A parallel plate air capacitor has capacity C and plate area A. Now the space between its plates is filled by two slabs of thickness d/2 each, having dielectric constants K₁ and K₂. Distance of separation between plates is d. The capacity of the capacitor is ( ε₀ = permittivity of free space) (a) (2ε₀ d)/(A)((K₁+K₂)/(K₁K₂)) (b) 2C((K₁K₂)/(K₁+K₂)) (c) (2ε₀ d)/(A)(K₁+K₂) (d) 2C((K₁+K₂)/(K₁K₂))

Answer: (b). Two layers of thickness d/2 are in series: C'=(ε₀A)/(d/2K₁+d/2K₂) C'=(2ε₀A)/(d)·(K₁K₂)/(K₁+K₂), with C=(ε₀A)/(d) So the answer is (b) 2C((K₁K₂)/(K₁+K₂)).

A parallel plate air capacitor has capacity C and plate area A. Now… | MHT CET 2026 (14 May, Shift 1)

Q18MHT CET 2026 · 14 MAY · SHIFT 1Medium

Dielectrics

A parallel plate air capacitor has capacity $C$ and plate area $A$ . Now the space between its plates is filled by two slabs of thickness $\frac{d}{2}$ each, having dielectric constants $K_1$ and $K_2$ . Distance of separation between plates is $d$ . The capacity of the capacitor is ( $\varepsilon_0$ = permittivity of free space)

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Answer: (b)

Two layers of thickness $\frac{d}{2}$ are in series: $C'=\frac{\epsilon_0A}{\frac{d}{2K_1}+\frac{d}{2K_2}}$
$C'=\frac{2\epsilon_0A}{d}\cdot\frac{K_1K_2}{K_1+K_2}$ , with $C=\frac{\epsilon_0A}{d}$
So the answer is (b) $2C\left(\frac{K_1K_2}{K_1+K_2}\right)$ .

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Dielectrics question from MHT CET 2026 (14 May, Shift 1)

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