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Charging of Capacitors

The work done...

Question

The work done in taking a unit positive charge from $P$ to $A$ is $W_{A}$ and from $P$ to $B$ is $W_{B}$ . Then :

W_{A} > W_{B}

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W_{A} < W_{B}

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W_{A} = W_{B}

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W_{A} + W_{B}

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Solution

The correct option is B $W_{A} = W_{B}$
As potential at $A$ and $B$ is the same $V_{A} = V_{B} = k Q / d$ .
and $W = q Δ V$
$Δ V$ in both the cases is same. So work done in both the cause will be the same.

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