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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Electric Field and Potential

  • question_answer
    Three particles, each having a charge of \[10\,\mu C\] are placed at the corners of an equilateral triangle of side \[10\,cm\]. The electrostatic potential energy of the system is (Given \[\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N-{{m}^{2}}/{{C}^{2}}\])                                                           [MP PMT 1994]

    A)                                                               Zero

    B)                                                 Infinite

    C)                                                               \[27\,J\]

    D)                                                 \[100\,J\]

    Correct Answer: C

    Solution :

                         For pair of charge \[U=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{1}}{{q}_{2}}}{r}\] \[{{U}_{System}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{10\times {{10}^{-6}}\times 10\times {{10}^{-6}}}{10/100} \right.\]\[\left. +\frac{10\times {{10}^{-6}}\times 10\times {{10}^{-6}}}{10/100}+\frac{10\times {{10}^{-6}}\times 10\times {{10}^{-6}}}{10/100} \right]\]       \[=3\times 9\times {{10}^{9}}\times \frac{100\times {{10}^{-12}}\times 100}{10}=27J\]


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