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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

$(a)\;0\qquad(b)\;V_{A}\qquad(c)\;V_{A}-V_{B}\qquad(d)\;V_{B}$

Answer : (c) $\;V_{A}-V_{B}$
Explanation :
$V_{A}=\large\frac{k\;q_A}{a} + \large\frac{k\;q_B}{b}$
$V_{B}=\large\frac{k\;(q_{A}+q_{B})}{b}$
Now the charge on B becomes $\;q_{B^{|}}$
$V^{|}_{B}=\large\frac{k\;(q_{A}+q_{B^{|}})}{b}=0\quad\;q_{B^{|}}=-q_{A}$
$V^{|}_{A}=\large\frac{k\;q_{A}}{a} + \large\frac{k\;q_{B^{|}}}{b}$
$V^{|}_{A}=\large\frac{k\;q_A}{a} - \large\frac{k\;q_B}{b}$
$=\large\frac{k\;q_{a}}{ab}\;(b-a)$
$V^{|}_{A}=V_{A}-V_{B}\;.$