Question

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 theatmosphere and the surface results (due to the low conductivity ofthe lower atmosphere) in a current of only 1800 A over the entireglobe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise theearth’s surface? (This never happens in practice because there is amechanism to replenish electric charges, namely the continualthunderstorms and lightning in different parts of the globe). (Radiusof earth = 6.37 × 106 m.)

Answer 1

Given: The surface charge density of the earth’s surface 10 −9  C/ m 2 , the current over the entire globe is 1800 A and the radius of the earth is 6.37× 10 6  m.

The surface area of earth is given as,

A=4π r 2

Where, the radius of the earth is r.

By substituting the given values in the above formula, we get

A=4π ( 6.37× 10 6 ) 2 A=5.09× 10 14   m 2

The charge on the earth surface is given as,

q=ρ×A

Where, the surface charge density of the earth’s surface is ρ.

By substituting the given values in the above formula, we get

q= 10 −9 ×5.09× 10 14 =5.09× 10 5  C

The current is given as,

I= q t

Where, the time taken to neutralize the earth’s surface is t.

By substituting the given values in the above formula, we get

1800= 5.09× 10 5 t t= 5.09× 10 5 1800 t=282.77 s

Thus, the time taken to neutralize the earth’s surface is 282.77 s.