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Question

# If the atmospheric electric field is approximately 150 V/m and the radius of the earth is 6400 km, then the total charge of the earth is:-

A
6.8×105 C
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B
6.8×106 C
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C
6.8×108 C
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D
6.8×109 C
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Solution

## The correct option is A 6.8×105 CGiven that, Atmospheric electric field, →E=150 V Radius of earth, R=6400 km=6400×103 m Let us draw a spherical Gaussian surface of radius R passing through the atmosphere, such that the total charge of the earth gets enclosed within it. Let qin be the total charge enclosed and dA be an elemental area of the Gaussian surface. Applying Gauss's law, ∮→E.−→dA=qinϵo Since the atmospheric electric field will be perpendicular to the Gaussian surface, ⇒EA=qinϵo ⇒E×4πR2=qinϵo ⇒150×4π×(6400×103)2=qinϵo ⇒qin=150×4π×(6400×103)2×8.8×10−12 ∴qin=6.79×105 C≈6.8×105 C Hence, option (a) is correct. Why this question? Note: The approximate charge enclosed in earth suggests that the earth is a very huge reservair of charge

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