Problem — Fermi estimate: capacitance of the Earth

Problem

Fermi estimate: Earth-Moon capacitance. Estimate the electric capacitance of the Earth treated as an isolated sphere ($R_T = 6{,}37\cdot 10^6\ \text{m}$), using $C = 4\pi\varepsilon_0 R$. Comment on the order of magnitude compared with common laboratory capacitors.

Solution

Capacitance of an isolated sphere. With $\varepsilon_0 = 8{,}85\cdot 10^{-12}\ \text{F/m}$ and $R_T = 6{,}37\cdot 10^6\ \text{m}$:

$C_T = 4\pi\varepsilon_0 R_T = 4\pi \cdot 8{,}85\cdot 10^{-12} \cdot 6{,}37\cdot 10^6 = 1{,}112\cdot 10^{-10} \cdot 6{,}37\cdot 10^6 \approx 7{,}08\cdot 10^{-4}\ \text{F}$

That is, $C_T \approx 7\cdot 10^{-4}\ \text{F} \sim 700\ \mu\text{F}$.

Comment. Surprisingly this capacitance is of the same order as a large electrolytic laboratory capacitor (hundreds of $\mu\text{F}$): despite the enormous radius, the Earth as an isolated sphere has a modest capacitance, because $C$ grows only linearly with the radius.

$\ev{C_T \approx 7\cdot 10^{-4}\ \text{F} \sim 700\ \mu\text{F}}$

Linked atoms

Topics: Campo elettrico e potenziale Concepts: Capacità e condensatore Skills: Stime di Fermi Objects: Sfera carica