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? (?₀ = 8.85 × 10^-12C²/N ? m²)

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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* ε₀ = 8.85 × 10^-12 C²/N ? m²Plugging these values into the equation, we get:$$C = \frac{(8.85 imes 10^{-12} \mathrm{C}^2/\mathrm{N} \cdot \mathrm{m}^2)(0.0010 \mathrm{m}^2)}{5.0 imes 10^{-5} \mathrm{m}} = 1.77 imes 10^{-10} \mathrm{F}$$Converting to picofarads (pF), we get:$$C = 1.77 imes 10^{-10} \mathrm{F} = 177 \mathrm{pF}$$Therefore, the capacitance of the capacitor is 177 pF, which is closest to choice A, 180 pF.

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