Question

Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.50 cm² and a plate separation of 0.0100 mm.

Answer 1

The capacitance of the Teflon-filled parallel-plate capacitor is 0.13275 F.

Step-by-step explanation:

To determine the capacitance of the Teflon-filled parallel-plate capacitor, we can use the formula:

C = (ε₀ * A) / d

Where:

• C is the capacitance
• ε₀ is the vacuum permittivity, which is approximately 8.85 x 10^-12 F/m
• A is the area of the plates in square meters (convert from cm² to m²)
• d is the separation distance between the plates in meters (convert from mm to m)

Let's plug in the values:

C = (8.85 x 10^-12 F/m * 1.50 x 10^-4 m²) / 1.00 x 10^-5 m

C = 0.13275 F