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

Question

Infinity Learn · Accepted Answer

Let ‘a’ and ‘b’ be the radii of the sphere and $$shellV=q4$$πε$$01a-1b$$ When charge ‘–$$3Q$$’ is given to the $$shellVsphere=14$$πε$$0qa+-3qb$$ $$Vshell=14$$πε$$0qb+-3qb$$ ∴$$Vsphere-Vshell=14$$πε$$0qa-qb=V$$

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