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Two concentric shells have radii R & 2R , charges $\;q_{A}\;and\;q_{B}\;$ and potential 2V and $\;\large\frac{3}{2} V\;$ respectively . Now shell B is earthed and let charges on them becomes $\;q_{A}^{|}\;and\;q_{B}^{|}\;.$ Then $\;\large\frac{q_{A}^{|}}{q_{A}}+\large\frac{q_{B}^{|}}{q_{B}}\;$ is

$(a)\;-\large\frac{1}{2}\qquad(b)\;1\qquad(c)\;\large\frac{1}{2}\qquad(d)\;-1$

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Answer : (c) $\;\large\frac{1}{2}$
Explanation :
$2V=\large\frac{k q_{A}}{R}+\large\frac{k q_{B}}{2R}$
$\large\frac{3V}{2}=\large\frac{k q_{A}}{2R}+\large\frac{k q_{B}}{2R}$
Substracting these two we get
$\large\frac{V}{2}=\large\frac{k\;q_{A}}{2R}$
$\large\frac{3V}{2}=\large\frac{V}{2}+\large\frac{k\;q_{B}}{2R}$
$\large\frac{k\;q_{B}}{2R}=V$
$\large\frac{q_{A}}{q_{B}}=\large\frac{1}{2} \quad \; q_{B}=2 q_{A}--(1)$
Now after earthing outer shell
$q_{A}^{|}=q_{A}$
And $\;V_{B}=0$
$\large\frac{kq_{A}^{|}}{2R}+\large\frac{k q_{B}^{|}}{2R}=0$
$\large\frac{q_{A}^{|}}{q_{B}^{|}}=-1$
$q_{B}^{|}=-q_{A}^{|}==-q_{A}$
$q_{B}^{|}=-q_{A}$
From equation (1) $\;q_{A}=\large\frac{q_{B}}{2}$
$q_{B}^{|}=-\large\frac{q_{B}}{2}$
$\large\frac{q_{B}^{|}}{q_{B}}=-\large\frac{1}{2}$
and $\;\large\frac{q_{A}^{|}}{q_{A}}=1$
$\large\frac{q_{A}^{|}}{q_{A}}+\large\frac{q_{B}^{|}}{q_{B}}=\large\frac{1}{2}$
answered Feb 19, 2014 by yamini.v
 

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