At a certain distance from a point charge, the magnitude of the electric field is $500 \mathrm{~V/m}

Question

At a certain distance from a point charge, the magnitude of the electric field is $500 \mathrm{~V/m}$ and the electric potential is $3000 \mathrm{~V}$. The distance of the point from the charge is

Professor PC Thomas Classes · Answer

(a) : The magnitude of the electric field due to a point charge is $E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2}$. The electric potential is $V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}$. Dividing the expression for potential by the expression for electric field, we get $\frac{V}{E} = \frac{r^2}{r} = r$. Therefore, the distance is $r = \frac{3000}{500} = 6 \mathrm{~m}$.