Air capacitor has capacitance C₁. The space between two plates of capacitor is filled with two dielectric as shown in figure. The new capacitance of the capacitor is C₂. Then (C₁)/(C₂) is [ d = distance between two plates, K₁ and K₂ are dielectric constants of first and second dielectrics respectively] (a) 11/30 (b) 8/30 (c) 7/30 (d) 13/30

Answer: (b). The two slabs, each d/2 thick, are in series. C₂=(ε₀A)/(d/2(5)+d/2(3))=(ε₀A)/(d)·60/16=15/4C₁ (C₁)/(C₂)=4/15=8/30 So the answer is (b) 8/30.

Air capacitor has capacitance C₁. The space between two plates of… | MHT CET 2026 (20 May, Shift 1)

Q17MHT CET 2026 · 20 MAY · SHIFT 1Medium

Dielectrics

Air capacitor has capacitance $C_1$ . The space between two plates of capacitor is filled with two dielectric as shown in figure. The new capacitance of the capacitor is $C_2$ . Then $\frac{C_1}{C_2}$ is [ $d$ = distance between two plates, $K_1$ and $K_2$ are dielectric constants of first and second dielectrics respectively]

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Answer: (b)

The two slabs, each $\frac{d}{2}$ thick, are in series.
$C_2=\frac{\epsilon_0A}{\frac{d}{2(5)}+\frac{d}{2(3)}}=\frac{\epsilon_0A}{d}\cdot\frac{60}{16}=\frac{15}{4}C_1$
$\frac{C_1}{C_2}=\frac{4}{15}=\frac{8}{30}$
So the answer is (b) $\frac{8}{30}$ .

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Dielectrics question from MHT CET 2026 (20 May, Shift 1)

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