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Two conducting concentric , hollow sphere A and B have radii a and b respectively with A inside B . Their common potentials is V . A is now given some charge such that its potential becomes zero . The potential of B will now be

$(a)\;0\qquad(b)\;V (1-\large\frac{a}{b})\qquad(c)\;\large\frac{V_{a}}{b}\qquad(d)\;\large\frac{V_{b}}{a}$

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Answer : (b) $\;V (1-\large\frac{a}{b})$
Explanation :
Let the charge on sphere A is $\;q_{A}\;$ charge on sphere B is $\;q_{B}\;$ initially.
Then , $\;V_{A}=\large\frac{k\;q_{A}}{a}+\large\frac{k\;q_{B}}{b}$
and $\;V_{B}=\large\frac{k\;q_{A}}{b}+\large\frac{k\;q_{B}}{b}$
But $\;V_{A}=V_{B}=V \quad\;\large\frac{k\;q_{A}}{a}+\large\frac{k\;q_{B}}{b}=\large\frac{k\;q_{A}}{b}+\large\frac{k\;q_{B}}{b}$
$q_{A}=0$
Now let the charge on A becomes q such that ite potential becomes zero
Then $\;V_{A}=\large\frac{k\;q}{a}+\large\frac{k\;q_{B}}{b}=0$
$q=-\large\frac{q_{B a}}{b}$
$V_{B}=\large\frac{k\;q}{b}+\large\frac{k\;q_{B}}{b}=\large\frac{k\;q_{B}}{b}-\large\frac{k\;q_{B}\;a}{b^2}$
$V_{B}=\large\frac{k\;q_{B}}{b}\;(1-\large\frac{a}{b})$
By $\;\large\frac{k\;q_{B}}{b}=V$
Therfore $\;V_{B}=V\;(1-\large\frac{a}{b})\;.$
answered Feb 18, 2014 by yamini.v
 

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