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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Self Evaluation Test - Electric Chages and Fields

  • question_answer
    A sphere of radius R carries charge density \[\rho \]proportional to the square of the distance from the center such that \[\rho =C{{R}^{2}}\], where C is a positive constant. At a distance R/2 from the center, the magnitude of the electric field is

    A) \[\frac{C{{R}^{3}}}{20{{\in }_{0}}}\]

    B)                    \[\frac{C{{R}^{3}}}{10{{\in }_{0}}}\]

    C) \[\frac{C{{R}^{3}}}{5{{\in }_{0}}}\]  

    D)        \[\frac{C{{R}^{3}}}{40{{\in }_{0}}}\]

    Correct Answer: D

    Solution :

    [d] For, \[r=R/2\] Using Gauss's law, we have \[\oint{\vec{E}.d\vec{A}=\frac{{{q}_{in}}}{{{\varepsilon }_{0}}}\text{ or }E\times 4\pi {{r}^{2}}=\int\limits_{0}^{R/2}{\frac{\rho 4\pi {{r}^{2}}dr}{\varepsilon {{ & }_{0}}}}}\] \[\text{or }E=\frac{C{{r}^{3}}}{5{{\in }_{0}}}=\frac{C{{R}^{3}}}{40{{\in }_{0}}}.\]


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