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

  • question_answer
    Two point charges \[\text{3}\times \text{1}{{0}^{\text{6}}}C\]and \[\text{8}\times \text{1}{{0}^{\text{6}}}C\] repel each other by a force of \[\text{6}\times \text{1}{{0}^{\text{3}}}N\]. If each of them is given an additional charge \[\text{ 6}\times \text{1}{{0}^{\text{6}}}C\], the force between them will be     [DPMT 2003]

    A)                    \[2.4\times \text{1}{{0}^{3}}N\] (attractive)      

    B)                    \[2.4\times \text{1}{{0}^{9}}N\] (attractive)

    C)                    \[\text{1}.\text{5}\times \text{1}{{0}^{\text{3}}}N\] (repulsive)                       

    D)                    \[\text{1}.\text{5}\times \text{1}{{0}^{\text{3}}}N\] (attractive)

    Correct Answer: D

    Solution :

                 \[F\propto {{Q}_{1}}{{Q}_{2}}\] Þ \[\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{{{Q}_{1}}{{Q}_{2}}}{{{Q}_{1}}'{{Q}_{2}}'}\]                    \[=\frac{3\times {{10}^{-6}}\times 8\times {{10}^{-6}}}{(3\times {{10}^{-6}}-6\times {{10}^{-6}})(8\times {{10}^{-6}}-6\times {{10}^{-6}})}=\frac{3\times 8}{-\,3\times 2}=-\frac{4}{1}\]                    Þ \[{{F}_{2}}=-\frac{{{F}_{1}}}{4}=-\frac{6\times {{10}^{-3}}}{4}=-1.5\times {{10}^{-3}}N\] (Attractive)


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