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Capacitance of Parallel Plate Capacitor

In an isolate...

Question

In an isolated 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 plates is:

A

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

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B

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

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C

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

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D

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

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Solution

The correct option is C $\frac{Q_{2} - Q_{3}}{2 C}$
$Q_{2} = - Q_{3}$

So, (A) or (C)

and $V = \frac{Q}{C}$

But (A) cannot follow this $Q_{2} = - Q_{3}$

and net $Q = 0$

So, $V = \frac{Q_{2} - Q_{3}}{2 C}$

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