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? (ε₀ = 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{K \cdot \varepsilon_0 \cdot A}{d} $. Substituting the given values: $ C = \frac{2.4 \cdot 8.85 \times 10^{-12} \cdot 6.0 \times 10^{-2}}{0.04} = 32 \times 10^{-12} $ F.

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

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