Question

For a point charge of + 3.50 uC what is the radius of the equipotential surface which is at a potential of 2.50 k V ?

Answer 1

Potential due to a point charge is given by :

$V=-\frac{kQ}{r}[k=\frac{1}{4\pi {ϵ}_0}]$

Given, $k=9\times {10}^{9}v=2.5\times {10}^{3}VQ=3.5\times {10}^{-6}C$ so,

$r=\frac{kQ}{v}=\frac{9\times {10}^9\times 3.5\times {10}^{-6}}{2.5\times {10}^3}=12.6m$

$\therefore$ Radius of the surface would be $12.6m$