A parallel plate capacitor of area A,plate se | vedclass

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(B) The capacitor can be viewed as a combination of three capacitors. The region with $K_1$ and $K_2$ are in parallel with each other,and this combina

Vedclass · Accepted Answer

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

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(B) The capacitor can be viewed as a combination of three capacitors. The region with $K_1$ and $K_2$ are in parallel with each other,and this combination is in series with the region having $K_3$.Capacitance of the first part: $C_1 = \frac{\varepsilon_0 (A/2) K_1}{d/2} = \frac{\varepsilon_0 A K_1}{d}$Capacitance of the second part: $C_2 = \frac{\varepsilon_0 (A/2) K_2}{d/2} = \frac{\varepsilon_0 A K_2}{d}$Capacitance of the third part: $C_3 = \frac{\varepsilon_0 A K_3}{d/2} = \frac{2 \varepsilon_0 A K_3}{d}$Since $C_1$ and $C_2$ are in parallel,their equivalent capacitance is $C_p = C_1 + C_2 = \frac{\varepsilon_0 A}{d} (K_1 + K_2)$.Now,$C_p$ and $C_3$ are in series,so the equivalent capacitance $C$ is given by:$\frac{1}{C} = \frac{1}{C_p} + \frac{1}{C_3} = \frac{d}{\varepsilon_0 A (K_1 + K_2)} + \frac{d}{2 \varepsilon_0 A K_3}$If a single dielectric $K$ is used,$C = \frac{K \varepsilon_0 A}{d}$,so $\frac{1}{C} = \frac{d}{K \varepsilon_0 A}$.Equating the two expressions for $\frac{1}{C}$:$\frac{d}{K \varepsilon_0 A} = \frac{d}{\varepsilon_0 A} \left( \frac{1}{K_1 + K_2} + \frac{1}{2K_3} \right)$Therefore,$\frac{1}{K} = \frac{1}{K_1 + K_2} + \frac{1}{2K_3}$.

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