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

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$40.5$

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(C) For an air-filled parallel plate capacitor, the capacitance is given by $C = \frac{A \varepsilon_0}{d} = 9 	ext{ pF}$.When the space is filled with dielectrics of thickness $d_1$ and $d_2$, the system acts as two capacitors in series.The capacitance of the first part is $C_1 = \frac{K_1 A \varepsilon_0}{d_1} = \frac{3 A \varepsilon_0}{d/3} = 9 \frac{A \varepsilon_0}{d} = 9C = 9 	imes 9 	ext{ pF} = 81 	ext{ pF}$.The capacitance of the second part is $C_2 = \frac{K_2 A \varepsilon_0}{d_2} = \frac{6 A \varepsilon_0}{2d/3} = 9 \frac{A \varepsilon_0}{d} = 9C = 9 	imes 9 	ext{ pF} = 81 	ext{ pF}$.Since the capacitors are in series, the equivalent capacitance $C_{	ext{eq}}$ is given by $\frac{1}{C_{	ext{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$.$C_{	ext{eq}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{81 	imes 81}{81 + 81} = \frac{81}{2} = 40.5 	ext{ pF}$.

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