Question

Problem 21.35 Enhanced with Feedback A science-fair radio uses a homemade capacitor made of two 45 cm x 45 cm sheets of aluminum foil separated by a 0.30-mm-thick sheet of paper. Assume dielectric constant of paper is 3.0. You may want to review (D ages 92-695 Part A What is its capacitance? Express your answer using two significant figures Submit My Answers Give Up

Answer 1

(1) Capacitance of a parallel plate capacitor with insulator/dielectric between the plates is given by,

C = ₀A / [d-t+(t/k)]

Here, A is area of plate, d is distance between plates, t is thickness of insulating plate, k is dielectric constant and ₀ is permittivity of free space.

For the given problem, t = d

Which reduces the above expression as, C = ₀Ak / t

Given, A = (45 x 45 x 10^-4) m² , k=3 , t = 0.30 x 10^-3 m and ₀ = 8.85 x 10^-12 F/m

Putting all the values we get,

C = [8.85 x 10^-12 F/m x (45 x 45 x 10^-4) m² x 3] / [0.30 x 10^-3 m]

C = 1.79 x 10^-8 Farad

(2) Time constant for discharge of capacitor is given by,

= RC

Given, R = 7 x 10³ ohm , C = 1.79 x 10^-8 Farad

Putting all the values we get,

= 7 x 10³ ohm x 1.79 x 10^-8 Farad

= 12.53 x 10^-5 s