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JEE Mains · Physics · STD 12 - Electric charges and fields
  • JEE Mains 2014
  • Hard
  • MCQ

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

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Answer & Solution

Correct Answer

(C) \( - 680\)

Step-by-step Solution

Detailed explanation

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

\(\mathrm{q}=-680\, \mathrm{Kc}\)

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