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The magnitude of the average electric field normally present in the atompshere just above the surface of the Earth is about 150 N/C, directed inward towards the centre of the Earth. This gives the total net surface charge carried by the Earth to be:
$(Given ε_{0} = 8 .85 \times 10^{- 12} C^{2} / N - m^{2}, R = 6 .37 \times 10^{6} m)$

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+670 kC

–670 kC

+680 kC

–680 kC

answer is D.

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Detailed Solution

$\frac{q}{4 {πε}_{0} R^{2}} = 150$

$q = 4 {πε}_{0} R^{2} \times 150 = 678 \times 10^{3} C ≃ 680 kC$

as it is towards earth it must be negative charge.

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