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Charges on Conductors

Initially the...

Question

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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V_{A}

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V_{A} - V_{B}

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V_{B}

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Solution

The correct option is A $V_{A} - V_{B}$
Before grounding, let charges on $A$ and $B$ be $q_{A}$ and $q_{B}$ respectively, and radius be $r_{A}$ and $r_{B}$ respectively.
So, $V_{A} = k q_{A} / r_{A} + k q_{B} / r_{B}$
and, $V_{B} = k q_{A} / r_{B} + k q_{B} / r_{B}$
where $k = 1 / 4 π ϵ_{o}$

After grounding of $B$ , let charges on $A$ and $B$ be $q_{A}^{^{'}}$ and $q_{B}^{^{'}}$ respectively.
As $A$ is isolated, $q_{A}^{^{'}} = q_{A}$

Now, new potential on spheres are:
$V_{A}^{^{'}} = k q_{A}^{^{'}} / r_{A} + k q_{B}^{^{'}} / r_{B} = k q_{A} / r_{A} + k q_{B}^{^{'}} / r_{B}$
and, $V_{B}^{^{'}} = k q_{A}^{^{'}} / r_{B} + k q_{B}^{^{'}} / r_{B} = k q_{A} / r_{B} + k q_{B}^{^{'}} / r_{B} = 0 \Rightarrow q_{B}^{^{'}} = - q_{A}$

So, $V_{A}^{^{'}} = k q_{A} / r_{A} - k q_{A} / r_{B} = k q_{A} / r_{A} + k q_{B} / r_{B} - (k q_{A} / r_{B} + k q_{B} / r_{B}) = V_{A} - V_{B}$

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Initially the spheres A and B are at potential VA and VB respectively. Now sphere B is earthed by closing the switch. The potential of A will now become :

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 :