Question:

A parallel-plate capacitor consists of two circular plates with a radius of 5 cm, 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:

Given data:

• Radius of the plates (r) = 5 cm = 0.05 m
• Distance between the plates (d) = 2 mm = 0.002 m
• Relative permittivity of the dielectric material (εᵣ) = 4

The capacitance (C) of a parallel-plate capacitor can be calculated using the formula:

C = (ε₀ * εᵣ * A) / d

where ε₀ is the vacuum permittivity (8.85 x 10⁻¹² F/m), εᵣ is the relative permittivity, A is the area of one plate, and d is the distance between the plates.

First, we need to calculate the area (A) of one plate using the given radius (r):

A = π * r²

A = π * (0.05 m)²

A = π * 0.0025 m²

A ≈ 0.00785398 m²

Next, we can substitute the known values into the capacitance formula:

C = (ε₀ * εᵣ * A) / d

C = (8.85 x 10⁻¹² F/m) * (4) * (0.00785398 m²) / (0.002 m)

C ≈ (3.54 x 10⁻¹⁰ F/m) * (0.00785398 m²) / (0.002 m)

C ≈ (2.78 x 10⁻¹² F) / (0.002 m)

C ≈ 1.39 x 10⁻⁹ F

Therefore, the capacitance of this parallel-plate capacitor is approximately 1.39 nanofarads (nF).