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JEE Main & Advanced Physics EM Waves Question Bank Self Evaluation Test - Electromagnetic Waves

  • question_answer
    An electromagnetic wave of frequency \[1\times {{10}^{14}}\]hertz is propagating along z-axis. The amplitude of electric field is\[4\text{ }V/m\]. If \[{{\varepsilon }_{0}}=8.8\times {{10}^{-12}}{{C}^{2}}/N-{{m}^{2}},\] then average energy density of electric field will be:                        

    A) \[35.2\times {{10}^{-10}}J/{{m}^{3}}\]

    B) \[35.2\times {{10}^{-11}}J/{{m}^{3}}\]

    C) \[35.2\times {{10}^{-12}}J/{{m}^{3}}\]

    D) \[35.2\times {{10}^{-13}}J/{{m}^{3}}\]

    Correct Answer: C

    Solution :

    [c] Given: Amplitude of electric field, \[{{E}_{0}}=4\,v/m\] Absolute permitivity, \[{{\varepsilon }_{0}}=8.8\times {{10}^{-12}}{{c}^{2}}/N-{{m}^{2}}\] Average energy density \[{{u}_{E}}=?\] Applying formula, Average energy density \[{{u}_{E}}=\frac{1}{4}{{\varepsilon }_{0}}{{E}^{2}}\] \[\Rightarrow \,\,{{u}_{E}}=\frac{1}{4}\times 8.8\times {{10}^{-12}}\times {{(4)}^{2}}=35.2\times {{10}^{-12}}\,J/{{m}^{3}}\]


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