Energy density in a thunderstorm

Problem

During a thunderstorm the electric field between cloud and ground can approach the dielectric breakdown strength of air, but under ordinary conditions it is around $E \approx 200\;\text{V/cm}$. Estimate the electric energy density $u = \tfrac12 \varepsilon_0 E^2$ present in the air under these conditions.

Solution

Converting the field to SI units. We need to express the field in V/m: $E = 200\;\text{V/cm} = \frac{200\;\text{V}}{10^{-2}\;\text{m}} = 2\cdot 10^{4}\;\text{V/m}$

Energy density formula. The energy per unit volume stored in an electric field in vacuum (or air) is $u = \tfrac12\,\varepsilon_0\,E^2$

Numerical substitution. With $E^2 = (2\cdot 10^4)^2 = 4\cdot 10^{8}\;\text{V}^2/\text{m}^2$: $u = \tfrac12\,(8{,}85\cdot 10^{-12})\,(4\cdot 10^{8})$

Result. $\ev{u \approx 1{,}77\cdot 10^{-3}\;\text{J/m}^3}$

This is a modest energy density: the field under ordinary conditions is far from air’s dielectric breakdown strength (~$3\cdot 10^6\;\text{V/m}$), beyond which a lightning discharge would be triggered.

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