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CET Karnataka Medical CET - Karnataka Medical 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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