Each plate of an ideal air-filled parallel-plate capacitor has an area of 0.0010 $m ^ { 2 }$ ,and the separation of the plates is $0.050 \mathrm {~mm}$ An $\vec { E }$ field of $7.4 \times 10^6 \mathrm {~V} / \mathrm { m }$ is present between the plates.What is the capacitance of this capacitor? (ε₀ = 8.85 × 10^-12C²/N ∙ m²)

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Quizplus · Accepted Answer

The capacitance of a parallel-plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$, where $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}$ (permittivity of free space), $A = 0.0010 \, \text{m}^2$ (area of the plates), and $d = 0.050 \, \text{mm} = 0.050 \times 10^{-3} \, \text{m}$ (distance between the plates). Plugging in these values gives $C = \frac{8.85 \times 10^{-12} \times 0.0010}{0.050 \times 10^{-3}} = 177 \, \text{pF}$, which rounds to 180 pF.

Each Plate of an Ideal Air-Filled Parallel-Plate Capacitor Has an Area