JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Critical Thinking

  • question_answer
    A piece of cloud having area \[25\times {{10}^{6}}{{m}^{2}}\] and electric potential of \[{{10}^{5}}\] volts. If the height of cloud is \[0.75\,km\], then energy of electric field between earth and cloud will be [RPET 1997]

    A)                                                      \[250\,J\]                               

    B)            \[750\,J\]

    C)                    \[1225J\]

    D)                                      \[1475J\]

    Correct Answer: D

    Solution :

               Energy \[=\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}\times (A\times d)\]\[=\frac{1}{2}{{\varepsilon }_{0}}\left( \frac{{{V}^{2}}}{{{d}^{2}}} \right)\,Ad\]            \[=\frac{1}{2}\times \frac{8.85\times {{10}^{-12}}\times {{({{10}^{5}})}^{2}}\times 25\times {{10}^{6}}}{0.75\times {{10}^{3}}}=1475J\]


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