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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 in the figure. The potential difference between the plates is :

A

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

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B

| \frac{Q_{2}}{C} |

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C

| \frac{Q_{3}}{C} |

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D

\frac{1}{C} [(Q_{1} + Q_{2}) - (Q_{3} - Q_{4})]

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Solution

The correct options are
B $| \frac{Q_{2}}{C} |$
C $| \frac{Q_{3}}{C} |$

For parallel plate capacitor, $Q_{2}$ and $Q_{3}$ must be same and equal in magnitude.i.e, $| Q_{2} | = | Q_{3} |$

So, potential difference between the plates, $V = \frac{| Q_{2} |}{C} = \frac{| Q_{3} |}{C}$

Ans: $(B), (C)$

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