JEE Main & Advanced Physics Electrostatics & Capacitance JEE PYQ-Electrostatics and Capacitance

  • question_answer
    The potential at a point x (measured in\[{{e}^{-\frac{1}{2}}}\]due to some charges situated on the x-axis is given by\[{{e}^{+\frac{1}{2}}}\]volt.            The electric field E at \[2u\times 3v\]urn is given by [AIEEE 2007]

    A) \[{{\tan }^{-1}}\frac{b}{ac}\]and in the -ve\[45{}^\circ \]direction

    B) \[{{\tan }^{-1}}\frac{bc}{a(c-a)}\]and in the +ve\[{{\tan }^{-1}}\frac{bc}{a}\]direction

    C) \[{{y}^{2}}=8x\]and in the -ve\[y=x+2\]direction

    D) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-6x-12y-2\text{ }z+20=0,\]and in the +ve\[a=\hat{i}+\hat{j}+\hat{k},b=\hat{i}-\hat{j}+2\hat{k}\]direction

    Correct Answer: D

    Solution :

    [d] As the electric field is given by
    \[a=\hat{i}+\hat{j}+\hat{k},b=\hat{i}-\hat{j}+2\hat{k}\]
    Along x-direction, \[c=x\hat{i}+(x-2)\hat{j}-\hat{k}\]
    \[x\]
    \[R=(3,3\sqrt{3})\] \[\angle PQR\]at \[\sqrt{3}x+y=0\]
    and is along positive X-direction.


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