Question

Two identical metal plates are given poistive charges `Q_1` and `Q_2` `(ltQ_1)` respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C, the potencial difference between them is
A. (a) `(Q_1+Q_2)//(2C)`
B. (b) `(Q_1+Q_2)//C`
C. (c) `(Q_1-Q_2)//C`
D. (d) `(Q_1-Q_2)//(2C)`

Answer 1

Correct Answer - D
Within the capacitor,
`E_1=(Q_1)/(2epislon_0A)`, `E_2=(Q_2)/(2epislon_0A)`
where A= area of each plate
d= separation between two plate
`E=E_1-E_2=(1)/(2epislon_0A)(Q_1-Q_2)`
Hence, `V=Ed`
`=1/2(d)/(epislon_0A)(Q_1-Q_2)=(Q_1-Q_2)/(2C)`