A hollow charged metal sphere has radius r. If the potential difference between its surface and a point
at a distance 3r from the centre is V, then electric field intensity at a distance 3r is

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$\frac{V}{3 r}$

$\frac{V}{2 r}$

$\frac{V}{4 r}$

$\frac{V}{6 r}$

answer is C.

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Detailed Solution

$V = \frac{q}{4 {πε}_{0} r} - \frac{q}{4 {πε}_{0} 3 r} = \frac{q}{6 {πε}_{0} r}$

Electric field intensity at a distance 3r is given by

$E = \frac{q}{4 {πε}_{0} (3 r)^{2}} = \frac{q}{4 {πε}_{0} \cdot 9 r^{2}}$

Thus, $\frac{E}{V} = \frac{q (6 {πε}_{0} r)}{4 {πε}_{0} 9 r^{2} (q)} = \frac{1}{6 r} \Rightarrow E = \frac{V}{6 r}$

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