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Question

Capacitance of a parallel plate capacitor becomes 43 times its original value if a dielectric slab of thickness t=d2 is inserted between the plates [d is the separation between the plates]. The dielectric constant of the slab is

A
4
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B
8
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C
2
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D
6
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Solution

The correct option is C 2
Original value of capacitance be Co=ϵ0Ad
Let dielectric constant of the slab is K.

Now, Consider two capacitor C1(with dielectric slab) and C2(without dielectric slab) in series.
C1=Kϵ0Ad2=2Kϵ0Ad
C2=ϵ0Ad2=2ϵ0Ad
Ceq=C1C2C1+C2 =(2KK+1)×ϵ0Ad =(2KK+1)×Co

Given:
Ceq=43Co
2KK+1=43
K=2

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