Question

in figure two identical capacitors C1 and C2 each of 1 microfarad capacitors connected to a battery of 6V, initially ,switch S is closed. after some time,S is Left Open and dielectric slab of dielectric constant equal to 3 are inserted to fill completely the space between the plates of two capacitor how will charge and potential difference between the Plates of capacitor be affected after the slabs are inserted

Answer 1

heya friend!!☺☺
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here's your answer!!☺☺

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

so,

C' = 3C0

(a)

now,

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

(b)

initially

V0 = q/C0

the new potential difference will be

V' = q/C'

or

V' = q0 / 3C0

thus,

V' = (1/3).V0

so, the potential difference will decrease threefold.

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hope it helps you!!☺☺