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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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$\frac{V}{2 r}$

$\frac{V}{3 r}$

$\frac{V}{6 r}$

$\frac{V}{4 r}$

answer is C.

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Detailed Solution

$v = \frac{q}{4 π E_{0} r} (1 - \frac{1}{3}) = \frac{q}{6 π E_{0} r} E = \frac{k q}{{(3 r)}^{2}} ∴ \frac{E}{v} = \frac{1}{6 r}$

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