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There are four concentric shells A , B , C ,D of radii a , 2a ,3a and 4a respectively . Shells B and D are given charges +q and -q respectively . Shell C is now earthed . The potential difference $\;V_{A}-V_{C}\;$ is

$(a)\;\large\frac{k\;q}{2a}\qquad(b)\;\large\frac{k\;q}{3a}\qquad(c)\;\large\frac{k\;q}{4a}\qquad(d)\;\large\frac{k\;q}{6a}$

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Answer : (d) $\;\large\frac{k\;q}{6a}$
Explanation :
$V_{C} = \large\frac{k\;q}{3a} +\large\frac{k\;q^{|}}{3a} -\large\frac{k\;q}{4a}$
$V_{C}=0$
$\large\frac{k\;q}{12a}+\large\frac{k\;q^{|}}{3a}=0$
$q^{|}=-\large\frac{q}{4}$
$V_{A}=\large\frac{k\;q}{2a}-\large\frac{\large\frac{k\;q}{4}}{3a}-\large\frac{k\;q}{4a}$
$V_{A}=\large\frac{k\;q}{4a}-\large\frac{k\;q}{12a}=\large\frac{k\;q}{6a}$
$V_{A}-V_{C}=\large\frac{k\;q}{6a}$
answered Feb 7, 2014 by yamini.v
 

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