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MCQ · Q2

Q.A slab of material of dielectric constant k has the same area A as the plates of a parallel plate capacitor and has thickness (3/4)d, where d is the separation of the plates. The change in capacitance when the slab is inserted between the plates is (A), (B), (C), (D) -- four algebraic options in ε0A/d and k, printed in the source but not reliably recoverable letter-by-letter from the extracted text.

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Using the general dielectric-slab formula from section 8.10.2, C=ϵ0A(d−t)+t/kC=\dfrac{\epsilon_0A}{(d-t)+t/k}, with t=34dt=\tfrac{3}{4}d: d−t=d4d-t=\tfrac{d}{4} and t/k=3d4kt/k=\tfrac{3d}{4k}, so the denominator is d4+3d4k=d4(1+3k)=d(k+3)4k\dfrac{d}{4}+\dfrac{3d}{4k}=\dfrac{d}{4}\left(1+\dfrac{3}{k}\right)=\dfrac{d(k+3)}{4k}. Hence C′=ϵ0Ad(k+3)/4k=4kϵ0Ad(k+3)C'=\dfrac{\epsilon_0A}{d(k+3)/4k}=\dfrac{4k\epsilon_0A}{d(k+3)} -- this is the new capacitance with the slab inserted. The CHANGE in capacitance the question asks for is ΔC=C′−C0=4kϵ0Ad(k+3)−ϵ0Ad=ϵ0Ad[4kk+3−1]=ϵ0Ad×4k−(k+3)k+3=3ϵ0A(k−1)d(k+3)\Delta C=C'-C_0=\dfrac{4k\epsilon_0A}{d(k+3)}-\dfrac{\epsilon_0A}{d}=\dfrac{\epsilon_0A}{d}\left[\dfrac{4k}{k+3}-1\right]=\dfrac{\epsilon_0A}{d}\times\dfrac{4k-(k+3)}{k+3}=\dfrac{3\epsilon_0A(k-1)}{d(k+3)}. The four lettered options as printed in the extracted source text carry fraction fragments ("3/4", "2/3", "3/2", "4/3") that the PDF-to-text extraction corrupted too badly to reconstruct a reliable letter-by-letter match against this derived formula, so the exact lettered choice is left unconfirmed here rather than guessed (per this platform's honesty rule) -- but the new capacitance C′=4kϵ0A/[d(k+3)]C'=4k\epsilon_0A/[d(k+3)] and the resulting change ΔC=3ϵ0A(k−1)/[d(k+3)]\Delta C=3\epsilon_0A(k-1)/[d(k+3)] are both derived directly and correctly from the section 8.10.2 formula. [!ANSWER] C' = 4kε0A / [d(k+3)]; ΔC = 3ε0A(k-1) / [d(k+3)].

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