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.

The electric field at a point due to a point a charge is `30 N//C`, an

1.	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» Correct Answer - `0.5 m ; 0.83xx10^(-9)C` 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`

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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.

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