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(C) The initial capacitance is $C = \frac{\epsilon_0 A}{d} = 9 \ pF$.The space is divided into two layers of thickness $d_1 = d/3$ and $d_2 = 2d/3$,wh

Vedclass · Accepted Answer

$40.5$

Vedclass · Answer

(C) The initial capacitance is $C = \frac{\epsilon_0 A}{d} = 9 \ pF$.The space is divided into two layers of thickness $d_1 = d/3$ and $d_2 = 2d/3$,which act as two capacitors in series.The capacitance of the first layer is $C_1 = \frac{K_1 \epsilon_0 A}{d_1} = \frac{3 \epsilon_0 A}{d/3} = 9 \left( \frac{\epsilon_0 A}{d} ight) = 9 	imes 9 = 81 \ pF$.The capacitance of the second layer is $C_2 = \frac{K_2 \epsilon_0 A}{d_2} = \frac{6 \epsilon_0 A}{2d/3} = \frac{18}{2} \left( \frac{\epsilon_0 A}{d} ight) = 9 	imes 9 = 81 \ pF$.Since $C_1$ and $C_2$ are in series,the equivalent capacitance $C_{eq}$ is given by:$C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{81 	imes 81}{81 + 81} = \frac{81 	imes 81}{162} = 40.5 \ pF$.

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