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JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Morning)

  • question_answer
    Three charges \[Q,+q\] and \[+q\] are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of Q is [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

    A) \[+q\]                                       

    B) \[-2q\]

    C) \[\frac{-\sqrt{2}q}{\sqrt{2}+1}\]

    D)                  \[\frac{-q}{1+\sqrt{2}}\]

    Correct Answer: C

    Solution :

    Let a be the length of two equal sides of the triangle. The net electrostatic energy of the system Is zero         \[k\left( \frac{Qq}{a}=\frac{{{q}^{2}}}{a}+\frac{Qq}{\sqrt{2}a} \right)=0\]\[\Rightarrow \]\[Q=-\frac{\sqrt{2}}{\sqrt{2}+1}q\]


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