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A metal sphere of radius $r_1$ is charged to potential $V_1$ and enclosed in thin spherical shell. If metal sphere is connected to shell by a conductor the potential of shell will be :

$(A)\;V_1 \\ (B)\;\frac{V_1r_2}{r_1} \\ (C)\; \frac{V_1r_1}{r_2} \\ (D)\;\frac{V_1r_1}{r_1+r_2} $

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1 Answer

Let the charge of metal sphere be $q_1$
$V_1= \large\frac{1}{4 \pi \in _0}$ $\frac{q_1}{r_1}$
=> $ q_1=V_1r_1(4 \pi \in _0)$
On connecting shell and metal sphere, entire charge will flow to the shell.
=> $V_2 =\large\frac{Kq_1}{r_2}=\frac{1}{4 \pi \in_0} \frac{V_1r_1}{r_2}$
$\qquad= \large\frac{V_1r_1}{r_2}$
Hence C is the correct answer.


answered Jan 8, 2014 by meena.p
edited Aug 1, 2014 by thagee.vedartham

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