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(C) The initial capacitance with air is $C = \frac{\varepsilon_0 A}{d} = 9 \, pF$.When the space is filled with two dielectrics of thicknesses $d_1 =

Vedclass · Accepted Answer

$40.5$

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(C) The initial capacitance with air is $C = \frac{\varepsilon_0 A}{d} = 9 \, pF$.When the space is filled with two dielectrics of thicknesses $d_1 = d/3$ and $d_2 = 2d/3$,the capacitor 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 $C'$ is given by:$C' = \frac{C_1 C_2}{C_1 + C_2} = \frac{81 	imes 81}{81 + 81} = \frac{81}{2} = 40.5 \, pF$.

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