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Two identical sheets of a metallic foil ...

Two identical sheets of a metallic foil are separated by `d` and capacitance of the system is `C` and charged to a potential difference `E`. Keeping the charge constant, the separation is increased by `'l'`. Then, the new capacitance and potential difference will be

A

`(epsilon_(0)A)/(d),E`

B

`(epsilon_(0)A)/((d+l)),E`

C

`(epsilon_(0)A)/((d+l)),(1+(l)/(d))E`

D

`(epsilon_(0)A)/(d),(1+(l)/(d))E`

Text Solution

Verified by Experts

`q=CV=C_(1)V_(1)`
where `C=(epsilon_(0)A)/(d)andC_(1)=(epsilon_(0)A)/((d+l))`
`q=(epsilon_(0)AE)/(d)=(epsilon_(0)AE_(1))/((d+l))`
`therefore (epsilon_(0)A)/(d)E=(epsilon_(0)A)/((d+l))E_(1)`
`therefore E_(1)=((d+l))/(d)E=(1+(l)/(d))E`
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