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Charges Q0 and 2Q0 are given to parallel...

Charges `Q_0` and `2Q_0` are given to parallel plates A and B, respectively, and they are separated by a small distance. The capacitance of the given arrangement is C. Now, plates A and B are connected to positive and negative terminals of battery of potential difference `V = 2Q_0//C` respectively, as shown, then the work done by the battery is

A

`(2Q_(0)^(2))/C`

B

`(4Q_(0)^(2))/C`

C

`(5Q_(0)^(2))/C`

D

`(6Q_(0)^(2))/C`

Text Solution

Verified by Experts

The correct Answer is:
C
c. After connection the charge of capacitor `Q'=CV=C(2Q_(0))/(C)=2Q_(0)`
The final charge in inner plate of `A` will be `2Q_(0)`. Hence charge supplied by battery
`DeltaQ=(2Q-((-Q_(0))/(2)))=(5Q_(0))/(2)`
Hence work done by battery
`W=(DeltaQ)V=(5Q_(0))/(2)xx(2Q_(0))/(C)=(5Q_(0)^(2))/(C)`
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