Question

At a point due to a point charge, the values of electric field intensity and potential are $32N{C}^{-1}$ and $16J{C}^{-1}$, respectively.
Calculate the magnitude of the charge

Answer 1

If r is the distance of the charge Q from the point of observation, then
$E=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r^2}=32..............\left(i\right)$
$V=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r}=16...............\left(ii\right)$
Dividing Eq.(ii) by Eq(i) We get r = 0.5 m and putting the value of r in Eq (ii) we get ,
$Q=\frac{16\times 0.5}{9\times {10}^9}=\left(8/9\right)\times {10}^{-9}C$