JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer A system of three charges are placed as shown in the figure : If \[D>>d,\]the potential energy of the system is best given by:             [JEE Main 9-4-2019 Morning]

    A) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ -\frac{{{q}^{2}}}{d}-\frac{qQd}{2{{D}^{2}}} \right]\]   

    B) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ +\frac{{{q}^{2}}}{d}+\frac{qQd}{{{D}^{2}}} \right]\]

    C) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ -\frac{{{q}^{2}}}{d}+\frac{2qQd}{{{D}^{2}}} \right]\] 

    D) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ -\frac{{{q}^{2}}}{d}-\frac{qQd}{{{D}^{2}}} \right]\]

    Correct Answer: D

    Solution :

              \[{{U}_{total}}={{U}_{self\,of\,dipole\,}}+{{U}_{\operatorname{int}eraction}}\] \[=-\frac{k{{q}^{2}}}{d}-\left( \frac{kQ}{{{D}^{2}}} \right)qd\] \[=-k\left[ \frac{{{q}^{2}}}{d}+\frac{qQd}{{{D}^{2}}} \right]\] Option [d]


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