Given, radius of earth R = 6.37 x
106 m
Negative surface charge density α =
10−9 C/
m2 Potential difference V = 400 kV = 400 x
103 V
Current on the globe I = 1800 A.
Surface area of earth A = 4π
R2 = 4 × 3.14 ×
(6.37×106)2 = 509.64 ×
1012m2 Charge on earth surface Q = Area of earth surface x surface charge density.
Q = Aσ = 509.64 ×
1012m2 = 509.64 ×
103 C
We know that Q = IT.
∴ Time required to neutralize earth’s surface.
t =
=
t = 283.1 s or t = 4 min 43 s
Thus, the time required to neutralize the earth’s surface is 283.1 s.