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 :

• A. V
• B. $V+\frac{Q}{C}$
• C. $V+\frac{Q}{2C}$
• D. $V-\frac{Q}{C}$, if Q < CV

Answer 1

Answer: (C)

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 $q^{\prime }=q+Q/2$ and charge on negative plate $-q^{\prime }$.

Now the potential difference between the plates is: $V^{\prime }=\frac{q^{\prime }}{C}=\frac{q+Q/2}{C}=\frac{CV+Q/2}{C}=V+\frac{Q}{2C}$