A parallel-plate capacitor is made from two aluminum-foil sheets, each 5.5 cm wide and 5.0 m long. Between the sheets is a Teflon strip of the same width and length that is 3.3*10-2 mm thick. What is the capacitance of this capacitor? (The dielectric constant of Teflon is 2.1.)

Capacitance = 1.55 * 10⁻⁷ F = 0.16 μF

Explanation:

Capacitance in the presence of a dielectric is given as C = κ ε₀A/d

Given:

Width = 0.055 m

Length = 0.05 m

Thickness of teflon = distance between the plates = d = 3.3*10⁻⁵ m

Dielectric constant of teflon = κ = 2.1

Area of cross section of the plates = A = 0.055 * 5 = 0.275 m²

Capacitance of the capacitor = C = κ ε₀ A/d

= (2.1) (8.85*10⁻¹²) (0.275) / (3.3*10⁻⁵) = 1.55 * 10⁻⁷ = 0.16 μF