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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. $\;q_{2}\;$ is

$(a)\;Q\qquad(b)\;Q (\large\frac{a}{b})\qquad(c)\;Q (\large\frac{b}{a})\qquad(d)\;zero$

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Answer : (b) $\;Q (\large\frac{a}{b})$
Explanation :
After earthing the solid sphere
$V_{a}^{|}=0$
$V_{a}^{|}=\large\frac{kq_{2}}{a}+\large\frac{kq_{1}}{b}$
$=\large\frac{kq_{2}}{a}-\large\frac{kQ}{b}=0$
$q_{2}=\large\frac{Qa}{b}\;.$
answered Feb 20, 2014 by yamini.v
 

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