In the given figure,a capacitor is formed by | vedclass

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(B) The capacitor consists of three dielectric slabs in series with thicknesses $d_1 = d$,$d_2 = 2d$,$d_3 = 3d$ and dielectric constants $K_1 = K$,$K_

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$\frac{15}{34} \frac{K \varepsilon_{0} A}{d}$

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(B) The capacitor consists of three dielectric slabs in series with thicknesses $d_1 = d$,$d_2 = 2d$,$d_3 = 3d$ and dielectric constants $K_1 = K$,$K_2 = 3K$,$K_3 = 5K$.The capacitance of each part is given by $C = \frac{K \varepsilon_0 A}{d_{thickness}}$.$C_1 = \frac{K \varepsilon_0 A}{d}$$C_2 = \frac{3K \varepsilon_0 A}{2d}$$C_3 = \frac{5K \varepsilon_0 A}{3d}$Since they are in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.$\frac{1}{C_{eq}} = \frac{d}{K \varepsilon_0 A} + \frac{2d}{3K \varepsilon_0 A} + \frac{3d}{5K \varepsilon_0 A}$$\frac{1}{C_{eq}} = \frac{d}{K \varepsilon_0 A} [1 + \frac{2}{3} + \frac{3}{5}] = \frac{d}{K \varepsilon_0 A} [\frac{15 + 10 + 9}{15}] = \frac{34d}{15K \varepsilon_0 A}$Therefore,$C_{eq} = \frac{15K \varepsilon_0 A}{34d}$.

In the given figure,a capacitor is formed by placing a compound dielectric between the plates of a parallel plate capacitor. The expression for the capacity of the said capacitor will be (Given area of plate $= A$):

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