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? (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 $\left.=6.37 \times 10^{6} m \right)$

• A. 273 s
• B. 263 s
• C. 283 s
• D. 205 s

Answer 1

Answer: (C)

Given, radius of earth $R=6.37 \times 10^{6} m$
Negative surface charge density $\sigma=10^{-9} C / m ^{2}$
Potential difference $V=400 \,kV =400 \times 10^{3} V$
Current on the globe $I=1800 \,A$
Surface area of earth $A=4 \pi R^{2}=4 \times 3.14 \times\left(6.37 \times 10^{6}\right)^{2}$
$=509.64 \times 10^{12} m ^{2}$
Charge on earth surface $Q=$ Area of earth surface $\times$ Surface charge density
$Q =A \sigma=509.64 \times 10^{12} \times 10^{-9} $
$=509.64 \times 10^{3}\, C$
We know that $Q=I t$
$\therefore $ 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 $283.1\, s$.