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(C) The initial capacitance of the air-filled parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d} = 9 \, pF$.When the space is filled

Vedclass · Accepted Answer

$40.5$

Vedclass · Answer

(C) The initial capacitance of the air-filled parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d} = 9 \, pF$.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 in series.The capacitance of the first part is $C_1 = \frac{K_1 \varepsilon_0 A}{d_1} = \frac{3 \varepsilon_0 A}{d/3} = 9 \frac{\varepsilon_0 A}{d} = 9 	imes 9 = 81 \, pF$.The capacitance of the second part is $C_2 = \frac{K_2 \varepsilon_0 A}{d_2} = \frac{6 \varepsilon_0 A}{2d/3} = 9 \frac{\varepsilon_0 A}{d} = 9 	imes 9 = 81 \, pF$.Since they are in series,the equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{81} + \frac{1}{81} = \frac{2}{81}$.Therefore,$C_{eq} = \frac{81}{2} = 40.5 \, pF$.

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