Question

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.

Answer 1

Let p be the point of zero potential be p at distance r from the charge 1..
$d=16cm$
Electric potential, $v=q_1/4\pi {ϵ}_{o}r+q_2/4\pi {ϵ}_o(d-r)$
for $v=0$,
$\frac{q_1}{r}=-\frac{q_2}{d-r}$
$\Rightarrow \frac{5\times {10}^{-8}}{r}=\frac{-3\times {10}^{-8}}{0.16-r}$
$r=10cm$
If, point p is outside,
$v=q_1/4\pi {ϵ}_{o}r+q_2/4\pi {ϵ}_o(r-d)$
so here, $r=40cm$