Question

The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about $150N/C$, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be : [Given ${ϵ}_0=8.85\times {10}^{-12}{C}^2/N-m^2,R_E=6.4\times {10}^{6}m]$

• A. $+660kC$
• B. $-660kC$
• C. $-680kC$
• D. $+680kC$

Answer 1

Answer: (C)

$-680kC$

From Gauss' Law,
the electric field entering the Earth's atmosphere is E. The flux of this field is,
$\int E.ds$ = $E\times 4\pi R^2$
This should be equal to the charge enclosed inside the Earth divided by $ϵ$
Hence,
$E\times 4\pi R^2$ = $\frac{q}{ϵ}$
Hence, putting back the value of $3$ knowns, the value of $q=-680\times {10}^{3}C$