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(D) The initial capacitance of the parallel plate capacitor is $C = \frac{\varepsilon_0 A}{d}$.When the space between the plates is filled with two di

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$\frac{2 K_1 K_2}{K_1 + K_2}$

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(D) The initial capacitance of the parallel plate capacitor is $C = \frac{\varepsilon_0 A}{d}$.When the space between the plates is filled with two dielectric layers of thickness $d/2$ each,they act as two capacitors in series.The capacitance of the first layer is $C_A = \frac{K_1 \varepsilon_0 A}{d/2} = \frac{2 K_1 \varepsilon_0 A}{d} = 2 K_1 C$.The capacitance of the second layer is $C_B = \frac{K_2 \varepsilon_0 A}{d/2} = \frac{2 K_2 \varepsilon_0 A}{d} = 2 K_2 C$.Since they are in series,the equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{C_A} + \frac{1}{C_B} = \frac{1}{2 K_1 C} + \frac{1}{2 K_2 C} = \frac{K_1 + K_2}{2 K_1 K_2 C}$.Therefore,$C_1 = \frac{2 K_1 K_2}{K_1 + K_2} C$.The ratio $C_1/C$ is $\frac{2 K_1 K_2}{K_1 + K_2}$.

$A$ parallel plate capacitor has capacitance $C$. If it is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$,its capacity becomes $C_1$. The ratio of $C_1$ to $C$ is

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