A capacitor is made by taking two sheets of aluminum foil,each 0.022 mm thick and placing between them a sheet of paper which comes from a ream of 580 sheets,the ream being 5.4 cm thick with sheets measuring 216 mm by 279 mm (the usual 8 1/2 by 11).What is the capacitance of the capacitor made this way if the dielectric constant of the paper is 3.7? (ε₀ = 8.85 × 10−¹² C²/N × m²)

Question

Quizplus · Accepted Answer

Firstly, we need to find the area of each foil sheet. The dimensions of each sheet are given as 216 mm by 279 mm. Therefore, the area of each sheet is:

A = (216 mm) x (279 mm) = 60,264 mm^2

Since we have two sheets, the total area is:

A_tot = 2 x 60,264 mm^2 = 120,528 mm^2

Next, we need to find the capacitance using the formula:

C = εA/d

where ε is the dielectric constant, A is the area of the plates, and d is the distance between the plates.

The distance between the plates is the thickness of the paper, which is given as 5.4 cm = 54 mm.

Substituting the values, we get:

C = (3.7)(120,528 mm^2)/(54 mm) = 8.2 nF

However, remember that we have two sheets of foil and a sheet of paper in between. Therefore, we need to multiply by 2 to account for the second foil sheet:

C_tot = 2 x 8.2 nF = 16.4 nF

Therefore, the closest answer choice is C) 21 nF.

A Capacitor Is Made by Taking Two Sheets of Aluminum