The two plates X and Y of a parallel plate ca | vedclass

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(D) Initially,both plates $X$ and $Y$ have a charge $Q$. The capacitor has capacitance $C$. When connected to a cell of $emf$ $E = Q/C$,the potential

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(D) Initially,both plates $X$ and $Y$ have a charge $Q$. The capacitor has capacitance $C$. When connected to a cell of $emf$ $E = Q/C$,the potential difference across the plates becomes $V = Q/C$.For plate $X$ connected to the positive terminal,the final charge $Q_X$ becomes $Q + Q = 2Q$.For plate $Y$ connected to the negative terminal,the final charge $Q_Y$ becomes $Q - Q = 0$.Since the capacitor is being charged by the cell,a charge of $Q$ flows from the positive terminal of the cell to the plate $X$ and from the plate $Y$ to the negative terminal of the cell. Inside the cell,this charge $Q$ flows from the negative terminal to the positive terminal.

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 $emf$ $E = Q/C$.

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