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 \({\varepsilon _0} = 8.85 \times {10^{ - 12}}\,{C^2}/N - {m^2},{R_E} = 6.37 \times {10^6}\,m\)]
Pick an option to check your answer.
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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