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JEE Main & Advanced Physics Electrostatics & Capacitance JEE PYQ-Electrostatics and Capacitance

  • 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:    [JEE ONLINE 09-04-2014] [Given\[{{\varepsilon }_{o}}=8.85\times {{10}^{-12}}{{C}^{2}}/N-{{m}^{2}},\]\[{{R}_{E}}=6.37\times {{10}^{6}}m\]]

    A) + 670 kC

    B) - 670 kC

    C) 680 kC

    D) + 680 kC

    Correct Answer: C

    Solution :

    [c] 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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