Question

Two point charges $5\times {10}^{-8}C$ and $-3\times {10}^{-8}C$ are located 16 cm apart. At what point on the line joining these charges the electric potential will be zero?

Answer 1

Let the distance of the point from charge $5\times {10}^{-8}C$ be $r$.

Hence the distance of this point from the other charge is $16-r$.

Since the potential at the point is zero,

$\frac{k(5\times {10}^{-8})}{r}+\frac{k(-3\times {10}^{-8})}{16-r}=0$

$⟹r=10cm$