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

  • question_answer
    The electric potential V at any point \[(x,y,z),\] all in metres in space is given by \[V=4{{x}^{2}}\] volt. The electric field at the point \[(1,0,2)\] in volt/metre is [AIPMT (M) 2011]

    A) 8 along positive .X-axis

    B) 16 along negative X-axis

    C) 16 along positive X-axis

    D) 8 along negative X-axis

    Correct Answer: D

    Solution :

    [d] We know that
    \[\mathbf{E}=-\left[ \mathbf{i}\frac{\partial V}{\partial x}+\mathbf{j}\frac{\partial V}{\partial y}+\mathbf{k}\frac{\partial V}{\partial z} \right]\]
                So,       \[\mathbf{E}=-\mathbf{i}\frac{\partial V}{\partial x}=-\mathbf{i}\frac{\partial }{\partial z}(4{{x}^{2}})\]
                            \[=-8x\,\mathbf{i}\,\text{V}\,{{\text{m}}^{-1}}\]
                            \[{{\mathbf{E}}_{(1,0,2)}}=-8\,\mathbf{i}\,\text{V}{{\text{m}}^{-1}}\]


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