The square plates of a 5000-pF parallel-plate capacitor measure 50 mm by 50 mm and are separated by a dielectric that is $0.23 \mathrm {~mm}$ thickand totally fills the region between the plates. The voltage rating (the maximum safe voltage) of the capacitor is $400 \mathrm {~V}$ What is the maximum energy that can be stored in this capacitor without damaging it? (ε₀ = 8.85 × 10^-12C²/N ∙ m²)

Question

Quizplus · Accepted Answer

The maximum energy stored in a capacitor is given by $ U = \frac{1}{2} C V^2 $. Substituting $ C = 5000 \times 10^{-12} \, \text{F} $ and $ V = 400 \, \text{V} $, we get $ U = \frac{1}{2} \times 5000 \times 10^{-12} \times 400^2 = 0.40 \, \text{mJ} $.

The Square Plates of a 5000-PF Parallel-Plate Capacitor Measure 50 0.23 mm0.23 \mathrm {~mm}0.23 mm

The Square Plates of a 5000-PF Parallel-Plate Capacitor Measure 50 $0.23 \mathrm {~mm}$