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MHT CET 2023 (10 May, Shift 2) · Dielectrics

Dielectrics question from MHT CET 2023 (10 May, Shift 2)

Q58MHT CET 2023 · 10 MAY · SHIFT 2Medium

Dielectrics

Two dielectric slabs having dielectric constant ' $K_1$ ' and ' $K_2$ ' of thickness $\frac{d}{4}$ and $\frac{3d}{4}$ are inserted between the plates as shown in figure. The net capacitance between $A$ and $B$ is $[\varepsilon_0$ is permittivity of free space]

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Answer: (d)

Layers between A and B are in series: $C=\frac{A\epsilon_0}{\frac{d}{4K_1}+\frac{3d}{4K_2}}$
$\frac{d}{4K_1}+\frac{3d}{4K_2}=\frac{d(K_2+3K_1)}{4K_1K_2}$
$C=\frac{4A\epsilon_0}{d}\left[\frac{K_1K_2}{3K_1+K_2}\right]$ . So the answer is (d).

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