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JEE Main & Advanced JEE Main Paper (Held On 19 May 2012)

  • question_answer
    The flat base of a hemisphere of radius a with no charge inside it lies in a horizontal plane. A uniform electric field E is applied at an angle\[\frac{\pi }{4}\]with the vertical direction. The electric flux through the curved surface of the hemisphere is     JEE Main  Online Paper (Held On 19  May  2012)

    A) \[\pi {{a}^{2}}E\]                             

    B)                        \[\frac{\pi {{a}^{2}}E}{\sqrt{2}}\]

    C)                        \[\frac{\pi {{a}^{2}}E}{2\sqrt{2}}\]                           

    D)                        \[\frac{\left( \pi +2 \right)\pi {{a}^{2}}E}{{{\left( 2\sqrt{2} \right)}^{2}}}\]

    Correct Answer: B

    Solution :

                    We know that,\[\phi =\oint{E.dS=E}\oint{dS}\cos {{45}^{o}}\] In case of hemisphere\[{{\phi }_{curved}}={{\phi }_{circular}}\] Therefore, \[{{\phi }_{curved}}=E\pi {{a}^{2}}\frac{1}{\sqrt{2}}=\frac{E\pi {{a}^{2}}}{\sqrt{2}}\]


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