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

  • question_answer
    A charge +q fixed at each of the points \[x={{x}_{0}},x=3{{x}_{0}},x=5{{x}_{0}}....\] up to \[\infty \] on X-axis and charge -q is fixed on each of the points \[x=2{{x}_{0}},x=4{{x}_{0}},....\] up to \[\infty \]. Here \[{{x}_{0}}\] is a positive constant. Take the potential at a point due to a charge Q at a distance r form it to be \[\frac{Q}{4\pi {{\varepsilon }_{0}}r},\] then the potential at the origin due to above system of charges will be:

    A) zero      

    B) infinite

    C) \[\frac{q}{8\,\pi \,{{\varepsilon }_{0}}{{x}_{0}}\,{{\log }_{e}}\,2}\]

    D) \[\frac{q\,{{\log }_{e}}\,2}{4\,\pi \,{{\varepsilon }_{0}}{{x}_{0}}}\]

    Correct Answer: D

    Solution :

    [d] \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{q}{{{x}_{0}}}+\frac{q}{3{{x}_{0}}}+\frac{q}{5{{x}_{0}}}+... \right]\]


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