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(B) The initial capacitance of the air capacitor is given by $C = \frac{\epsilon_0 A}{d} = 10 \ \mu F$.When the capacitor is divided into two equal pa

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$30$

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(B) The initial capacitance of the air capacitor is given by $C = \frac{\epsilon_0 A}{d} = 10 \ \mu F$.When the capacitor is divided into two equal parts as shown in the figure,the area of each part becomes $A/2$ while the distance $d$ remains the same. These two parts act as two capacitors connected in parallel.The capacitance of the first part is $C_1 = \frac{K_1 \epsilon_0 (A/2)}{d} = K_1 	imes (C/2) = 2 	imes (10/2) = 10 \ \mu F$.The capacitance of the second part is $C_2 = \frac{K_2 \epsilon_0 (A/2)}{d} = K_2 	imes (C/2) = 4 	imes (10/2) = 20 \ \mu F$.Since the two capacitors are connected in parallel,the equivalent capacitance is $C_{new} = C_1 + C_2$.Therefore,$C_{new} = 10 \ \mu F + 20 \ \mu F = 30 \ \mu F$.

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