The earth's surface has a negative surface charge density of . 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 g

Question

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 (Radius of earth = $6.37 \times 10^{6} m$ , assume the current is constant with time, despite the charge on the earth’s surface changing)

Smart Achievers · Accepted Answer

(3)

$Given, radius of earth R = 6.37 \times 10^{6} m$

$Negative surface charge density σ = 10^{- 9} C / m^{2}$

$Potential V = 400 k V = 400 \times 10^{3} V$

$Current on the globe I = 1800 A$

$Surface area of earth A = 4 π R^{2} = 4 \times 3.14 \times {(6.37 \times 10^{6})}^{2} = 509.64 \times 10^{12} m^{2}$

$Charge on earth surface Q = Area of earth surface \times Surface charge$

$Q = A σ = 509.64 \times 10^{12} \times 10^{- 9} = 509.64 \times 10^{3} C$

$We know that Q = I t$

$∴$ $Time required to neutralize earth's surface$

$t = \frac{Q}{I} = \frac{509.64 \times 10^{3}}{1800}$

$t = 283.1 s or t = 4 min 43 s$

$Thus, the time required to neutralize the earth's surface is 283.1 s .$

Solution

1 273 s

2 263 s

3 283 s

4 205 s

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