AMU Medical AMU Solved Paper-2008

  • question_answer
    Each of the two point charges are doubled and their distance is halved. Force of interaction becomes n times, where n is

    A)  4                                            

    B)  1

    C)  1/16                                     

    D)  16

    Correct Answer: D

    Solution :

                     From the formula,                 \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]                 \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2{{q}_{1}}\times 2{{q}_{2}}}{{{(r/2)}^{2}}}\]                 \[=\frac{4}{1/4}F\] \[\Rightarrow \]               \[F=16F\] \[\Rightarrow \]               \[nF=16F\] \[\therefore \]  n = 16


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