Previous Year NEET Question- Electrostatics L-3

Question: If a conducting sphere of radius $\mathrm{R}$ is charged. Then the electric field at a distance $r(r>R)$ from the centre of the sphere would be, $(V=$ potential on the surface of the sphere) (NEET-2023)#

A) $\frac{r V}{R^2}$

B) $\frac{R^2 V}{r^3}$

C) $\frac{R V}{r^2}$

D) $\frac{\mathrm{V}}{\mathrm{r}}$

Answer: $\frac{R V}{r^2}$#

Explanation

$\therefore V=\frac{K Q}{R}$

$E=\frac{K Q}{r^2}$

$E=\frac{V R}{r^2}$