Question

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

a. magnitude of the charge,and

b. distance of the charge from the point of observation.

Answer 1

If r is the distance of the charge Q from the point of observation, then

$E=14\pi {\epsilon }_0\frac{Q}{r^2}=32$ (i)

and $V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}=16$ (ii)

Dividing Eq (ii) br Eq. (i), we get r = 0.5m and putting the value of r in Eq. (ii), we get

$Q=\frac{16\times 0.5}{9\times {10}^9}=\left(8/9\right)\times {10}^{-9}C$