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Asked in JEE Main Online 2014 · Gauss's law and enclosed charge
Apply Gauss's law to a sphere just outside the Earth's surface: Q=ε₀ Φ=ε₀ E (4π R_E²), with the sign of Φ set by the field direction.
4π R_E²=4π(6.37×10⁶)²=5.10×10¹⁴ m².
Φ=150×5.10×10¹⁴=7.65×10¹⁶, and Q=8.85×10⁻¹²×7.65×10¹⁶=6.77×10⁵ C.
The field points inward, so the enclosed charge is negative: Q≈-680 kC.
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