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 electric 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:* C is the capacitance in farads (F)* ε₀ is the permittivity of free space (8.85 × 10^-12 C²/N ∙ m²)* A is the area of each plate in square meters (m²)* d is the distance between the plates in meters (m)In this problem, we have:* A = 0.0010 m²* d = 0.050 mm = 5.0 × 10^-5 m* E = 7.4 × 10⁶ V/mWe can use the equation above to calculate the capacitance:$$C = \frac{\varepsilon_0 A}{d} = \frac{(8.85 imes 10^{-12} ext{ C}^2/ ext{N} \cdot ext{m}^2)(0.0010 ext{ m}^2)}{5.0 imes 10^{-5} ext{ m}} = 180 ext{ pF}$$Therefore, the capacitance of the capacitor is 180 pF.

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