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

Explanation

Surface charge density of the earth, `a = 10^(-9) C m^(-2)`
Current over the entire globe, `I = 1800 A`
Radius of the earth, `r = 6.37 xx 10^(6)` m
Surface area of the earth,
`A = 4nr^(2)`
`=4nxx(6.37 xx 10^(6))^(2)`
`= 5.09 xx 10^(14) m^(2)`
Charge on the earth surface,
`q = sigma xx A`
`=10^(-9) xx 5.09 xx 10^(14)`
`= 5.09 xx 10^(5) C`
The taken to neutralize the earth's surface = t
Current, `I =(q)/(t)`
`t = (q)/(I)`
`= (5.09 xx 10^(5))/(1800)=282.77 s`
Therefore, the time taken to neutralize the earth's surfaceis `282.77 s`