Cloud-to-ground potential difference

Estimate the potential difference between a storm cloud (at $\sim 1$ km from the ground) and the ground before a lightning strike, knowing that the breakdown electric field of air is $\sim 3\cdot 10^6$ V/m.

Solution

Setup. We treat the cloud-to-ground region as an approximately uniform field (parallel-plate capacitor model). For a uniform field the potential difference is $\abs{\Delta V} \approx E\,d$ Just before the discharge the field reaches the breakdown value of air. Data (estimate). $E \approx 3\cdot 10^6 \ \text{V/m}$, $d \approx 1 \ \text{km} = 10^3$ m. Substitution. $\abs{\Delta V} \approx (3\cdot 10^6)(10^3) = 3\cdot 10^9 \ \text{V}$ Result. $\ev{\abs{\Delta V} \sim 10^9 \ \text{V}}$ Comment. The order of magnitude is a billion volts. The estimate is rough: the real field is not uniform and breakdown occurs from regions where the field is locally amplified (points, droplets), so the effective potential can be lower. The correct idea of the scale of the phenomenon nonetheless stands.

Links

Topics: Electric field and potential Concepts: Electric field · Electric potential Skills: Superposition principle Methods: Gradient method Objects: Infinite charged plane