Question

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

Answer 1

If r is the distance of the charge Q from the point of observation, then
$$ E = \dfrac {1}{ 4 \pi \epsilon_0} \dfrac {Q}{r^2} = 32 ..............(i) $$
$$ V = \dfrac {1}{ 4 \pi \epsilon_0} \dfrac {Q}{r} = 16 ...............(ii) $$
Dividing Eq.(ii) by Eq(i) We get r = 0.5 m and putting the value of r in Eq (ii) we get ,
$$ Q = \dfrac { 16 \times 0.5}{ 9 \times 10^9 }= (8/9) \times 10^{-9} C $$