JEE Main & Advanced JEE Main Paper Phase-I (Held on 09-1-2020 Evening)

  • question_answer
    An electric field \[\vec{E}=4\text{x}\hat{i}-({{y}^{2}}+1)\hat{j}\,\,N/C\] passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as \[{{\phi }_{l}}\] and \[{{\phi }_{ll}}\] respectively. The difference between \[({{\phi }_{l}}-{{\phi }_{ll}})\] is (in\[N{{m}^{2}}/C\]) ______. [JEE MAIN Held on 09-01-2020 Evening]

    Answer:

    -48


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