Question

The rectangular plates in a parallel-plate capacitor are 0.063 m × 5.4 m. a distance of 3.5 × 10–5 m separates the plates. the plates are separated by a dielectric made of teflon, which has a dielectric constant of 2.1. what is the capacitance of the capacitor? a. 0.12 µf b. 0.18 µf c. 1.2 µf d. 1.8 µf

Answer 1

The capacitance of the parallel-plate capacitor with Teflon as a dielectric is 0.18 µF.

Step-by-step explanation:

The capacitance of a parallel-plate capacitor with a dielectric material can be calculated using the formula:

C = (ε₀ * A) / d

Where C is the capacitance, ε₀ is the permittivity of free space (ε₀ = 8.85 × 10-12 F/m), A is the area of the plates, and d is the distance between the plates. In this case, the plates have dimensions of 0.063 m × 5.4 m, and the dielectric material is Teflon with a dielectric constant (κ) of 2.1.

Using the formula, we have:

C = (8.85 × 10-12 F/m * 0.063 m * 5.4 m) / 3.5 × 10-5 m) * 2.1 = 0.18 µF

Therefore, the capacitance of the capacitor is 0.18 µF (option b).