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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Charge and Coulombs Law

  • question_answer
    Three charges \[4q,\,Q\] and \[q\] are in a straight line in the position of 0, \[l/2\] and \[l\] respectively. The resultant force on \[q\] will be zero, if \[Q=\]                                                                         [CPMT 1980]

    A)                    ? q                                    

    B)                    \[-\,2q\]

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

    D)                    \[4q\]

    Correct Answer: A

    Solution :

                       The force between 4q and q; \[{{F}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{4q\times q}{{{l}^{2}}}\]                    The force between Q and q; \[{{F}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{Q\times q}{{{(l/2)}^{2}}}\]                    We want \[{{F}_{1}}+{{F}_{2}}=0\] or \[\frac{4{{q}^{2}}}{{{l}^{2}}}=-\frac{4Qq}{{{l}^{2}}}\] Þ \[Q=-q\]


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