Q85 / 160 · JIPMER 2006 · Physics

Two similar charges having mass $m$ and $2 m$ are placed in an electric field. The ratio of their kinetic energy is :

  • A
    $4: 1$
  • B
    $1: 1$
  • Correct answer
    C
    $2: 1$
  • D
    $1: 2$

Solution

Force on charged particle is electric field $F=q E$ $\therefore$ Acceleration $a=\frac{F}{m}=\frac{q E}{m}$ $\therefore$ Velocity $\begin{aligned} v & =u+a t \\ & =0+\frac{q E}{m} t\end{aligned}$ So, kinetic energy $\begin{aligned} & =\frac{\frac{1}{2} m_1\left(\frac{q E}{m_1} t\right)^2}{\frac{1}{2} m_2\left(\frac{q E}{m_2} t\right)^2} \\ & =\frac{m_2}{m_1}=\frac{2 m}{m} \\ & =\frac{2}{1} \\ & =2: 1\end{aligned}$

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

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