Each plate of a parallel-plate air-filled capacitor has an area of 0.0040 $m ^ { 2 }$ , and the separation of the plates is $0.020 \mathrm {~mm} .$ An electric field of $8.2 \times 10 ^ { 6 } \mathrm {~N} / \mathrm { C }$ is present between the plates. What is the energy density in the region between the plates? (?₀ = 8.85 × 10^-12C²/N ? m²)

Question

Quizplus · Accepted Answer

The energy density in a capacitor is given by:$$u = \frac{1}{2} \varepsilon_0 E^2$$where:* u is the energy density in J/m^3* ε0 is the permittivity of free space in C^2/Nm^2* E is the electric field strength in V/mPlugging in the given values, we get:$$u = \frac{1}{2} (8.85 	imes 10^{-12} \mathrm{C}^2/\mathrm{Nm}^2) (8.2 	imes 10^6 \mathrm{V/m})^2 = 300 \mathrm{J/m}^3$$Therefore, the energy density in the region between the plates is 300 J/m^3.

Each Plate of a Parallel-Plate Air-Filled Capacitor Has an Area m2m ^ { 2 }m2

Each Plate of a Parallel-Plate Air-Filled Capacitor Has an Area $m ^ { 2 }$