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 4.5, and the thickness of one sheet of it is 0.20 mm. If the sheets paper measures 27 cm x 36 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?

Capacitance of capacitor C = εA/d where C = 1.3 nF = 1.3 ×10⁻⁹ F, ε = permittivity of paper = 4ε₀, ε₀ = 8.854 × 10⁻¹² C²N⁻¹m⁻¹ , A =area of paper sheets = 27 cm x 36 cm = 972 cm ² = 0.0972 m² and d = distance between aluminium foils = number of sheets,n × thickness of sheet, t = nt , t = 0.20 mm = 0.2 × 10⁻³ m

C = εA/d = 4ε₀A/nt

n = 4ε₀A/Ct = 4 × 8.854 × 10⁻¹² C²N⁻¹m⁻¹ × 0.0972 m²/(1.3 ×10⁻⁹ F × 0.2 × 10⁻³ m) = 13.24 ≅ 13 sheets

abidemiokin abidemiokin · Answer 2 · 2020-04-09T05:54:28+02:00

Answer:

15sheets

Explanation:

Formula fie calculating the capacitance of a capacitor is expressed as;

C =εA/d where;

C is the capacitance of the capacitor = 1.3nF = 1.3 × 10^-9F

ε is the permittivity of the paper

If ε₀ = permittivity of free space = 8.854 × 10⁻¹² C²N⁻¹m⁻¹

ε = 4.5ε₀ = 4.5×8.854 × 10⁻¹² C²N⁻¹m⁻¹

ε = 39.843× 10⁻¹² C²N⁻¹m⁻¹

A is the area of the plates

A = 27cm×36cm

A = 0.27×0.36

A = 0.0972m²

d is the distance between the plates.

d = number of paper between the plates n × thickness t

d = nt

If t = 0.2mm

t = 0.0002m

d = 0.0002n

Substituting the given datas into the formula to get n, we have;

1.3×10^-9 = 39.843 × 10⁻¹²(0.0972)/0.0002n

1.3×10^-9 × 0.0002n = 39.843 × 10⁻¹² × 0.0972

2.6×10^-13n = 3.87×10⁻¹²

n = 3.87×10⁻¹²/2.6×10^-13

n = 1.49×10¹

n = 14.9/approximately 15sheets