm$

Therefore, the potential is zero at a distance of $10 \;cm$ from the positive charge between the charges. Suppose point $P$ is outside the system of two charges at a distance s from the negative charge, where potential is zero, as shown in the following figure.

For this arrangement, potential is given by,

Where, $V=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1}}{s}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{2}}{(s-d)} \ldots (ii)$

$\varepsilon_{0}=$ Permittivity of free space For $V=0,$ equation (ii) reduces to $0=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1}}{s}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{2}}{(s-d)}$

$\Rightarrow \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1}}{s}=-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{2}}{(s-d)}$

$\Rightarrow \frac{q_{1}}{s}=-\frac{q_{2}}{(s-d)}$

$\Rightarrow \frac{5 \times 10^{-8}}{s}=-\frac{\left(-3 \times 10^{-8}\right)}{(s-0.16)}$

$\Rightarrow 5(s-0.16)=3 \,s$

$\Rightarrow 0.8=2 \,s \Rightarrow s=0.4 \,m =40\, cm$

Therefore, the potential is zero at a distance of $40 \,cm$ from the positive charge outside the system of charges.