Each plate of a parallel-plate air-filled capacitor has an area of $0.0040 \mathrm {~m} ^ { 2 }$, 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? $\left( \varepsilon _ { 0 } = 8.85 \times 10 ^ { - 12 } \mathrm { C } ^ { 2 } / \mathrm { N } \cdot \mathrm { m } ^ { 2 } \right)$

Question

Quizplus · Accepted Answer

The energy density $u$ in the region between the plates of a capacitor can be calculated using the formula $u = \frac{1}{2} \varepsilon_0 E^2$, where $\varepsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \mathrm{C}^2/\mathrm{N}\cdot\mathrm{m}^2$) and $E$ is the electric field ($8.2 \times 10^6 \mathrm{N}/\mathrm{C}$). Substituting the given values, $u = \frac{1}{2} \times 8.85 \times 10^{-12} \times (8.2 \times 10^6)^2 \approx 300 \mathrm{J}/\mathrm{m}^3$.

Each Plate of a Parallel-Plate Air-Filled Capacitor Has an Area 0.0040 m20.0040 \mathrm {~m} ^ { 2 }0.0040 m2

Each Plate of a Parallel-Plate Air-Filled Capacitor Has an Area $0.0040 \mathrm {~m} ^ { 2 }$