VMMC VMMC Medical Solved Paper-2014

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

    A)  4                            

    B) 1

    C) 18                          

    D) 16

    Correct Answer: D

    Solution :

     From the formula. \[f=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]and\[f'=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2{{q}_{1}}\times 2{{q}_{2}}}{{{(r/2)}^{2}}}\] So,\[f'=16f\] \[nf=16f\] \[n=16\]


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