[${\varepsilon _0} = 8.85 \times {10^{ - 12}}\,{C^2}/N - {m^2},{R_E} = 6.37 \times {10^6}\,m$]
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
\(\mathrm{q}=-680\, \mathrm{Kc}\)