In an isolated parallel plate capacitor of capacitance `C`, the plates have charge `Q_(1)` and `Q_(2)`. The potential difference between the plates is

Options

A. `(Q_(1)+Q_(2))/(2C)`
B. `(Q_(1)-Q_(2))/(2C)`
C. `(Q_(1))/(2C)`
D. `(-Q_(2))/(2C)`

Answer

B

Explanation

The outer surface of plates will have charges `(Q_(1)+Q_(2))/(2)`.Then charge distribution will be
Facing surfacing form a capacitor,
`p.d. V_(1)-V_(2)=V=((Q_(1)-Q_(2))//2)/(c)`
`V=(Q_(1)-Q_(2))/(2C)`