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Electric Field Due to Disk and Infinitely Large Sheet

A parallel pl...

Question

A parallel plate capacitor having capacitance C has two plates of same area A and thickness t. The figure shows the charges available on the four surfaces of the plates. The potential difference V between the two plates is given by :

A

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

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B

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

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C

\frac{q_{1} - q_{4}}{2 C}

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D

\frac{q_{1} - q_{4}}{C}

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Solution

The correct option is A $\frac{q_{2} - q_{3}}{2 C}$
As the total charge of a capacitor is zero, the charge distribution on the inner faces of left and right plates are $+ Q = (q_{2} - q_{3}) / 2$ and $- Q = - (q_{2} - q_{3}) / 2$ respectively.
Thus the potential difference between the plates is $V = \frac{Q}{C} = \frac{(q_{2} - q_{3})}{2 C}$

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