Browse Questions

# 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.

$(a)\;\large\frac{Q(a+b)}{(a-b)}\qquad(b)\;\large\frac{Qa^2}{b}\qquad(c)\;\large\frac{Q(a-b)}{b}\qquad(d)\;-\large\frac{Qb}{a}$

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 .
Then
$V_{a}=\large\frac{k(\large\frac{Qa}{b}-q)}{a}+\large\frac{k(q-Q)}{b}$
$V_{a}=\large\frac{kQ}{b}-\large\frac{kq}{a}+\large\frac{kq}{b}-\large\frac{kQ}{b}$
$V_{a}=\large\frac{kq(a-b)}{ab}$
$V_{b}=\large\frac{k(\large\frac{Qa}{b}-q)}{b}+\large\frac{k(q-Q)}{b}$
Since they are connected
$V_{a}=V_{b}$
$\large\frac{Qa}{b}-q=0$
$q=\large\frac{Qa}{b}$
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 .
edited Feb 20, 2014 by yamini.v