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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

AI Generated Solution

To solve the problem step by step, we will analyze the situation involving a capacitor, its capacitance, and the electric field. ### Step 1: Understand the Initial Conditions - A capacitor of capacitance \( C \) is connected to a cell with an electromotive force (emf) \( V \). - When fully charged, the charge \( Q \) on the capacitor can be calculated using the formula: \[ Q = C \cdot V \] ...
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