Question

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: 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?

Express your answer as a whole number.

Part B:

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?

Express your answer using two significant figures.

Part C: 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?

1.She will need a larger area of Teflon than of posterboard.

2.She will need a smaller area of Teflon than of posterboard

Answer 1