The earth’s surface has a negative surface charge density of 10^ …

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 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 $= 6.37 \times 10^6 m.$)

✓

Answer

Surface charge density of the earth, $\sigma=10^{-9}\ \text{C m}^{-2}$
Current over the entire globe, $\text{I}=1800\ \text{A}$
Radius of the earth, $\text{r}=6.37\times10^6\ \text{m}$
Surface area of the earth,
$\text{A}=4\pi\text{r}^2$
$=4\pi\times(6.37\times10^6)^2$
$=5.09\times10^{14}\ \text{m}^2$
Charge on the earth surface,
$\text{q}=\sigma\times\text{A}$
$=10^{-9}\times5.09\times10^{14}$
$=5.09\times10^{5}\ \text{C}$
Time taken to neutralize the earth's surface = t
$\text{Current},\ \text{I}=\frac{\text{q}}{\text{t}}$
$\text{t}=\frac{\text{q}}{\text{I}}$
$=\frac{5.09\times10^5}{1800}=282.77\text{s}$
Therefore, the time taken to neutralize the earth's surface is 282.77s.

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