A parallel-plate capacitor consists of two parallel,square plates having dimensions 1.0 cm by 1.0 cm.The plates are separated by $1.0 \mathrm {~mm} \text {, }$ 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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Quizplus · Accepted Answer

The capacitance of a parallel-plate capacitor is given by $C = \frac{\kappa \epsilon_0 A}{d}$, where $\kappa$ is the dielectric constant, $\epsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \, \text{F/m}$), $A$ is the area of the plates, and $d$ is the separation between the plates. Given $\kappa = 2.1$, $A = (1.0 \, \text{cm} \times 1.0 \, \text{cm}) = 1.0 \times 10^{-4} \, \text{m}^2$, and $d = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m}$, substituting these values gives $C = \frac{2.1 \times 8.85 \times 10^{-12} \times 1.0 \times 10^{-4}}{1.0 \times 10^{-3}} = 1.8655 \times 10^{-12} \, \text{F} = 1.9 \, \text{pF}$.

A Parallel-Plate Capacitor Consists of Two Parallel,square Plates Having Dimensions 1.0 mm, 1.0 \mathrm {~mm} \text {, }1.0 mm,

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