Electrostatics
Intermediate
Capacitor
Source: High school physics (Chinese)
Problem
Treat the Earth as an isolated spherical conductor of radius $6400$ km, and let the measured electric field at the surface be $100$ V/m.
What is the total charge carried by the Earth?
$Q \approx 4.55 \times 10^{5}$ C (negative), i.e. $\approx -4.55 \times 10^{5}$ C.
For an isolated spherical conductor in equilibrium, the surface field obeys Coulomb's law as if all charge sat at the center:
$$E = \dfrac{kQ}{R^2} \quad \Longrightarrow \quad Q = \dfrac{E R^2}{k}.$$Substituting $R = 6.4 \times 10^{6}$ m, $E = 100$ V/m, $k = 9.0 \times 10^{9}\ \mathrm{N \cdot m^2/C^2}$:
$$Q = \dfrac{100 \times (6.4 \times 10^{6})^2}{9.0 \times 10^{9}} \approx 4.55 \times 10^{5}\ \mathrm{C}.$$Since the measured surface field at the Earth points downward (into the surface), the Earth carries net negative charge.