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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 17

  • question_answer
    The potential at a point x (measured in\[\mu \,m\]) due to some charges situated on the x-axis is given by \[V(x)=20/({{x}^{2}}-4)\]Volt The electric field E' at \[x=4\,\mu \,m\] is given by

    A) \[(10/9)\,Volt/\mu \,m\] Volt/ |li m and in the \[+\text{v}e\,x\]x direction

    B) \[\left( 5/3 \right)\text{ }Volt/\mu \,m\]and in the \[-\text{v}ex\]direction

    C) \[\left( 5/3 \right)\text{ }Volt/\mu \,m\]and in the \[+\text{v}e\text{ }x\]direction

    D) \[\left( 10/9 \right)\text{ }Volt/\mu \text{ }m\]and in the \[-\text{v}e\text{ }x\]direction

    Correct Answer: A

    Solution :

    Here, \[V(x)=\frac{20}{{{x}^{2}}-4}\]volt We know that \[E=-\frac{dV}{dx}=-\frac{d}{dx}\left( \frac{20}{{{x}^{2}}-4} \right)\] or         \[E=+\frac{40x}{{{({{x}^{2}}-4)}^{2}}}\] At     \[x=4\,\mu m,\] \[E=+\frac{40\times 4}{{{({{4}^{2}}-4)}^{2}}}=+\frac{160}{144}=+\frac{10}{9}Volt/\mu m\]. Positive sign indicates that \[\vec{E}\] is in \[+ve\]x-direction.


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