An air capacitor has capacitance C_1. The spa | vedclass

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(D) For an air capacitor,$C_1 = \frac{A \varepsilon_0}{d}$.When two dielectrics are inserted as shown,the arrangement acts as two capacitors in series

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

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(D) For an air capacitor,$C_1 = \frac{A \varepsilon_0}{d}$.When two dielectrics are inserted as shown,the arrangement acts as two capacitors in series,each with plate separation $d/2$ and area $A$.The capacitance of the first part is $C_{a} = \frac{K_1 A \varepsilon_0}{d/2} = \frac{2 K_1 A \varepsilon_0}{d} = 2 K_1 C_1$.The capacitance of the second part is $C_{b} = \frac{K_2 A \varepsilon_0}{d/2} = \frac{2 K_2 A \varepsilon_0}{d} = 2 K_2 C_1$.Since they are in series,the equivalent capacitance $C_2$ is given by:$\frac{1}{C_2} = \frac{1}{C_a} + \frac{1}{C_b} = \frac{1}{2 K_1 C_1} + \frac{1}{2 K_2 C_1} = \frac{1}{2 C_1} \left( \frac{1}{K_1} + \frac{1}{K_2} \right) = \frac{1}{2 C_1} \left( \frac{K_1 + K_2}{K_1 K_2} \right)$.Therefore,$C_2 = \frac{2 C_1 K_1 K_2}{K_1 + K_2}$.The ratio $\frac{C_1}{C_2} = \frac{C_1}{\frac{2 C_1 K_1 K_2}{K_1 + K_2}} = \frac{K_1 + K_2}{2 K_1 K_2}$.

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