A particle of charge q and mass m is subjected to an electric field


A particle of charge $q$ and mass $m$ is subjected to an electric field $E=E_{0}\left(1-a x^{2}\right)$ in the $x$-direction, where $a$ and $E_{0}$ are constants. Initially the particle was at rest at $x=0$. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is :

  1. (1) $a$

  2. (2) $\sqrt{\frac{2}{a}}$

  3. (3) $\sqrt{\frac{3}{a}}$

  4. (4) $\sqrt{\frac{1}{a}}$

Correct Option: , 3


(3) Given,

Electric field, $E=E_{0}\left(1-x^{2}\right)$

$\therefore$ Force, $F=q E=q E_{0}\left(1-x^{2}\right)$

Also, $F=m a=m v \frac{d v}{d x}$              $\left(\because a=v \frac{d v}{d x}\right)$

$\therefore m v \frac{d v}{d x}=q E_{0}\left(1-x^{2}\right)$

$\Rightarrow v d v=\frac{q E_{0}\left(1-x^{2}\right) d x}{m}$

Integrating both sides we get,

$\Rightarrow \int_{0}^{v} v d v=\int_{0}^{x} \frac{q E_{0}\left(1-x^{2}\right) d x}{m}$

$\Rightarrow \frac{v^{2}}{2}=\frac{q E_{0}}{m}\left(x-\frac{9 x^{3}}{3}\right)=0$

$\Rightarrow x=\sqrt{\frac{3}{a}}$


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