Correct Answer - Option 4 : 0.7765 m
Concept:
The capacitance of a parallel plate capacitor is given by
\(C = \frac{{\varepsilon A}}{d} = \frac{{{\varepsilon _0}{\varepsilon _r}A}}{d}\)
Where εr 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