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A 500pF parallel-plate capacitor has ±40nC charges on its plates. The plates are separated by 0.6 mm. Find: the area of each plate

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Final answer:

The area of each plate is approximately 33.9 x 10^-3 m².

Step-by-step explanation:

First, let's use the formula for the capacitance of a parallel-plate capacitor:

C = (ε₀ * A) / d

Where C is the capacitance, ε₀ is the permittivity of free space (8.85 x 10^-12 F/m), A is the area of each plate, and d is the separation between the plates.

Given that C = 500 pF, ε₀ = 8.85 x 10^-12 F/m, and d = 0.6 mm = 0.6 x 10^-3 m, we can rearrange the formula to solve for A:

A = (C * d) / ε₀

Plugging in the values, we have:

A = (500 x 10^-12 F * 0.6 x 10^-3 m) / 8.85 x 10^-12 F/m

Calculating the above expression gives:

A ≈ 33.9 x 10^-3 m²

So, the area of each plate is approximately 33.9 x 10^-3 m².

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