A slab of material of dielectric constant k has the same area as the plates of the parallel plate capacitor but has thickness 3d/4, where d is the separation of the plates. How is the capacitance changed, when the slab is inserted between the plates? ( C = original capacitance) (a) (4kC)/((k+3)) (b) ((k+3)C)/(4k) (c) ((k-3)C)/(4k) (d) (4kC)/((k-3))

Answer: (a). C' = (ε₀ A)/(d - t + t/k) with t = 3d/4. d - t + t/k = d/4 + 3d/4k = (d(k+3))/(4k). C' = 4k/k+3·(ε₀ A)/(d) = (4kC)/(k+3). So the answer is (a).

A slab of material of dielectric constant k has the same area as the… | MHT CET 2026 (13 May, Shift 2)

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Dielectrics

A slab of material of dielectric constant $k$ has the same area as the plates of the parallel plate capacitor but has thickness $\frac{3d}{4}$ , where $d$ is the separation of the plates. How is the capacitance changed, when the slab is inserted between the plates? ( $C$ = original capacitance)

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Answer: (a)

$C' = \frac{\epsilon_0 A}{d - t + \frac{t}{k}}$ with $t = \frac{3d}{4}$ .
$d - t + \frac{t}{k} = \frac{d}{4} + \frac{3d}{4k} = \frac{d(k+3)}{4k}$ .
$C' = \frac{4k}{k+3}\cdot\frac{\epsilon_0 A}{d} = \frac{4kC}{k+3}$ .
So the answer is (a).

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Dielectrics question from MHT CET 2026 (13 May, Shift 2)

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