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

Options

A. `V`
B. `V+Q/C`
C. `V+Q/(2C)`
D. `V-Q/C` if `QltCV`

Answer

C

Explanation

`q_1=q_4=q_("total")/2=(CV-CV+Q)/2=Q/2`
`q_2=(Q+CV)-Q/2=(Q/2+CV)`
`q_3=-q_2=-(Q/2+CV)`
Electric field between two plates and hence the potentiall difference is due to `q_2` and `q_3` only.
`PD=q_2/C=V+Q/(2C)`