Question

The earths surface has a negative surface charge density of ${10}^{-9}C{m}^{-2}$. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earths surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = $6.37\times {10}^6$ m.)

Answer 1

Surface charge density of the earth, $\sigma ={10}^{-9}C{m}^{-2}$
Current over the entire globe, $I=1800A$
Radius of the earth, $r=6.37\times {10}^{6}m$
Surface area of the earth,

$A=4\pi r^2=5.09\times {10}^{14}{m}^2$

Charge on earth surface, $q=\sigma A=5.09\times {10}^{5}C$

Time taken to neutralize the earth’s surface = t.

Current $I=\frac{q}{t}$

$⟹t=5.09\times {10}^5/1800=283s$