A parallel-plate capacitor has a plate area of 100 cm^2 and a separation of 1.0 cm between its plates. It is connected across a battery with an emf of 24 volts. Determine the force of attraction between the plates.

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Quantrex Academy · Accepted Answer

Correct answer: B. 2.5 × 10^ -7 N

The force of attraction between the plates of a parallel-plate capacitor is given by:

F = \dfrac{1}{2} \varepsilon_0 E^2 A

Since the electric field

E = \dfrac{V}{d}

, the formula becomes:

F = \dfrac{\varepsilon_0 A V^2}{2 d^2}

Given values are:

A = 100 \text{ cm}^2 = 10^{-2} \text{ m}^2

d = 1.0 \text{ cm} = 10^{-2} \text{ m}

V = 24 \text{ V}

\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m}

Substituting these values into the formula:

F = \dfrac{8.85 \times 10^{-12} \times 10^{-2} \times (24)^2}{2 \times (10^{-2})^2}

F = \dfrac{8.85 \times 10^{-14} \times 576}{2 \times 10^{-4}}

F = 8.85 \times 288 \times 10^{-10} \text{ N}

F = 2548.8 \times 10^{-10} \text{ N}

F \approx 2.5 \times 10^{-7} \text{ N}

A parallel-plate capacitor has a plate area of 100 cm^2 and a...

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