Question

At s 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
a. magnitude of the charge, and
b. distance of the charge from the point of observation.

Answer 1

Correct Answer - a. `(8)/(9) xx 10^-9 C`
b. 0.5 m.
If r is the distance of the charge Q from the point of observation, then
`E = (1)/(4 pi epsilon_0) Q/(r^2) = 32` …(i)
and `V = (1)/(4 pi epsilon_0) 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 = (16 xx 0.5)/(9 xx 10^9) = (8//9) xx 10^-9 C`.