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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.$\;q_{1}\;$ is

$(a)\;-Q\qquad(b)\;-Q\;(\large\frac{a}{b})\qquad(c)\;-4Q\qquad(d)\;zero$

Can you answer this question?
 
 

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

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