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 520 sheets,the ream being 5.3 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) A) 33 nF B) 45 nF C) 19 nF D) 1.5 nF E) 5 nF

Question

Quizplus · Accepted Answer

The thickness of one sheet of paper is 5.3 cm / 520 = 0.0102 cm = 0.102 mm. The area of the paper is 216 mm × 279 mm = 60264 mm² = 0.060264 m². The capacitance C is given by C = (ε₀ * ε_r * A) / d, where ε₀ is the permittivity of free space, ε_r is the dielectric constant, A is the area, and d is the separation. Substituting the values, C = (8.85 × 10⁻¹² * 3.7 * 0.060264) / 0.000102 = 19 nF.

A Capacitor Is Made by Taking Two Sheets of Aluminum