Fermi estimate — electrostatic shock

Problem

Fermi estimate: electrostatic shock. A person standing on a dry wooden floor builds up an electrostatic charge of $\sim 1\;\mu\text{C}$ by rubbing their feet. Estimate the potential difference between the person and the floor, modelling them as a conducting sphere of $35\;\text{cm}$ radius. Compare with the dielectric strength of air ($\sim 3$ kV/mm) to explain the classic everyday spark.

Solution

Potential of the sphere. The potential on the surface of a conducting sphere of radius $R$ with charge $q$ is that of a point charge concentrated at the centre: $V = \frac{kq}{R} = \frac{9\cdot 10^9 \cdot 1\cdot 10^{-6}}{0{,}35} \approx \ev{V \approx 25\;\text{kV}}.$ Tens of thousands of volts: a surprisingly high value for such a small charge.

Comparison with the dielectric strength. Air ionises (sparks) when the field exceeds $\sim 3\;\text{kV/mm}$. Bringing a finger close to a metal handle, the air gap shrinks to fractions of a millimetre; with $25$ kV available, an air gap of only $d \sim \frac{V}{E_\text{max}} = \frac{25\;\text{kV}}{3\;\text{kV/mm}} \approx 8\;\text{mm}$ is enough for the field to reach the discharge threshold. That is why the spark jumps before you even touch the object: the air breaks down and conducts, dissipating the accumulated charge. It is the same physics as lightning, on a domestic scale.

Connections

Topics: Electric field and potential Concepts: Electric potential · Conductors in electrostatic equilibrium Skills: Applying Gauss’s theorem Methods: Gauss’s theorem Objects: Charged sphere