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
A$\frac{{{Q_2} - {Q_3}}}{{2C}}$
B$\frac{{{Q_2} + 2{Q_3}}}{{2C}}$
C$\frac{{{Q_1} + {Q_4}}}{{2C}}$
D$\frac{{{Q_1} + {Q_2} + {Q_3} + {Q_4}}}{{2C}}$
Medium
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A$\frac{{{Q_2} - {Q_3}}}{{2C}}$
a 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
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