Q146 / 170 · Manipal (MET) 2017 · Physics

A charged particle of mass $m$ and charge $q$ is released from rest in a electric field of magnitude $E$. The kinetic energy of the particle after $t$ second is

  • A
    $\frac{q E t}{2 m}$
  • B
    $\frac{q^2 E^2 t}{2 m}$
  • C
    $\frac{q^2 E^2 t^2}{m}$
  • Correct answer
    D
    $\frac{q^2 E^2 t^2}{2 m}$

Solution

Force on a particle with charge $q$ in electric field $E$ is $\begin{aligned} F & =q E=m a \\ \therefore a & =\frac{q E}{m}\end{aligned}$ here $a=\frac{v}{t}$ $\therefore \frac{v}{t}=\frac{q E}{m} \Rightarrow v=\frac{q E t}{m}$ Also, kinetic energy is due to velocity $(v)$ is $K=\frac{1}{2} m v^2=\frac{1}{2} m\left(\frac{q E t}{m}\right)^2=\frac{q^2 E^2 t^2}{2 m}$

Hence, the correct answer is option: 'D'.

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