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(B) $1$. Initial state: The capacitor has a dielectric of thickness $d/2$ and constant $K$. The capacitance $C_i$ is given by the series combination o

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$V \left( \frac{2K}{K+1} \right)$

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(B) $1$. Initial state: The capacitor has a dielectric of thickness $d/2$ and constant $K$. The capacitance $C_i$ is given by the series combination of two capacitors: one with dielectric $(C_1 = \frac{2K\epsilon_0 A}{d})$ and one with air $(C_2 = \frac{2\epsilon_0 A}{d})$. $C_i = \frac{C_1 C_2}{C_1 + C_2} = \frac{\frac{2K\epsilon_0 A}{d} \cdot \frac{2\epsilon_0 A}{d}}{\frac{2\epsilon_0 A}{d}(K+1)} = \frac{2K\epsilon_0 A}{d(K+1)}$. $2$. Charge on the capacitor: $Q = C_i V = \frac{2K\epsilon_0 A V}{d(K+1)}$. $3$. After disconnecting the battery,the charge $Q$ remains constant. $4$. Final state: The dielectric is removed,so the capacitor becomes an air-filled parallel plate capacitor with capacitance $C_f = \frac{\epsilon_0 A}{d}$. $5$. Final potential difference $V_f = \frac{Q}{C_f} = \frac{2K\epsilon_0 A V}{d(K+1)} \cdot \frac{d}{\epsilon_0 A} = V \left( \frac{2K}{K+1} \right)$.

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