WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    A charge q is located at the centre of a cube. The electric flux through any face is :

    A)  \[\frac{4\pi q}{6(4\pi {{\varepsilon }_{0}})}\]                     

    B)  \[\frac{\pi q}{6(4\pi {{\varepsilon }_{0}})}\]

    C)  \[\frac{q}{6(4\pi {{\varepsilon }_{0}})}\]                              

    D)  \[\frac{2\pi q}{6(4\pi {{\varepsilon }_{0}})}\]

    Correct Answer: A

    Solution :

                                    According to Gausss theorem, electric flux through the cube (closed surface), \[\text{o }\!\!|\!\!\text{ =}\frac{q}{{{\varepsilon }_{0}}}\] Since, cube has six surfaces and all the faces are symmetrical, therefore electric flux through any face \[\text{o }\!\!|\!\!\text{  }\!\!\!\!\text{ }\,\text{=}\frac{q}{6{{\varepsilon }_{0}}}=\frac{4\pi q}{6(4\pi {{\varepsilon }_{0}})}\]


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