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JEE Main 2014 · 9 Apr · Shift 1, Physics question 16

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 ϵo=8.85×10−12\epsilon_o=8.85\times10^{-12} C2^2/N-m2^2, RE=6.37×106R_E=6.37\times10^6 m]

Choose one option
Answer and worked solution

Official answer: C

By Gauss's law with the field pointing inward, the charge is negative:
Q=−ϵ0E⋅4πR2=−8.85×10−12×150×4π(6.37×106)2≈−6.8×105 C=−680 kC.Q=-\epsilon_0E\cdot4\pi R^2=-8.85\times10^{-12}\times150\times4\pi(6.37\times10^6)^2\approx-6.8\times10^5\text{ C}=-680\text{ kC}.

Topics:Electric flux and Gauss's law
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