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Two identical metal plates are given poi...

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

`(Q_(1)+Q_(2))/(2C)`

B

`(Q_(1)+Q_(2))/(C)`

C

`(Q_(1)-Q_(2))/(C)`

D

`(Q_(1)-Q_(2))/(2C)`

Text Solution

Verified by Experts

The correct Answer is:
D

Resultant field `=(Q_(1)-Q_(2))/(2in_(0)A),V=((Q_(1)-Q_(2))/(2in_(0)A))d=(Q_(1)-Q_(2))/(2C)`
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