A capacitor of capacitance C is charged to a | vedclass

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(C) Initially,the capacitor has charges $+CV$ and $-CV$ on its plates. After adding a charge $+Q$ to the positive plate,the total charge on the system

Vedclass · Accepted Answer

$V + \frac{Q}{2C}$

Vedclass · Answer

(C) Initially,the capacitor has charges $+CV$ and $-CV$ on its plates. After adding a charge $+Q$ to the positive plate,the total charge on the system of two plates is $(+CV) + (-CV) + Q = Q$.When a charge is distributed on two parallel plates,the outer surfaces of the plates must have equal charges,each equal to half of the total charge. Therefore,the charge on the outer surface of each plate is $Q_{outer} = \frac{Q}{2}$.Using the conservation of charge on the positive plate:Charge on the inner surface of the positive plate = (Total charge on positive plate) - (Charge on outer surface) = $(CV + Q) - \frac{Q}{2} = CV + \frac{Q}{2}$.The charge on the inner surface of the negative plate will be equal and opposite,i.e.,$-(CV + \frac{Q}{2})$.The potential difference $V'$ across the capacitor is given by the charge on the inner surface divided by the capacitance $C$:$V' = \frac{CV + Q/2}{C} = V + \frac{Q}{2C}$.

$A$ 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:

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