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 condutivity 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 earth's surface (This never happens in parctice there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth `= 6.3 xx 10^(6) m`)

Explanation

Here `r = 6.37 xx 10^(6) m, Q = 10^(-9) cm^(2), 1 = 1800 A`
Area of the globe A `= 4 pi r^(2) = 4 xx 3.14 xx (6.37 xx 10^(6))^(2)`
`t = (Q)/(I) = (509.64 xx 10^(3))/(1800 = 283.1 S`