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 C2 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