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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......$kC$ [Given ${\varepsilon _0} = 8.85 \times {10^{ - 12}}\,{C^2}/N - {m^2},{R_E} = 6.37 \times {10^6}\,m$]
$ + 670$
$ - 670$
$ - 680$
$ + 680$
Solution
Given,
Electric field $\mathrm{E}=150\, \mathrm{N} / \mathrm{C}$
Total surface charge carried by earth $q=?$
According to Gauss's law.
$\phi=\frac{q}{\epsilon_{0}}=E A$
$ \text { or, } q=\epsilon_{0} \mathrm{EA}$
$=\epsilon_{0} \mathrm{E} \pi r^{2} $
$=8.85 \times 10^{-12} \times 150 \times\left(6.37 \times 10^{6}\right)^{2} $
$=680 \,\mathrm{Kc} $
As electric field directed inward hence
$\mathrm{q}=-680\, \mathrm{Kc}$
