[\dfrac{2}{k} = \dfrac{3}...] A parallel– plate capacitor of area A, plate separation d and capacitance C is filled with four diel

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A parallel– plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constant k₁, k₂, k₃ and k₄ as shown in the figure below. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by parallel plate capacitor

$\dfrac{1}{k} = \dfrac{1}{k_1} + \dfrac{1}{k_2} +\dfrac{1}{k_3}+ \dfrac{3}{2k_4}$

$k = k_1 + k_2 + k_3 + 3k_4$

$k = \dfrac{2}{3}(k_1 + k_2 + k_3) + 2k_4$

$\dfrac{2}{k} = \dfrac{3}{(k_1 + k_2 + k_3)} + \dfrac{1}{k_4}$

Explanation:
Ans: 4 or bonus parallel plate capacitor explanation

$\dfrac{1}{C_1} + \frac{3}{C_4} = \frac{3d}{2k_1\epsilon_0A} + \dfrac{3d}{2k_4\epsilon_0A} = \dfrac{3d}{2\epsilon_0A}\left\{\dfrac{1}{k_1} + \dfrac{1}{k_4}\right\}$

$C_{eq} = \dfrac{K\epsilon_0A}{d} = \dfrac{2\epsilon_0A}{3d}\left\{\dfrac{k_1k_4}{k_1 + k_4} + \dfrac{k_2k_4}{k_2 + k_4} + \dfrac{k_3k_4}{k_3 + k_4}\right\}$

$k = \dfrac{2}{3}\left\{\dfrac{k_1k_4}{k_1 + k_4} + \dfrac{k_2k_4}{k_2 + k_4} + \dfrac{k_3k_4}{k_3 + k_4}\right\}$

Alter:

Wrong solution seems to be correct as per the options given.

parallel plate capacitor explanation part 2

$\dfrac{2}{3}\dfrac{A}{D}\epsilon_0 = \dfrac{2A}{3}\dfrac{\epsilon_0}{d}(k_1 + k_2 + k_3) = C_A$

$\dfrac{2}{3}\dfrac{A}{D}\epsilon_0 = \dfrac{A\epsilon_0}{d}k_4 = C_B$

$\dfrac{1}{C_{eq}} = \dfrac{1}{C_A} + \dfrac{1}{C_B}$

$\dfrac{d}{A\epsilon_0K_{eq}} = \dfrac{3d}{2A\epsilon_0}(k_1 + k_2 + k_3) + \dfrac{d}{2A\epsilon_0k_4}$

$\dfrac{1}{K_{eq}} = \dfrac{3}{2}(k_1 + k_2 + k_3) + \dfrac{1}{2k_4}$

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