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.The plates are separated by 1.0 mm.The space between the plates is filled with teflon.The dielectric constant for teflon is 2.1.The permittivity of free space is 8.85 × 10^-12 C^2/N ∙ m^2.First, we need to convert the dimensions of the plates and the distance between the plates to meters:1.0 cm = 0.01 m1.0 mm = 0.001 mNext, we can calculate the area of each plate:A = (0.01 m)^2 = 1.0 × 10^-4 m^2Now, we can calculate the capacitance of the capacitor:C = εA/dC = (2.1 × 8.85 × 10^-12 C^2/N ∙ m^2)(1.0 × 10^-4 m^2)/(0.001 m)C = 1.9 × 10^-12 FC = 1.9 pFTherefore, the capacitance of the capacitor is 1.9 pF.

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