A parallel-plate capacitor with plate separation of 4.0 cm has a plate area of 6.0 × $10 ^ { - 2 }$ $m ^ { 2 }$ .What is the capacitance of this capacitor if a dielectric material with a dielectric constant of 2.4 is placed between the plates, completely filling the space? (?₀ = 8.85 × 10^-12C²/N ? m²)

Question

Quizplus · Accepted Answer

The capacitance of a parallel-plate capacitor with a dielectric is given by $ C = \frac{{\kappa \varepsilon_0 A}}{d} $. Substituting the given values: $ \kappa = 2.4 $, $ \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 $, $ A = 6.0 \times 10^{-2} \, \text{m}^2 $, and $ d = 0.04 \, \text{m} $, we get $ C = \frac{{2.4 \times 8.85 \times 10^{-12} \times 6.0 \times 10^{-2}}}{0.04} = 32 \times 10^{-12} \, \text{F} $.

A Parallel-Plate Capacitor with Plate Separation of 4 10−210 ^ { - 2 }10−2

A Parallel-Plate Capacitor with Plate Separation of 4 $10 ^ { - 2 }$