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 answerD$\frac{q^2 E^2 t^2}{2 m}$
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'.