Fermi estimate — energy of a storm cloud

Problem

Fermi estimate: energy of a storm cloud. Estimate the energy stored in a storm cloud, modelled as a charged slab of side $L = 5\;\text{km}$, height $h = 5\;\text{km}$ above the ground, with average electric field $E = 10^5\;\text{V/m}$. How many days of average household electricity consumption does this correspond to? Also estimate the instantaneous discharge power assuming $\Delta t \sim 1$ ms.

Solution

Field energy density. The energy per unit volume of an electric field is $u = \frac{1}{2}\varepsilon_0 E^2 = \frac{1}{2}\cdot 8{,}85\cdot 10^{-12}\cdot (10^5)^2 \approx 4{,}4\cdot 10^{-2}\;\text{J/m}^3.$

Volume of the field region. Between the cloud and the ground the field occupies a volume $\mathcal{V} = L^2\,h = (5\cdot 10^3)^2\cdot 5\cdot 10^3 = 1{,}25\cdot 10^{11}\;\text{m}^3.$

Total energy. $U = u\,\mathcal{V} \approx 4{,}4\cdot 10^{-2}\cdot 1{,}25\cdot 10^{11} \approx \ev{U \sim 5\cdot 10^{9}\;\text{J}}.$ In kilowatt-hours: $U \approx 5\cdot 10^9 / 3{,}6\cdot 10^6 \approx \ev{1500\;\text{kWh}}$. With an average household consumption of $\sim 10$ kWh per day, this amounts to roughly $150$ days, i.e. a few months of a household’s electricity.

Discharge power. If all this energy is released in a lightning strike of duration $\Delta t \sim 1\;\text{ms}$: $P = \frac{U}{\Delta t} \approx \frac{5\cdot 10^9}{10^{-3}} \approx \ev{5\cdot 10^{12}\;\text{W} \sim 5\;\text{TW}}.$ A peak power of the order of terawatts — comparable to the electric power of the entire planet — but delivered only for an instant. This is the typical signature of lightning: enormous power, modest total energy.

Connections

Topics: Electric field and potential Concepts: Field energy density · Electric field Skills: Applying Gauss’s theorem Methods: Gauss’s theorem Objects: Infinite charged plane