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In an isolated parallel plate capacitor of capacitance $C$ , the four surface have charges $Q_{1}, Q_{2}, Q_{3} and Q_{4}$ as shown. The potential difference between the plates is

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$\frac{Q_{2} + Q_{3}}{2 C}$

$\frac{Q_{1} + Q_{4}}{2 C}$

$\frac{Q_{1} + Q_{2} + Q_{3} + Q_{4}}{2 C}$

$\frac{Q_{2} - Q_{3}}{2 C}$

answer is C.

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Detailed Solution

Plane conducting surfaces facing each other must have equal and opposite charge densities. Here as the plate areas are equal, $Q_{2} = - Q_{3} .$

The charge on a capacitor means the charge on the inner surface of the positive plate (here it is Q₂)

Potential difference between the plates $= \frac{charge}{capacitance} = \frac{Q_{2}}{C} = \frac{2 Q_{2}}{2 C} = \frac{Q_{2} - (- Q_{2})}{2 C} = \frac{Q_{2} - Q_{3}}{2 C} .$

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