A capacitor with capacitance C is charged to the potential difference (V) from a cell and then disconnected from it. A charge + Q is now given to its positive plate. The potential difference across the capacitor is now

Answer: 

The initial charge of the capacitor (q)= CV

Therefore, the positive plate of the capacitor will get a charge of +q while the negative plate will get -q

As the battery is disconnected the charge will remain constant. 

By applying charge Q to the positive plate, it will distribute uniformly between the two plates.

Thus, the charge on the positive plate

q’ = q+ Q/2 

and charge on the negative plate,

-q’