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

Let t be the time required to neutralise the earth.s surface. Then
`t= ("Charge")/("Current")`
Also we know that:
Charge density, `sigma = ("Charge")/("Area")`
Here, radius of earth, `R= 6.37 xx 10^(6)m`
`rArr` Surface area, `A= 4pi R^(2)`
Charge density `sigma= 10^(-9) C m^(-2)`
`rArr` Charge `= sigma xx A`
`=10^(-9) xx 4pi (6.37 xx 10^(6))^(2)C`
`:. t= ("Charge")/("Current")`
`-(10^(-9) xx 4pi (6.37 xx 10^(6))^(2))/(1,800)`
=283.4s