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 over 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 x 10⁶m).

Solution

Given,
Surface charge density of earth's surface = 10^–9 C/m²
Current flowing across the surface of earth, I = 1800 A
The radius of earth,r= 6370 km = 6.37 x 10⁶ m

Charge on entire surface of the earth, q= $σ . A$

Area of earth's surface, A = 4 ${πr}^{2}$

This implies,
q=4 $π$ (6.37 x 10⁶)² x 10^–9 C

Using the formula , I= $\frac{q}{t}$ we get,

t= $\frac{q}{I}$
Therefore,
Time required for the flow of entire charge is,

$= \frac{4 \times 3.14 \times {(6.37 \times 10^{6})}^{2} \times 10^{- 9}}{1800} = 283 seconds .$

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