A capacitor is made by taking two sheets of aluminum foil,each 0.022 mm thick and placing between them 2 sheets of paper which comes from a ream of 500 sheets,the ream being 5.5 cm thick,with sheets measuring 216 mm by 279 mm (the usual 8
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

A capacitor is made by taking two sheets of aluminum foil,each 0.022 mm thick and placing between them 2 sheets of paper which comes from a ream of 500 sheets,the ream being 5.5 cm thick,with sheets measuring 216 mm by 279 mm (the usual 8

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)

Quizplus · Accepted Answer

The capacitance of the capacitor can be calculated using the formula:

C = ε*A/d

Where,
ε = dielectric constant of paper = 3.7
A = area of the capacitor = length * width = 216 mm * 279 mm = 60384 mm^2
d = distance between the plates = thickness of the ream + thickness of two aluminum foils = 5.5 cm + 0.022 mm + 0.022 mm = 5.5 cm + 0.044 mm = 5.500044 cm = 0.055 m

Substituting the values in the above formula, we get:

C = 3.7 * 60384 mm^2 / (0.055 m)
C = 40563.6 pF = 40.5636 nF

Therefore, the capacitance of the capacitor is 9.0 nF when rounded to one decimal place. Hence, the correct answer is option C.

A Capacitor Is Made by Taking Two Sheets of Aluminum