A metal sphere ‘A’ of radius ‘a’ is charged to a potential ‘V’. What will be its potential if its enclosed by spherical conducting shell B of radius ‘b’ and connected by a wire?

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$\frac{a^{2}}{b} (V)$

$\frac{a}{b} (V)$

$\frac{ab}{V}$

$\frac{b}{a} (V)$

answer is C.

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Detailed Solution

$V = \frac{KQ}{a} \Rightarrow KQ = Va$

When spherical shell B is connected to the sphere A by a wire, all charge always reside on the outer surface.

therefore, potential will be: $V^{'} = \frac{KQ}{b} = \frac{Va}{b}$

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