JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Electric Field and Potential

  • question_answer
    A sphere of radius \[1\,cm\] has potential of \[8000\,V\], then energy density near its surface will be     [RPET 1999]

    A)            \[64\times {{10}^{5}}J/{{m}^{3}}\]

    B)                                      \[8\times {{10}^{3}}J/{{m}^{3}}\]

    C)                    \[32\,J/{{m}^{3}}\]

    D)                                      \[2.83\,J/{{m}^{3}}\]

    Correct Answer: D

    Solution :

                 Energy density \[{{u}_{e}}=\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}=\frac{1}{2}\times 8.86\times {{10}^{-12}}\times {{\left( \frac{V}{r} \right)}^{2}}\] \[=2.83J/{{m}^{3}}\]


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