Fri December 28, 2012 By: Meghana Anil

The Figure below shows two identical Capacitors, C1 and C2 , each of 1microfarad capacitance, connected to a battery of 6V . Initially switch 'S ' is closed . After sometime 'S ' is left open and dielectric constant K=3 are inserted to fill completely the space between the plates of the two capacitors. How will the (1)charge and (2)potential difference between the plates of the capacitor be affected after the slabs are inserted?

Expert Reply
Fri December 28, 2012

We know that by inserting a dielectric of K >1 inside the parallel plates of a capacitor the capacitance will increase, the new capacitance will be given as

C' = C0K

here C0 is the original capacitance


C' = 3C0



the charge on the capacitor plates will not be affected when the battery is disconnected (open switch).



V0 = q/C0

the new potential difference will be

V' = q/C'


V' = q/ 3C0


V' = (1/3).V0

so, the potential difference will decrease threefold

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