The two plates $x$ and $y$ of a parallel plate capacitor of capacitance $C$ are given a charge of amount $Q$ each. $x$ is now joined to the positive terminal and $y$ to the negative terminal of a cell of emf $V = \frac{Q}{C}$ , then select the correct statements

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The total charge on the plate $x$ will be $2 Q$

The energy supplied by the cell is $C V^{2}$

The total charge on the plate $y$ will be zero

A charge of amount $Q$ will flow from the posistive terminal to the negative terminal of the cell through the capacitor

answer is A, B, C, D.

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Detailed Solution

Let the charge flowing from the positive terminal to the negative terminal of the battery be $q$ , then

$q = C V = C (\frac{Q}{C}) = Q$

Total charge on plate $x$ is $Q + q = Q + Q = 2 Q$

Total charge on plate $y$ is $Q - q = Q - Q = 0$

The energy supplied by the cell is $W_{b a t t e r y} = (\frac{Q}{C}) q = \frac{Q^{2}}{C} = C V^{2}$

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