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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2008

  • question_answer
    The charges Q, +q and +q are placed at the vertices of an equilateral triangle of side \[l\]. If the net electrostatic potential energy of the system is zero, then Q is equal to

    A)  \[-\frac{q}{2}\]                                                

    B)  \[-q\]

    C)  \[\frac{+q}{2}\]                                               

    D)  Zero

    Correct Answer: A

    Solution :

    Potential energy of the system                 \[U=\frac{KQq}{l}+\frac{K{{q}^{2}}}{l}+\frac{KqQ}{l}=0\] \[\Rightarrow \]               \[\frac{Kq}{l}(Q+q+Q)=0\] \[\Rightarrow \]                                               \[Q=-\frac{q}{2}\]


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