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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 ϵ0=8.85×10−12C2/N−m2,RE=6.4×106m]

A
+660kC
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B
660kC
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C
680kC
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D
+680kC
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Solution

The correct option is D −680kCFrom Gauss' Law,the electric field entering the Earth's atmosphere is E. The flux of this field is,∫E.ds = E×4πR2This should be equal to the charge enclosed inside the Earth divided by ϵHence,E×4πR2 = qϵHence, putting back the value of 3 knowns, the value of q=−680×103C

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