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JEE Main & Advanced JEE Main Paper (Held On 9 April 2014)

  • question_answer
    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: [Given \[{{\varepsilon }_{o}}=8.85\times {{10}^{-12}}{{C}^{2}}/N-{{m}^{2}},{{R}_{E}}=6.37\times {{10}^{6}}m\]]   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) + 670 kC                              

    B) - 670 kC

    C) 680 kC  

    D) + 680 kC

    Correct Answer: C

    Solution :

    Given, Electric field E = 150 N/C Total surface charge carried by earth q = ? According to Gauss?s law. \[\phi =\frac{q}{{{\in }_{0}}}=EA\]or\[q={{\in }_{0}}EA\] \[={{\in }_{0}}E\pi {{r}^{2}}.\] \[=8.85\times {{10}^{-12}}\times 150\times {{(6.37\times {{10}^{6}})}^{2}}.\] \[\simeq 680Kc\] As electric field directed inward hence q = ? 680 Kc                


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