Question

A parallel-plate capacitor is made from two aluminum-foil sheets, each 5.9 cm wide and 5.6 m long. Between the sheets is a Teflon strip of the same width and length that is 3.1×10^−2 mm thick. What is the capacitance of this capacitor? (The dielectric constant of Teflon is 2.1.)

Answer 1

1.98 x 10⁻⁷ F

Step-by-step explanation:

w = width of the sheet = 5.9 cm = 0.059 m

L = length of the sheet = 5.6 m

Area of the sheet is given as

A = L w = (5.6) (0.059) = 0.3304 m²

d = distance between the sheets = 3.1 x 10⁻⁵ m

k = dielectric constant of teflon = 2.1

Capacitance is given as

`C = (k\epsilon _(o)A)/(d)`

`C = ((2.1)(8.85* 10^(-12))(0.3304))/(3.1* 10^(-5))`

C = 1.98 x 10⁻⁷ F