Dimensional check of the energy density

Problem

Starting from $\varepsilon_0 = 8{,}85\cdot 10^{-12}\;\text{C}^2/(\text{N}\cdot\text{m}^2)$ and $E$ in V/m, verify dimensionally that the quantity $u = \tfrac12 \varepsilon_0 E^2$ does indeed have the units of energy per unit volume. Use the relations $1\;\text{V} = 1\;\text{J/C}$ and $1\;\text{N}\cdot\text{m} = 1\;\text{J}$.

Solution

Units of the factors. The numerical factor $\tfrac12$ is dimensionless, so the units of $u$ are the product of the units of $\varepsilon_0$ and $E^2$: $[\varepsilon_0] = \frac{\text{C}^2}{\text{N}\cdot\text{m}^2}, \qquad [E^2] = \left(\frac{\text{V}}{\text{m}}\right)^2 = \frac{\text{V}^2}{\text{m}^2}$

Product. $[u] = \frac{\text{C}^2}{\text{N}\cdot\text{m}^2}\cdot\frac{\text{V}^2}{\text{m}^2} = \frac{\text{C}^2\,\text{V}^2}{\text{N}\cdot\text{m}^4}$

Substituting $1\;\text{V} = 1\;\text{J/C}$. Thus $\text{C}^2\,\text{V}^2 = \text{C}^2\cdot(\text{J/C})^2 = \text{J}^2$: $[u] = \frac{\text{J}^2}{\text{N}\cdot\text{m}^4}$

Substituting $1\;\text{J} = 1\;\text{N}\cdot\text{m}$. One of the two joules in the numerator is written as $\text{N}\cdot\text{m}$: $[u] = \frac{\text{J}\cdot(\text{N}\cdot\text{m})}{\text{N}\cdot\text{m}^4} = \frac{\text{J}\cdot\text{N}\cdot\text{m}}{\text{N}\cdot\text{m}^4} = \frac{\text{J}}{\text{m}^3}$

Result. $\ev{[u] = \text{J/m}^3}$

The quantity $\tfrac12\varepsilon_0 E^2$ indeed has the dimensions of energy per unit volume: it is an energy density.

Links

Topics: Campo elettrico e potenziale Concepts: Densità di energia del campo