CET Karnataka Medical CET - Karnataka Medical Solved Paper-2003

  • question_answer
    A charged particle of mass m and charge q is released from rest in an uniform electric field E neglecting the effect of gravity, the kinetic energy of the charged particle after t second is:

    A)  \[\frac{2{{E}^{2}}{{t}^{2}}}{mq}\]

    B)  \[\frac{E{{q}^{2}}m}{2{{t}^{2}}}\]

    C)  \[\frac{Eqm}{t}\]

    D)  \[\frac{{{E}^{2}}{{q}^{2}}{{t}^{2}}}{2m}\]

    Correct Answer: D

    Solution :

    Electrostatic force on charged particle  \[F=qE\] So, by Newtons law of motion   \[F=ma=qE\]  or   \[a=\frac{qE}{m}\] Velocity attained by particle \[\upsilon =0+\frac{qE}{m}t\] So, kinetic energy \[K=\frac{1}{2}m{{\upsilon }^{2}}\] \[K=\frac{1}{2}\frac{m{{q}^{2}}{{E}^{2}}}{{{m}^{2}}}{{t}^{2}}=\frac{{{E}^{2}}{{q}^{2}}{{r}^{2}}}{2m}\]


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