# 3.14 The earth’s surface has a negative surface charge density of 10-9 C m-2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A cover the entire globe. If there were no mechanism of sustaining atmospheric 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 thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 10 6 m.)

Given,

Surface Charge density of earth $\rho$ = $\dpi{100} 10^{-9}C m^{-2}$

Current over the entire globe = 1800 A

Radius of earth, r = 6.37 x $\dpi{100} 10^6$ m

$\dpi{100} \therefore$ The surface area of earth A = $\dpi{100} 4\pi r^2$

=  $\dpi{80} 4\pi (6.37\times10^6 )^2$   = $\dpi{100} 5.09 \times 10^{14} m^2$

Now, Charge on the earth surface,   $q=\rho\times A$

Therefore,

$q=\dpi{100} 5.09 \times 10^{5} C$

Let the time taken to neutralize earth surface be  t

$\dpi{100} \therefore$ Current I = q / t

$\dpi{100} \implies$  t = 282.78 s.

Therefore, time take to neutralize the Earth's surface is 282.78 s

Exams
Articles
Questions