Q122 / 200 · NEET 2023 (Manipur) · Physics

If a conducting sphere of radius $R$ 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 $)$

Correct answer
A
$\frac{R V}{r^2}$
B
$\frac{V}{r}$
C
$\frac{r V}{R^2}$
D
$\frac{R^2 V}{r^3}$

Solution

For $r>R$, Electric field $E=\frac{K q}{r^2} \ldots$ and potential at surface of sphere $V=\frac{K q}{R}$ $\therefore$ From equation (1) $V=\frac{E r^2}{R}$ $\Rightarrow E=\frac{R V}{r^2}$

Hence, the correct answer is option: 'A'.

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