Question

A capacitor of capacitance C is charged to a 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 1

Answer:

Explanation:

Initially charge on the capacitor is q=CV.

Thus, positive plate of the capacitor will get charge +q and negative plate get charge −q.

As battery is disconnected so charge is constant. Thus, when charge Q is given to positive plate, the will equally distribute into two plates. So the charge on positive plate becomes q

′

=q+Q/2 and charge on negative plate −q

′

.

Now the potential difference between the plates is: V

′

=

C

q

′

​

=

C

q+Q/2

​

=

C

CV+Q/2

​

=V+

2C

Q

​