A parallel-plate capacitor consists of two parallel, square plates that have dimensions 1.0 cm by 1.0 cm. If the plates are separated by 1.0 mm, and the space between them is filled with teflon, what is the capacitance of this capacitor? (The dielectric constant for teflon is 2.1, and ε₀ = 8.85 × 10^-12 C²/N ∙ m².) A) 1.9 pF B) 0.44 pF C) 2.1 pF D) 0.89 pF

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The capacitance of a parallel-plate capacitor is given by the formula:C = εA/dwhere:C is the capacitance in farads (F)ε is the permittivity of the dielectric material between the plates in farads per meter (F/m)A is the area of each plate in square meters (m^2)d is the distance between the plates in meters (m)In this problem, we are given the following information:The plates are square with dimensions 1.0 cm by 1.0 cm. Therefore, the area of each plate is:A = (1.0 cm)^2 = 1.0 × 10^-4 m^2The plates are separated by 1.0 mm. Therefore, the distance between the plates is:d = 1.0 mm = 1.0 × 10^-3 mThe space between the plates is filled with teflon, which has a dielectric constant of 2.1. Therefore, the permittivity of the dielectric material is:ε = ε0 * k = 8.85 × 10^-12 F/m * 2.1 = 1.86 × 10^-11 F/mNow we can plug these values into the formula for capacitance:C = εA/d = (1.86 × 10^-11 F/m)(1.0 × 10^-4 m^2)/(1.0 × 10^-3 m) = 1.86 × 10^-12 F = 1.9 pFTherefore, the capacitance of the capacitor is 1.9 pF.

A Parallel-Plate Capacitor Consists of Two Parallel, Square Plates That