JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Electric Field and Potential

  • question_answer
    Electric potential at any point is \[V=-5x+3y+\sqrt{15}z\], then the magnitude of the electric field is         [MP PET 2002]

    A)            \[3\sqrt{2}\]                         

    B)            \[4\sqrt{2}\]

    C)            \[5\sqrt{2}\]                         

    D)            7

    Correct Answer: D

    Solution :

                         \[{{E}_{x}}=-\frac{dV}{dx}=-(-5)=5;\]\[{{E}_{y}}=-\frac{dV}{dy}=-3\] and \[{{E}_{z}}=-\frac{dV}{dz}=-\sqrt{15}\] \[{{E}_{net}}=\sqrt{E_{x}^{2}+E_{y}^{2}+E_{z}^{2}}=\sqrt{{{(5)}^{2}}+{{(-3)}^{2}}+{{(-\sqrt{15})}^{2}}}=7\]


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