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A solid metallic sphere of radius a is surrounded by a conducting spherical shell of radius b $\;(b >a)\;$ . The solid sphere is given a charge Q . Potential at the surface of the solid sphere is $\;V_{a}\;$ and at the surface of spherical shell is $\;V_{b}\;$. If we put -4Q charge on the shell the potential differance between the sphere and shell becomes $\;\bigtriangleup V\;.$ Now we earthed the outer shell and the charge on the ouer shell is $\;q_{1}\;$.Now we removed the earthing connection from the shell and earthed the inner solid sphere. Charge on solid sphere becomes $\;q_{2}\;$. Answer the following question based on the passage.Now if we connected the shell and solid sphere after removing earthing from the solid sphere charge on outer surface of shell becomes.


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Answer : (c) $\;\large\frac{Q(a-b)}{b}$
Explanation :
Initially before connecting charge on solid sphere $\;q_{2}=\large\frac{Qa}{b}$
Charge on shell = $\;-Q$
After connecting let q amount of charge flow from solid sphere to shell .
Since they are connected
Therefore now charge on solid sphere = $\;\large\frac{Qa}{b}-q=0$
Charge on outer shell = $\;\large\frac{Qa}{b}-Q$
Distribution of charges is shown in figure .
answered Feb 20, 2014 by yamini.v
edited Feb 20, 2014 by yamini.v

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