JEE Main & Advanced Physics Electrostatics & Capacitance JEE PYQ-Electrostatics and Capacitance

  • question_answer
    A point charge of magnitude + \[1\mu C\] is fixed at (0, 0, 0). An isolated uncharged spherical conductor, is fixed with its center at (4, 0, 0). The potential and the induced electric field at the sphere is: [JEE ONLINE 22-04-2013]

    A)             \[1.8\times {{10}^{5}}\] and \[-5.625\times {{10}^{6}}\,V/m\]

    B)             \[0\operatorname{V}\] and \[0\operatorname{V}/\operatorname{m}\]

    C)             \[2.25\times {{10}^{5}}\operatorname{V}\] and \[-5.625\times {{10}^{6}}\operatorname{V}/\operatorname{m}\]

    D)             \[2.25\times {{10}^{5}}\operatorname{V}\] and V/m

    Correct Answer: C

    Solution :

    [c] \[q=1\mu C=1\times {{10}^{-6}}C\]
    \[r=4\,cm=4\times {{10}^{-2}}m\]
    Potential \[V=\frac{kq}{r}\]
    \[=\frac{9\times {{10}^{9}}\times {{10}^{-6}}}{4\times {{10}^{-2}}}\]
                \[=2.25\times {{10}^{5}}V.\]
    Induced electric field \[E=\frac{kq}{{{r}^{2}}}\]
    \[=\frac{9\times {{10}^{9}}\times 1\times {{10}^{-6}}}{16\times {{10}^{-4}}}=-5.625\times {{10}^{6}}\,V/m\]


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