NEET AIPMT SOLVED PAPER MAINS 2011

  • question_answer
    The electric potential V at any point (.x, y, z), all in metres in space is given by V = 4x2 volt. The electric field at the point (1, 0, 2) in volt/metre is                                                                

    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 :

    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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