In an insulated parallel-plate capacitor of capacitance $C$, the four surfaces have charges $Q_1, Q_2, Q_3$ and $Q_4$ as shown. The potential difference between the plate is
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The facing surfaces of plates of capacitor would have equal and opposite charges, Hence, $\mathrm{Q}_{2}=-\mathrm{Q}_{3},$ potential difference between the plats
${=\frac{Q_{2}}{C}=\frac{2 Q_{2}}{2 C}} $
${=\frac{Q_{2}-\left(-Q_{2}\right)}{2 C}=\frac{Q_{2}-Q_{3}}{2 C}}$
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