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(D) The initial capacitance of the air-filled capacitor is $C_0 = \frac{\epsilon_0 A}{d} = 9 \ pF$.When the space is filled with two dielectrics of th

Vedclass · Accepted Answer

$40.5$

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(D) The initial capacitance of the air-filled capacitor is $C_0 = \frac{\epsilon_0 A}{d} = 9 \ pF$.When the space is filled with two dielectrics of thicknesses $d_1 = d/3$ and $d_2 = 2d/3$,the system acts as two capacitors connected in series.The capacitance of the first part is $C_1 = \frac{k_1 \epsilon_0 A}{d_1} = \frac{3 \epsilon_0 A}{d/3} = 9 \frac{\epsilon_0 A}{d} = 9 C_0 = 9 	imes 9 = 81 \ pF$.The capacitance of the second part is $C_2 = \frac{k_2 \epsilon_0 A}{d_2} = \frac{6 \epsilon_0 A}{2d/3} = 9 \frac{\epsilon_0 A}{d} = 9 C_0 = 9 	imes 9 = 81 \ pF$.Since they are in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$.$\frac{1}{C_{eq}} = \frac{1}{81} + \frac{1}{81} = \frac{2}{81}$.Therefore,$C_{eq} = \frac{81}{2} = 40.5 \ pF$.

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