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JEE Main & Advanced AIEEE Solved Paper-2003

  • question_answer
    Resultant force is zero, as three forces acting on the particle can be represented in magnitude and direction by three sides of a triangle in same order. Hence, by Newton's 2nd law \[\left( F=m\frac{d\,\,v}{dt} \right)\], particle velocity (v) will be same. (\[\because \] if F = 0, v = constant) If the electric flux entering and leaving an enclosed surface respectively is \[{{\phi }_{1}}\] and \[{{\phi }_{2}}\], the electric charge inside the surface will be     AIEEE  Solved  Paper-2003

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

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

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

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

    Correct Answer: A

    Solution :

    From Gauss's law, \[\frac{Charge\text{ }enclosed}{{{\varepsilon }_{0}}}=\] Flux leaving the surface                     \[\frac{q}{{{\varepsilon }_{0}}}={{\phi }_{2}}-{{\phi }_{1}}\]          (i.e, net flux) \[\Rightarrow \]               \[q=({{\phi }_{2}}-{{\phi }_{1}}){{\varepsilon }_{0}}\]


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