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(B) The system consists of two parts in series: the top part (containing $K_1$ and $K_2$ in parallel) and the bottom part (containing $K_3$).$1$. The

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

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(B) The system consists of two parts in series: the top part (containing $K_1$ and $K_2$ in parallel) and the bottom part (containing $K_3$).$1$. The top part has two capacitors in parallel,each with area $A/2$ and distance $d/2$. Their equivalent capacitance is $C_{top} = \frac{\epsilon_0 (A/2) K_1}{d/2} + \frac{\epsilon_0 (A/2) K_2}{d/2} = \frac{\epsilon_0 A}{d} (K_1 + K_2)$.$2$. The bottom part is a single capacitor with area $A$ and distance $d/2$,filled with dielectric $K_3$. Its capacitance is $C_{bottom} = \frac{\epsilon_0 A K_3}{d/2} = \frac{2 \epsilon_0 A K_3}{d}$.$3$. Since these two parts are in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_{top}} + \frac{1}{C_{bottom}}$.$4$. Substituting the values: $\frac{1}{C_{eq}} = \frac{d}{\epsilon_0 A (K_1 + K_2)} + \frac{d}{2 \epsilon_0 A K_3} = \frac{d}{\epsilon_0 A} \left[ \frac{1}{K_1 + K_2} + \frac{1}{2K_3} \right]$.$5$. The equivalent capacitance of the whole system is $C_{eq} = \frac{K \epsilon_0 A}{d}$.$6$. Equating the two expressions for $1/C_{eq}$: $\frac{d}{K \epsilon_0 A} = \frac{d}{\epsilon_0 A} \left[ \frac{1}{K_1 + K_2} + \frac{1}{2K_3} \right]$.$7$. Therefore,$\frac{1}{K} = \frac{1}{K_1 + K_2} + \frac{1}{2K_3}$.

If the equivalent dielectric constant of the given system is $K$,then...

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