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A solid conducting sphere, having a charge Q, is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of -4Q, the new potential difference between the same two surfaces is

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-2V

answer is D.

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Detailed Solution

When charge Q is on inner solid conducting sphere Electric field between spherical surface $E = \frac{KQ}{r^{2}} so, \int \bar{E} \cdot \bar{dr} = V given$

Now when a charge -4Q is given to hollow shell Electric filed between surface remain

unchanged $\bar{E} = \frac{KQ}{r^{2}}$

As filed inside the hollow spherical shell =0

$∴$ potential difference between them remain unchanged $i.e \int \bar{E} \bar{dr} = V$

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