Question:

A parallel-plate capacitor consists of two plates with an area of 0.05 m² each, separated by a distance of 2 mm. The space between the plates is filled with a dielectric material with a relative permittivity of 4. Determine the capacitance of this capacitor.

Answer:

The capacitance of a parallel-plate capacitor can be determined using the formula:

C = ε₀ * (A / d)

where C is the capacitance, ε₀ is the vacuum permittivity (8.85 × 10⁻¹² F/m), A is the area of either plate, and d is the separation distance between the plates.

Given: Area of each plate, A = 0.05 m² Separation distance, d = 2 mm = 0.002 m Relative permittivity, εᵣ = 4

To find the capacitance, substitute the given values into the formula:

C = (8.85 × 10⁻¹² F/m) * (0.05 m² / 0.002 m) * (4)

C = (8.85 × 10⁻¹² F/m) * 25 * 4

C = 8.85 × 10⁻¹² * 100

C = 8.85 × 10⁻¹⁰ F

Therefore, the capacitance of the parallel-plate capacitor is 8.85 × 10⁻¹⁰ F.