A parallel-plate capacitor of area A, plate s | vedclass

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Question

(A) The capacitor can be modeled as a combination of four smaller capacitors. The top portion of thickness $d/2$ is divided into three parts,each of a

Vedclass · Accepted Answer

$\frac{2}{K} = \frac{3}{K_1 + K_2 + K_3} + \frac{1}{K_4}$

Vedclass · Answer

(A) The capacitor can be modeled as a combination of four smaller capacitors. The top portion of thickness $d/2$ is divided into three parts,each of area $A/3$. These three capacitors $(C_1, C_2, C_3)$ are in parallel.$C_1 = \frac{\epsilon_0 K_1 (A/3)}{d/2} = \frac{2\epsilon_0 K_1 A}{3d}$$C_2 = \frac{2\epsilon_0 K_2 A}{3d}$$C_3 = \frac{2\epsilon_0 K_3 A}{3d}$Their equivalent capacitance $C_p = C_1 + C_2 + C_3 = \frac{2\epsilon_0 A}{3d}(K_1 + K_2 + K_3)$.This parallel combination is in series with the bottom capacitor $C_4$ of thickness $d/2$ and area $A$.$C_4 = \frac{\epsilon_0 K_4 A}{d/2} = \frac{2\epsilon_0 K_4 A}{d}$.The total equivalent capacitance $C$ is given by $\frac{1}{C} = \frac{1}{C_p} + \frac{1}{C_4}$.Substituting the values: $\frac{1}{C} = \frac{3d}{2\epsilon_0 A(K_1 + K_2 + K_3)} + \frac{d}{2\epsilon_0 K_4 A}$.For a single dielectric $K$,$C = \frac{K\epsilon_0 A}{d}$,so $\frac{1}{C} = \frac{d}{K\epsilon_0 A}$.Equating the two: $\frac{d}{K\epsilon_0 A} = \frac{d}{\epsilon_0 A} [\frac{3}{2(K_1 + K_2 + K_3)} + \frac{1}{2K_4}]$.$\frac{1}{K} = \frac{3}{2(K_1 + K_2 + K_3)} + \frac{1}{2K_4}$.Multiplying by $2$: $\frac{2}{K} = \frac{3}{K_1 + K_2 + K_3} + \frac{1}{K_4}$.

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