A solid conducting sphere having a charge 'Q' is surrounded by an uncharged concentric conducting hollow spherical shell. Let the P.D 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 –3Q. The new PD between the same two surfaces is

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-2V

answer is A.

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Detailed Solution

Let ‘a’ and ‘b’ be the radii of the sphere and shellV=q4πε01a-1b When charge ‘–3Q’ is given to the shellVsphere=14πε0qa+-3qb Vshell=14πε0qb+-3qb ∴Vsphere-Vshell=14πε0qa-qb=V

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