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(D) Initially,the capacitors are in series with capacitance $C_A = C$ and $C_B = KC$. The equivalent capacitance is $C_{net} = \frac{C_A C_B}{C_A + C_

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$\frac{Q_B^{\prime}}{Q_B}=\frac{K+1}{2K}$

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(D) Initially,the capacitors are in series with capacitance $C_A = C$ and $C_B = KC$. The equivalent capacitance is $C_{net} = \frac{C_A C_B}{C_A + C_B} = \frac{C \cdot KC}{C + KC} = \frac{KC}{K+1}$.The charge on each capacitor in series is $Q_A = Q_B = C_{net}E = \frac{KCE}{K+1}$.After removing the dielectric,$C_A = C$ and $C_B = C$. The new equivalent capacitance is $C_{net}^{\prime} = \frac{C \cdot C}{C + C} = \frac{C}{2}$.The new charge on each capacitor is $Q_A^{\prime} = Q_B^{\prime} = C_{net}^{\prime}E = \frac{CE}{2}$.Calculating the ratio: $\frac{Q_B^{\prime}}{Q_B} = \frac{CE/2}{KCE/(K+1)} = \frac{K+1}{2K}$.

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