The earth’s 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 neutralize the earth’s 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 × 106 m.)
Surface Charge density of earth d = 10-9 Coulomb per meter square
Current over the entire globe = 1800 Ampere
Radius of earth r = 6.37 × 106 meter
Surface area of earth A = 4 × π × radius × radius
A = 4 × π × 6.37 × 106 × 6.37 × 106
A = 5.09 × 1014 m2
Charge on the earth surface q = d × A
q = 10-9 × 5.09 × 1014 m2
q = 5.09 × 105 Coulomb
Let the time taken to neutralize earth surface = t
Current 
t = 5.09 × 105C/1800 A
t = 282.78 seconds.