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Electric Field Due to Disk and Infinitely Large Sheet

An isolated p...

Question

An isolated parallel plate capacitor of capacitance $C$ has four surfaces with charges $Q_{1}, Q_{2}, Q_{3}$ and $Q_{4}$ as shown in the figure below. The potential difference across plates is (In the given diagram, $Q_{1}$ and $Q_{4}$ are the charges on the outer surfaces of the left & right plates respectively)

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{Q_{2} + Q_{3}}{C} ∣ ∣ ∣

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Solution

The correct option is C $∣ ∣ ∣ \frac{Q_{3}}{C} ∣ ∣ ∣$
For the parallel plate capacitor, $Q = C V$ Where,
$Q$ is the magnitude of the charges on the inner surface of the plates.

$C$ is the capacitance and $V$ is the potential difference across the plate.

Due to induction, charges on inner surfaces facing each other are equal and opposite.

So, $Q_{2} = - Q_{3}$

Thus, $V = ∣ ∣ ∣ \frac{Q_{2}}{C} ∣ ∣ ∣ = ∣ ∣ ∣ \frac{Q_{3}}{C} ∣ ∣ ∣$

Hence, options (b) and (c) are the correct answers.

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Electric Field Due to Disk and Infinitely Large Sheet

Standard XII Physics

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