A rolled-paper capacitor of value 0.02 μF is to be constructed using two strips of Aluminium of width 6 cm, and wax impregnated paper of thickness 0.06 mm, whose relative permittivity is 3.

The length of foil strips should be:

Concept:

The capacitance of a parallel plate capacitor is given by

\(C = \frac{{\varepsilon A}}{d} = \frac{{{\varepsilon _0}{\varepsilon _r}A}}{d}\)

Where ε_r is the relative permittivity

ε₀ is the absolute permittivity

A = area of plates

d = distance between the two plates.

Calculation:

Rolled paper capacitor C = 0.02 μF

Width of Aluminium strips (w) = 6 cm = 0.06 m

Rolled paper capacitor means an Aluminium strip is rolled with parallel capacitor plates.

Therefore, the area of that parallel plate = Area of Al

Area of Al. (A) = ω × ℓ = 0.06 × ℓ m²

Distance between two parallel plates = thickness of Al. strips

= 0.06 mm = 0.06 × 10^-3 m

Now, \(C = \frac{{{\varepsilon _r}{\varepsilon _0}A}}{d} = \frac{{3 \times 8.85 \times {{10}^{ - 12}} \times .06 \times \ell }}{{0.06 \times {{10}^{ - 3}}}}\)

⇒ 0.02 × 10^-6 = 3 × 8.85 × 10^-9 × ℓ

⇒ ℓ = 0.7765 m