Mastering Physics: Exercise 24.38

Mastering Physics: Exercise 24.38

A budding electronics hobbyist wants to make a simple 1.3 nF capacitor for tuning her crystal radio, using two sheets of aluminum foil as plates, with a few sheets of paper between them as a dielectric. The paper has a dielectric constant of 4.6, and the thickness of one sheet of it is 0.18 mm .

Part A

Question: If the sheets of paper measure 28 cm × 38 cm and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance?

Answer: 

Area = 28 cm × 38 cm

Area = 1064 cm^2

Area = 0.1064 m^2

C = K× ϵ_0 × A / d

d = K× ϵ_0 × A / C

d = 4.6 × 8.85×10^(−12) × 0.1064 / (1.3×10^(−9))

d = 3.332×10^(−3)

Sheets = 3.332×10^(−3)  (0.18×10^(−3))

Sheets = 18.5

n =  19 sheets 

Part B

Question: Suppose for convenience she wants to use a single sheet of posterboard, with the same dielectric constant but a thickness of 13.0 mm , instead of the paper. What area of aluminum foil will she need for her plates to get her 1.3 nF of capacitance?

Answer: 

C = K× ϵ_0 × A / d

A = d × C / (K× ϵ_0)

A = 13×10^(−3) × 1.3×10^(−9) / (4.6× 8.85×10^(−12))

A =  0.42  m^2

Part C

Question: Suppose she goes high-tech and finds a sheet of Teflon of the same thickness as the posterboard to use as a dielectric. Will she need a larger or smaller area of Teflon than of posterboard?

Answer:  She will need a larger area of Teflon than of posterboard.

Part D

Question: Explain.

Answer: Teflon has a smaller dielectric constant (2.1) than the posterboard, so she will need more area to achieve the same capacitance.