The electric field at a point due to a point a charge is `30 N//C`, and the electric potential at that point is `15 J//C`. Calcualte the distance of the point from the charge and the magnitude of the charge.

Answer

`0.5 m ; 0.83xx10^(-9)C`

Explanation

Here, `E = (q)/(4pi in_(0) r^(2)) = 30 NC^(-1)` …(i)
`V = (q)/(4pi in_(0) r) = 15 JC^(-1)` …(ii)
Divide (ii) by (i), we get
`r = (1)/(2) = 0.5m`
Put in (ii), `9xx10^(9) (q)/(1//2) = 15`
`q = (15)/(18) xx10^(-9) = 0.83xx10^(-9) C`