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

  • question_answer
    In order to establish an instantaneous displacemet current of \[1\text{ }mA\] in the space between the plates of \[2\mu F\] parallel plate capacitor, the potential difference need to apply is    

    A) \[100\,\,V{{s}^{-1}}\]        

    B)        \[200\,\,V{{s}^{-1}}\]

    C) \[300\,\,V{{s}^{-1}}\]       

    D)        \[500\,\,V{{s}^{-1}}\]

    Correct Answer: D

    Solution :

    [d] \[{{I}_{d}}=1\,\]   \[mA={{10}^{-3}}\,A\] \[C=2\mu F=2\times {{10}^{-6}}F\] \[{{I}_{D}}={{I}_{C}}=\frac{d}{dt}(CV)=C\frac{dV}{dt}\] Therefore, \[\frac{dV}{dt}=\frac{{{I}_{D}}}{C}=\frac{{{10}^{-3}}}{2\times {{10}^{-6}}}=500\,V{{s}^{-1}}\] Therefore, applying a varying potential difference of \[500\,V\,{{s}^{-1}}\] would produce a displacement current of desired value.


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