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JEE Main 2014 (9 Apr, Shift 1) · Electric Flux & Gauss's Law

Electric Flux & Gauss's Law question from JEE Main 2014 (9 Apr, Shift 1)

Q115JEE Main 2014 · 9 APR · SHIFT 1Medium

Electric Flux & Gauss's Law

The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about 150 N/C, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be:

[Given $\epsilon$ 0 = 8.85 $\times$ 10 $-$ 12 C2/N-m2, RE = 6.37 $\times$ 106 m]

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Answer: (c)

Earth acts like a charged sphere: $E = \frac{Q}{4\pi\varepsilon_0 R_E^2}$ , so $Q = 4\pi\varepsilon_0 R_E^2 E$ .
$Q = 4\pi (8.85 \times 10^{-12})(6.37 \times 10^{6})^2 (150) \approx 6.8 \times 10^{5}$ C
The field points inward, so the charge is negative.
So the answer is (c) $-680$ kC.

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