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Initially the spheres A and B are at potential $\;V_{A}\;$ and $\;V_{B}\;$ respectively . Now sphere B is earthed by closing the switch . The potential of A will now become


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Answer : (c) $\;V_{A}-V_{B}$
Explanation :
$V_{A}=\large\frac{k\;q_A}{a} + \large\frac{k\;q_B}{b}$
Now the charge on B becomes $\;q_{B^{|}}$
$V^{|}_{A}=\large\frac{k\;q_{A}}{a} + \large\frac{k\;q_{B^{|}}}{b}$
$V^{|}_{A}=\large\frac{k\;q_A}{a} - \large\frac{k\;q_B}{b}$
answered Feb 7, 2014 by yamini.v

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