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(B) The initial capacitance is $C_0 = \frac{K \epsilon_0 A}{d}$.When the material is replaced,the capacitor acts as two capacitors in parallel. The fi

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$\frac{4}{3}$

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(B) The initial capacitance is $C_0 = \frac{K \epsilon_0 A}{d}$.When the material is replaced,the capacitor acts as two capacitors in parallel. The first capacitor has area $A_1 = \frac{1}{3}A$,dielectric constant $K_1 = 2K$,and separation $d$. Its capacitance is $C_1 = \frac{K_1 \epsilon_0 A_1}{d} = \frac{(2K) \epsilon_0 (A/3)}{d} = \frac{2}{3} \frac{K \epsilon_0 A}{d} = \frac{2}{3} C_0$.The second capacitor has area $A_2 = \frac{2}{3}A$,dielectric constant $K_2 = K$,and separation $d$. Its capacitance is $C_2 = \frac{K_2 \epsilon_0 A_2}{d} = \frac{K \epsilon_0 (2A/3)}{d} = \frac{2}{3} \frac{K \epsilon_0 A}{d} = \frac{2}{3} C_0$.Since they are in parallel,the total capacitance is $C = C_1 + C_2 = \frac{2}{3} C_0 + \frac{2}{3} C_0 = \frac{4}{3} C_0$.Therefore,$\frac{C}{C_0} = \frac{4}{3}$.

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