The earth's surface has-a negative surface charge density of `10^(-9) Cm^(-2)`. The potential difference of 400kV between the top of the atmosphere and the surface results in a current of only 1800A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time would be required to neutralize the earth's surface charge? Take radius of earth `= 6.37xx 10^(6)` m.

Explanation

Given `sigma = 10^(-9) Cm^(-2) , V = 400 xx 10^(3) V, I= 1800 A, t =? , R = 6.37 xx 10^(6)`m
Surface area of the earth `=4piR^(2)`
i.e., `A=4xx3.142xx(6.37)^(2)xx10^(12)m^(2)`
`A=5.0964xx10^(14)m^(2)`
Total charge on the earth Q = A
i.e., `Q=5.0964xx10^(14)xx10^(-9)C`
i.e., `Q=5.096xx10^(5)C`
Time required for total discharge
`t=Q/I=(5.096xx10^(5))/1800`
i.e., `t=2.83xx10^(2)s`
Thus, the time required to neutralize the earth.s surface charge is `2.83xx10^(2)s`