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EAMCET Medical EAMCET Medical Solved Paper-1999

  • question_answer
    Two equal unlike charges, placed 3 cm. Apart in air. attract each other with a force of 40 N. The magnitude of each charge in micro coulombs is, take \[\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}\,SI\]units:

    A)  0.2                                        

    B)  2

    C)  20                                         

    D)  200

    Correct Answer: B

    Solution :

                     Force of attraction between two point charges \[{{q}_{1}}\] and \[{{q}_{2}}\] at separation r is given by \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]                 (Here: \[r=3cm=3\times {{10}^{-2}}m,\] \[F=40N,\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}\] ) \[40=\frac{9\times {{10}^{9}}\times q\times q}{9\times {{10}^{-4}}}\] \[{{q}^{2}}=\frac{9\times 40\times {{10}^{-4}}}{9\times {{10}^{9}}}\] \[{{q}^{2}}=40\times {{40}^{-13}}=4\times {{10}^{-12}}\] =\[q=2\times {{10}^{-6}}=2\mu C\]        


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