A parallel-plate capacitor consists of two parallel, square plates having dimensions 1.0 cm by 1.0 cm. The plates are separated by $1.0 \mathrm {~mm}$ and the space between them is filled with Teflon, which has a dielectric constant of 2.1. What is the capacitance of this capacitor? (ε₀ = 8.85 × 10^-12C²/N ∙ m²)

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The capacitance of a parallel-plate capacitor is given by:$$C = \frac{\epsilon_0 A}{d}$$where:* C is the capacitance in farads (F)* ε₀ is the permittivity of free space (8.85 × 10^-12 C²/N ∙ m²)* A is the area of each plate in square meters (m²)* d is the distance between the plates in meters (m)In this problem, the dimensions of each plate are 1.0 cm by 1.0 cm, which is equal to 1.0 × 10^-2 m by 1.0 × 10^-2 m. The distance between the plates is 1.0 mm, which is equal to 1.0 × 10^-3 m. The dielectric constant of Teflon is 2.1.Substituting these values into the equation, we get:$$C = \frac{(8.85 imes 10^{-12} ext{ C}^2/ ext{N} \cdot ext{m}^2)(1.0 imes 10^{-2} ext{ m})^2}{1.0 imes 10^{-3} ext{ m}} = 1.9 imes 10^{-12} ext{ F} = 1.9 ext{ pF}$$Therefore, the capacitance of the capacitor is 1.9 pF.

A Parallel-Plate Capacitor Consists of Two Parallel, Square Plates Having 1.0 mm1.0 \mathrm {~mm}1.0 mm

A Parallel-Plate Capacitor Consists of Two Parallel, Square Plates Having $1.0 \mathrm {~mm}$