A capacitor is made with two strips of metal foil, each 2.5 cm wide by 50 cm long, with a 0.70-?m thick strip of paper ? = 3.7) sandwiched between them. The capacitor is rolled up to save space. What is the capacitance of this device? The permittivity of free space $\varepsilon$₀ = 8.85 × 10^-12 F/m.)

Question

Quizplus · Accepted Answer

The capacitance of a parallel plate capacitor is given by the formula:

C = εA/d

Where ε is the permittivity of the dielectric material between the plates, A is the area of each plate, and d is the distance between the plates.

In this case, we have two metal foils as the plates with a strip of paper κ = 3.7 between them. The width of each plate is 2.5 cm and the length is 50 cm. So the area of each plate is:

A = 2.5 cm x 50 cm = 125 cm^2 = 0.0125 m^2

The distance between the plates is the thickness of the paper, which is 0.70 μm = 7.0 x 10^-7 m.

Therefore, the capacitance of the device is:

C = (3.7 x 8.85 x 10^-12 F/m) (0.0125 m^2) / (7.0 x 10^-7 m) = 0.0005795 F = 0.58 μF

So the answer is C.

A Capacitor Is Made with Two Strips of Metal Foil ε\varepsilonε

A Capacitor Is Made with Two Strips of Metal Foil $\varepsilon$