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# There are two concentric spherical shells of radii r and 2r . Initially a charge Q is given to the inner shell . Now , switch $\;S_{1}\;$ is closed and opened then $\;S_{2}\;$ is closed and opened and the process is repeated n times for both the keys alternatively . Find the final potential difference between the shells

$(a)\;\large\frac{kQ}{2^{n} r}\qquad(b)\;\large\frac{kQ}{2^{n-1}r}\qquad(c)\;\large\frac{kQ}{r2^{n+1}}\qquad(d)\;\large\frac{kQ}{2r}$

Can you answer this question?

Answer : (c) $\;\large\frac{kQ}{r2^{n+1}}$
Explanation :
When $\;S_{1}\;$ is closed
Let charge $\;q_{1}\;$ comes from the earth on outer shell
$V_{outer}=0$
$k\;(\large\frac{Q}{2r}+\large\frac{q_{1}}{2r})=0\;$ $\quad \; [q_{i}\;$ is charge on outer sphere after i times ]
$q_{1}=-Q$
When $\;S_{2}\;$ is closed
$k\;(\large\frac{q_{i}^{|}}{r}-\large\frac{Q}{2r})=0$ $\quad\;[q_{i}^{|}\;$ is charge on inner sphere after i times]
$q_{i}^{|}=\large\frac{Q}{2}$
Now again when switch $\;S_{1}\;$ is closed
$k\;(\large\frac{Q}{2\times2r}+\large\frac{q_{2}}{2r})=0$
$q_{2}=\large\frac{Q}{2}$
Now when $\;S_{1}\;$ opened and $\;S_{2}\;$ closed
$k\;(\large\frac{q_{2}^{|}}{r}-\large\frac{\large\frac{Q}{2}}{2r})=0$
Proceeding similarly we get after n times
$q_{n}=-\large\frac{Q}{2^{n-1}} \quad$ & $\;q_{n}^{|}=\large\frac{Q}{2^{n}}$
Potential difference between shells after n times
$V_{AB}=k\;q_{n}^{|}\;[\large\frac{1}{r}-\large\frac{1}{2r}]$
$V_{AB}=\large\frac{kQ}{2^{n}\times2}=\large\frac{kQ}{2^{n+1} r}\;.$
answered Feb 18, 2014 by

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