Question

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.)

Answer 1

Capacitance = 1.55 × 10⁻⁷ F = 0.16 μF

Step-by-step 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