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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Critical Thinking

  • question_answer
    Electric potential is given by \[V=6x-8x{{y}^{2}}-8y+6yz-4{{z}^{2}}\]            Then electric force acting on \[2C\] point charge placed on origin will be                                     [RPET 1999]

    A)                    \[2N\]

    B)                                      \[6N\]

    C)                    \[8N\]                             

    D)            \[20\,N\]

    Correct Answer: D

    Solution :

               \[{{E}_{x}}=-\frac{dV}{dx}=-(6-8{{y}^{2}}),\]\[{{E}_{y}}=-\frac{dV}{dy}=-(-\,16xy-8+6z)\] \[{{E}_{z}}=-\frac{dV}{dz}=-\,(6y-8z)\] At origin x = y = z = 0 so,\[{{E}_{x}}=-\,6,\,{{E}_{y}}=8\]and\[{{E}_{z}}=0\] Þ \[E=\sqrt{E_{x}^{2}+E_{y}^{2}}=10\,N/C\]. Hence force \[F=QE=2\times 10=20N\]


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