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

Here, `r = 6.37 xx 10^(6)m, sigma = 10^(-9) C m^(-2), I = 1800 A`
Area of the globe, `A = 4 pi r^(2) = 4 xx 3.14 xx (6.37 xx 10^(6))^(2) = 509.64 xx 10^(12) m^(2)`
As charge, `Q = sigma xx A = 10^(-9) xx 509.64 xx 10^(12) = 509.64 xx 10^(3)C`
`:. t = Q/I = (509.64 xx 10^(3))/1800 = 283.1s`