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Q.A parallel plate capacitor has plate area AA and separation dd. A metallic slab of thickness t<dt < d is inserted parallel to the plates. The capacitance becomes

a
ϵ0A/(dt)\epsilon_0 A / (d - t)
b
ϵ0A/d\epsilon_0 A / d
c
ϵ0At/d\epsilon_0 A t / d
d
ϵ0A(dt)/d\epsilon_0 A (d - t) / d

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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