Question
A capacitor of capacitance $C$ is charged to a potential difference $V$ from a cell and then disconnected from it. A charge $+Q$ is now given to its positive plate. The potential difference across the capacitor is now $Q$
if $V < CV$
✓
Answer
After redistribution half of total charge remains on outer surface and then apply
conservation of charge on each plate.
Charge on outer plates $=\frac{Q+C V-C V}{2}=\frac{Q}{2}$
Charge of innerface of first plate,
$=Q+C V-\frac{Q}{2}=\frac{Q}{2}+C V$
Charge on innerface of second plate $=-\left(\frac{Q}{2}+C V\right)$
$V^{\prime}=\frac{Q / 2+C V}{C}=V+\frac{Q}{2 C}$
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