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 :
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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