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(C) 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

$81$

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(C) 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 $d_1 = \frac{d}{3}$ and $d_2 = \frac{2d}{3}$,the capacitor acts as two capacitors in series.The capacitance of the first part is $C_1 = \frac{K_1 \epsilon_0 A}{d_1} = \frac{6 \epsilon_0 A}{d/3} = 18 \frac{\epsilon_0 A}{d} = 18 	imes 9 = 162 \ pF$.The capacitance of the second part is $C_2 = \frac{K_2 \epsilon_0 A}{d_2} = \frac{12 \epsilon_0 A}{2d/3} = \frac{36 \epsilon_0 A}{2d} = 18 \frac{\epsilon_0 A}{d} = 18 	imes 9 = 162 \ pF$.Since they are in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{162} + \frac{1}{162} = \frac{2}{162} = \frac{1}{81}$.Therefore,$C_{eq} = 81 \ pF$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

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