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A capacitor of capacitance C is connecte...

A capacitor of capacitance C is connected to a cell of emf V and when fully charged, it is disconnected. Now the separation between the plates is doubled. The change in flux of electric field through a closed surface enclosing the capacitor is

A

Zero

B

`(CV)/(epsilon_(0))`

C

`(CV)/(2epsilon_(0))`

D

`(2CV)/(epsilon_(0))`

Text Solution

Verified by Experts

Flux = `(q_("in"))/(2epsilon_(0))`
The two plates of the capacitor have equal and opposite charges.
Hence, net charge enclosed by the given surface = 0
`therefore` Flux is zero in both cases.
Hence change in flux = 0
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