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MGIMS WARDHA MGIMS WARDHA Solved Paper-2014

  • question_answer
    If electric flux entering and leaving an enclosed surface is\[{{\phi }_{1}}\]and\[{{\phi }_{2}}\]respectively, the electric charge inside the enclosed surface will be

    A)  \[({{\phi }_{1}}+{{\phi }_{2}}){{\varepsilon }_{0}}\]                         

    B)  \[({{\phi }_{2}}-{{\phi }_{1}}){{\varepsilon }_{0}}\]

    C)  \[({{\phi }_{1}}+{{\phi }_{2}}){{\varepsilon }_{0}}\]                         

    D)  \[({{\phi }_{2}}-{{\phi }_{1}})/{{\varepsilon }_{0}}\]

    Correct Answer: B

    Solution :

                    Applying Gauss theorem, \[\phi =\frac{q}{{{\varepsilon }_{0}}};\] \[q={{\varepsilon }_{0}}\phi \] Since,    \[\phi =({{\phi }_{2}}-{{\phi }_{1}})\]                 \[q=({{\phi }_{2}}-{{\phi }_{1}}){{\varepsilon }_{0}}\]


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