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.
(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 Cℰ2 amount of energy.
