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(C) The initial capacitance of the air-filled parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d} = 10 \mu F$.When the area is divided

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$30$

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(C) The initial capacitance of the air-filled parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d} = 10 \mu F$.When the area is divided into two equal halves and filled with dielectrics $K_1$ and $K_2$,the two parts act as two capacitors connected in parallel.The area of each part is $A' = \frac{A}{2}$ and the distance between the plates remains $d$.The capacitance of the first part is $C_1 = \frac{K_1 \varepsilon_0 A'}{d} = \frac{K_1 \varepsilon_0 A}{2d} = \frac{K_1}{2} C$.The capacitance of the second part is $C_2 = \frac{K_2 \varepsilon_0 A'}{d} = \frac{K_2 \varepsilon_0 A}{2d} = \frac{K_2}{2} C$.Since they are in parallel,the equivalent capacitance is $C_{	ext{eq}} = C_1 + C_2 = \frac{C}{2} (K_1 + K_2)$.Substituting the given values: $C_{	ext{eq}} = \frac{10 \mu F}{2} (2 + 4) = 5 \mu F 	imes 6 = 30 \mu F$.

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