The magnitude of the average electric field n | vedclass

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(C) Given:Electric field $E = 150\, N/C$ (directed inward,so $E = -150\, N/C$)Radius of Earth $R_E = 6.37 	imes 10^6\, m$Permittivity of free space $

Vedclass · Accepted Answer

$-680$

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(C) Given:Electric field $E = 150\, N/C$ (directed inward,so $E = -150\, N/C$)Radius of Earth $R_E = 6.37 	imes 10^6\, m$Permittivity of free space $\epsilon_0 = 8.85 	imes 10^{-12}\, C^2/(N \cdot m^2)$According to Gauss's law,the electric flux $\phi$ through the surface of the Earth is $\phi = \frac{q}{\epsilon_0} = E \cdot A$,where $A = 4\pi R_E^2$ is the surface area of the Earth.Thus,$q = \epsilon_0 \cdot E \cdot 4\pi R_E^2$Substituting the values:$q = (8.85 	imes 10^{-12}) 	imes (-150) 	imes 4 	imes 3.14159 	imes (6.37 	imes 10^6)^2$$q \approx -6.80 	imes 10^5\, C$Converting to $kC$ $(1\, kC = 10^3\, C)$:$q \approx -680\, kC$Since the electric field is directed inward,the charge is negative.

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