Question

A budding electronics hobbyist wants to make a simple 1.4 nFnF 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 4.8, and the thickness of one sheet of it is 0.18 mmmm .

A)If the sheets of paper measure 25 cmcm ×× 39 cmcm 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.

b) Suppose for convenience she wants to use a single sheet of posterboard, with the same dielectric constant but a thickness of 15.0 mmmm , instead of the paper. What area of aluminum foil will she need for her plates to get her 1.4 nFnF of capacitance? Express your answer using two significant figures.

Answer 1

Given: C=1.4 nF=1.4*10^-9 F

Dielectric constant K=4.8

A)Area A=25cm*39 cm=975cm²=975*10^-4 m²

Thickness of single sheet x=0.18 mm=0.18*10^-3 m

Let's say number of sheet=n

Therefore total thickness of sheet d=n*x

We know the formula of capacitance is,

d=2.9584*10^-3 m

We know number of sheet n=d/x

So,

n=16 sheets

b) Given: Thickness d= 15mm=15*10^-3 m

C=1.4nF=1.4*10^-9 F

K=4.8

We know

A=0.4944 m²

A=0.49 m²