Question

The two plate X and Y of a parallel-plates 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 $\epsilon =Q/C$.

• A. Charge of amount Q will flow from the positive terminal to the negative terminal of the cell through the capacitor.
• B. The total charge on the plate X will be $2Q$.
• C. The total charge on the plate Y will be zero.
• D. The cell will supply $C{\epsilon }^2$amount of energy.

Answer 1

Answer: (A), (B), (C)

The correct options are
A Charge of amount Q will flow from the positive terminal to the negative terminal of the cell through the capacitor.
B The total charge on the plate X will be $2Q$.
C The total charge on the plate Y will be zero.

As the emf of cell is $ϵ=Q/C,$ charge of amount Q will flow from the positive terminal to the negative terminal of the cell through the capacitor.
As the Y is connected to negative terminal of cell so it is ground and the total charge on plate Y will be zero.
Here cell charge the capacitor plate X and its total charge $=(Q+Q)=2Q$
The energy will supply by cell is $U=\frac{1}{2}C{ϵ}^2$
Ans: (A),(B),(C)