For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is 3/4 d, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C0) is given by the following relation: (a) C' = 3+K/4KC₀ (b) C' = 4+K/3C₀ (c) C' = 4K/K+3C₀ (d) C' = 4/3+KC₀

Answer: (c). C'=(ε₀A)/((d-t)+t/K) with t=3d/4. C'=(ε₀A)/(d/4+3d/4K)=(ε₀A)/(d)·4K/K+3 So the answer is (c) C'=4K/K+3C₀.

For changing the capacitance of a given parallel plate capacitor, a… | JEE Main 2021 (16 Mar, Shift 1)

Q80JEE Main 2021 · 16 MAR · SHIFT 1Medium

Dielectrics

For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4}$ d, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C0) is given by the following relation:

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Answer: (c)

$C'=\frac{\epsilon_0A}{(d-t)+\frac{t}{K}}$ with $t=\frac{3d}{4}$ .
$C'=\frac{\epsilon_0A}{\frac{d}{4}+\frac{3d}{4K}}=\frac{\epsilon_0A}{d}\cdot\frac{4K}{K+3}$
So the answer is (c) $C'=\frac{4K}{K+3}C_0$ .

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Dielectrics question from JEE Main 2021 (16 Mar, Shift 1)

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