The two plates of a parallel plate capacitor | vedclass

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Question

(C) Initially,the charges on the inner and outer surfaces of the plates are determined by the total charge. Let the plates be $A$ and $B$ with charges

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Potential difference across the capacitor plates is $\frac{Q_1}{C}$.

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(C) Initially,the charges on the inner and outer surfaces of the plates are determined by the total charge. Let the plates be $A$ and $B$ with charges $Q_1$ and $Q_2$ respectively.When the switch $(S)$ is closed,plate $B$ is connected to the earth,so its potential becomes $0 \ V$.The charge on the inner surface of plate $A$ will be $Q_{in} = \frac{Q_1 - Q_2}{2}$ and on the inner surface of plate $B$ will be $-Q_{in} = -\frac{Q_1 - Q_2}{2}$.However,a simpler way to view this is that the charge on the inner face of plate $A$ becomes $Q_1$ (if we consider the system as a capacitor with plate $A$ at potential $V$ and plate $B$ at $0 \ V$).Specifically,when plate $B$ is earthed,the charge on its outer surface becomes $0$. The charge on the inner surface of plate $B$ becomes $-Q_1$ to balance the charge $+Q_1$ on plate $A$.The total charge initially on plate $B$ was $Q_2$. After earthing,the final charge on plate $B$ is $-Q_1$.Therefore,the charge that flows to the earth is $\Delta Q = Q_{initial} - Q_{final} = Q_2 - (-Q_1) = Q_1 + Q_2$.The potential difference across the capacitor is $V = \frac{Q_1}{C}$.

The two plates of a parallel plate capacitor are given charges $Q_1$ and $Q_2$. The capacity of the capacitor is $C$. When the switch $(S)$ is closed,mark the correct statement. (Assume both $Q_1$ and $Q_2$ to be positive)

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