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
(1) V (2) $V+\frac{Q}{C}$ (3) $V+\frac{Q}{2C}$ (4) $V-\frac{Q}{C}$

Answer 1

Dear Student,


$$\begin{gathered} chargeonthecapacitor \\ q=CV \\ totalcharge=\frac{Q}{2}+q=\frac{Q}{2}+CV \\ PD=\frac{Q/2+CV}{C}=V+\frac{Q}{2C} \\ option3 \\ Regards \end{gathered}$$