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 ($U$) stored in a capacitor can be calculated using the formula $U = \frac{1}{2}CV^2$, where $C$ is the capacitance in farads and $V$ is the voltage in volts. Given that the capacitance ($C$) is $5000 \times 10^{-12}$ F (or 5000 pF) and the maximum safe voltage ($V$) is 400 V, substituting these values into the formula gives: $U = \frac{1}{2} \times 5000 \times 10^{-12} \times 400^2 = 0.4 \times 10^{-3}$ J, or 0.40 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 $0.23 \mathrm {~mm}$