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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Self Evaluation Test - Electrostatic Potential and Capacitance

  • question_answer
    If the electrostatic potential were given by \[\phi ={{\phi }_{0}}({{x}^{2}}+{{y}^{2}}+{{z}^{2}}),\]where is constant, then the charge density giving rise to the above potential would be.

    A) 0

    B) \[-6{{\phi }_{0}}{{\varepsilon }_{0}}\]

    C) \[-2{{\phi }_{0}}{{\varepsilon }_{0}}\]

    D) \[-\frac{6{{\phi }_{0}}}{{{\varepsilon }_{0}}}\]

    Correct Answer: B

    Solution :

    [b] \[E=-\nabla \phi =-{{\phi }_{0}}2\,\,[x\hat{i}+y\hat{i}+z\hat{x}]\] \[={{\varepsilon }_{0}}\nabla .E=-2{{\varepsilon }_{0}}{{\phi }_{0}}\nabla .(x\hat{i}+y\hat{i}+z\hat{x})\] \[n=-6{{\phi }_{0}}{{\varepsilon }_{0}}\]


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