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NEET Sample Paper NEET Sample Test Paper-15

  • question_answer
    The electric field in a region is given by E\[E=\frac{3}{5}{{E}_{0}}i+\frac{4}{5}{{E}_{0}}j\]with\[{{E}_{0}}=2.0\times {{10}^{3}}N/C.\] The flux of this field through a rectangular surface area of \[0.2\,{{m}^{2}}\]parallel to the Y-Z plane is

    A) \[60\,N{{m}^{2}}/C\]                     

    B) \[120\,N{{m}^{2}}/C\]

    C) \[180\,N{{m}^{2}}/C\]                   

    D) \[240\,N{{m}^{2}}/C\]   

    Correct Answer: D

    Solution :

    The area vector \[\overrightarrow{dA}=0.2\,\hat{i}\] \[\therefore \]    flux \[=\vec{E}.\overrightarrow{dA}=\left( \frac{3}{2}{{E}_{0}}\hat{i}+\frac{4}{5}{{E}_{0}}\hat{j} \right).0.2\hat{i}\] \[=0.2\times \frac{3}{5}\times 2\times {{10}^{3}}\] \[=240\,N{{m}^{2}}/C\] Hence, the correction option is [d].


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