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A hollow charged metal sphere has radius r. If the potential difference between its surface and a point at distance 2r from the centre is $ν$ , then the electric field intensity at distance 3r from the centre is

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$\frac{v}{6 r}$

$\frac{v}{4 r}$

$\frac{v}{9 r}$

$\frac{2 v}{9 r}$

answer is D.

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Detailed Solution

$V = \frac{q}{4 π ε_{0}} (\frac{1}{r} - \frac{1}{2 r}) = \frac{1}{4 π ε_{0}} \frac{1}{2 r}$

$E = \frac{1}{4 π ε_{0}} \frac{q}{9 r^{2}} \Rightarrow E = \frac{2 V}{9 r}$

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