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 $= 6.37 \times 10^6 m.)$
Exercise
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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 $= t$
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 $282.77s.$
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 $= t$
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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