An ideal air-filled parallel-plate capacitor consists of two circular plates, each of radius $0.30 \mathrm {~mm}$ How far apart should the plates be for the capacitance to be 300.0-pF? (ε₀ = 8.85 × 10^-12C²/N ∙ m²)

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The capacitance of a parallel-plate capacitor is given by:$$C = \frac{\epsilon_0 A}{d}$$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 are given the capacitance (C = 300.0 pF = 300.0 × 10^-12 F) and the radius of each plate (r = 0.30 mm = 3.0 × 10^-4 m). We can use the formula above to solve for the distance between the plates (d):$$d = \frac{\epsilon_0 A}{C}$$$$d = \frac{(8.85 imes 10^{-12} ext{ C}^2/ ext{N} \cdot ext{m}^2) \pi (3.0 imes 10^{-4} ext{ m})^2}{300.0 imes 10^{-12} ext{ F}}$$$$d = 8.35 imes 10^{-6} ext{ m} = 8.35 \mu ext{m}$$Therefore, the plates should be 8.35 μm apart for the capacitance to be 300.0 pF.

An Ideal Air-Filled Parallel-Plate Capacitor Consists of Two Circular Plates 0.30 mm0.30 \mathrm {~mm}0.30 mm

An Ideal Air-Filled Parallel-Plate Capacitor Consists of Two Circular Plates $0.30 \mathrm {~mm}$