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A capacitor C is charged to a potential ...

A capacitor `C` is charged to a potential difference `V` and battery is disconnected. Now if the capacitor plates are brough close slowely by some distance:

A

some `+ve` work is done by external agent

B

enegry of capacitor will decrease

C

enegry of capacitor will increase

D

none of the above

Text Solution

Verified by Experts

The correct Answer is:
B
`Q =` constant so `U_(i) = (Q^(2))/(2C) & U_(f) = (Q^(2))/(2KC)`
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