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

  • question_answer
    Three identical particles, each possessing the mass m and charge +q, are placed at the corners of an equilateral triangle with side \[{{r}_{0}}.\] The particles are simultaneously set free and start flying apart symmetrically due to Coulomb?s repulsion foces. The work performed by Coulomb?s forces acting on to a very large distance is \[(\text{where }k=1/4\pi {{\varepsilon }_{0}}.)\]

    A) \[\frac{3k{{q}^{2}}}{{{r}_{0}}}\]

    B) \[\frac{k{{q}^{2}}}{{{r}_{0}}}\]

    C) \[\frac{3k{{q}^{2}}}{2{{r}_{0}}}\]

    D) \[\frac{k{{q}^{2}}}{2{{r}_{0}}}\]

    Correct Answer: B

    Solution :

    [b] Since the given system is closed, the increase in KE is equal to decrease in P.E. \[\Rightarrow \frac{3}{2}m{{v}^{2}}=\frac{2k{{q}^{2}}}{{{r}_{0}}}-\frac{3k{{q}^{2}}}{r}\] \[\Rightarrow v=\sqrt{\frac{2k{{q}^{2}}\left( r-{{r}_{0}} \right)}{mr{{r}_{0}}},}\] \[\Rightarrow {{v}_{\max }}=\sqrt{\frac{2k{{q}^{2}}}{m{{r}_{0}}}}\] The work performed by the interaction force during the variation of the system?s configuration is equal to the decrease in the potential energy \[W={{U}_{1}}-{{U}_{2}}=\frac{3k{{q}^{2}}}{{{r}_{0}}}\] \[\therefore \] Work done per particle \[=\frac{k{{q}^{2}}}{{{r}_{0}}}\]


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