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VMMC VMMC Medical Solved Paper-2004

  • question_answer
    A particle of mass m and charge q is placed at rest in a uniform electric field E and then released, the kinetic energy attained by the particle after moving a distance y, will be

    A) \[{{q}^{2}}Ey\]                                 

    B) \[qEy\]                

    C)        \[q{{E}^{2}}y\]                 

    D)        \[qE{{y}^{2}}\]

    Correct Answer: B

    Solution :

    Force on charged particle in a uniform electric field is \[F=ma=Eq\] or            \[a=\frac{Eq}{m}\]                                                          ?(1) From the equation of motion, we have \[{{\upsilon }^{2}}={{u}^{2}}+2ay=0+2\times \frac{Eq}{m}\times y\] \[=\frac{2Eqy}{m}\] Now kinetic energy of the particle \[K=\frac{1}{2}m{{\upsilon }^{2}}=\frac{m}{2}\times \frac{2Eqy}{m}=Eq\,y\]


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