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? (?₀ = 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)* \epsilon_0 is the permittivity of free space (8.85 × 10^-12 C^2/N m^2)* A is the area of each plate in square meters (m^2)* d is the distance between the plates in meters (m)We are given that the capacitance is 300.0 pF, which is equal to 300.0 × 10^-12 F. We are also given that the radius of each plate is 0.30 mm, which is equal to 3.0 × 10^-4 m. The area of each plate is:$$A = \pi r^2 = \pi (3.0 	imes 10^{-4} \mathrm{~m})^2 = 2.83 	imes 10^{-7} \mathrm{~m}^2$$Substituting these values into the equation for capacitance, we get:$$300.0 	imes 10^{-12} \mathrm{~F} = \frac{8.85 	imes 10^{-12} \mathrm{~C}^2/\mathrm{N} \cdot \mathrm{m}^2 \cdot 2.83 	imes 10^{-7} \mathrm{~m}^2}{d}$$Solving for d, we get:$$d = \frac{8.85 	imes 10^{-12} \mathrm{~C}^2/\mathrm{N} \cdot \mathrm{m}^2 \cdot 2.83 	imes 10^{-7} \mathrm{~m}^2}{300.0 	imes 10^{-12} \mathrm{~F}} = 8.3 	imes 10^{-6} \mathrm{~m}$$Therefore, the plates should be 8.3 µm apart for the capacitance to be 300.0 pF.

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

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