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 $\vec { E }$ field with a magnitude 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? (ε₀ = 8.85 × 10^-12C²/N ∙ m²)

Question

Quizplus · Accepted Answer

The energy density in a capacitor is given by:$$u = \frac{1}{2} \epsilon_0 E^2$$where:* u is the energy density in J/m^3* ε₀ is the permittivity of free space in C^2/N ∙ m^2* E is the electric field strength in N/CPlugging in the given values, we get:$$u = \frac{1}{2} (8.85 imes 10^{-12} \mathrm{C}^2/\mathrm{N} \cdot \mathrm{m}^2) (8.2 imes 10^6 \mathrm{N}/\mathrm{C})^2 = 300 \mathrm{J}/\mathrm{m}^3$$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 $m ^ { 2 }$