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? (?₀ = 8.85 × 10^-12C²/N ? m²)

Question

Quizplus · Accepted Answer

The maximum energy that can be stored in a capacitor is given by:$$U=\frac{1}{2}CV^2$$where:* U is the energy stored in joules (J)* C is the capacitance in farads (F)* V is the voltage across the capacitor in volts (V)In this problem, we are given the following information:* C = 5000 pF = 5 × 10^-12 F* V = 400 VSubstituting these values into the equation above, we get:$$U=\frac{1}{2}(5 	imes 10^{-12} 	ext{ F})(400 	ext{ V})^2 = 0.4 	ext{ J} = 400 	ext{ mJ}$$Therefore, the maximum energy that can be stored in this capacitor without damaging it is 400 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 $0.23 \mathrm {~mm}$