A dielectric slab of thickness t and dielectric constant K is placed between the plates of a parallel-plate capacitor of separation d and area A. The new capacitance is:
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Question

A dielectric slab of thickness t and dielectric constant K is placed between the plates of a parallel-plate capacitor of separation d and area A. The new capacitance is:

G1 Physics · Accepted Answer

The air thickness is d−t and dielectric contributes equivalent thickness t/K. Hence dₑff=(d−t)+t/K = d−t(1−1/K). Therefore C=ε₀A/dₑff. Correct answer: D.