A particle of charge q and mass m is subjected to an electric field $E=E_{0}(1-ax^{2})$ 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: Option: 1 a Option: 2 Option: 3 Option: 4

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

As we know that force acting on a charge in an electric field is qE and the acceleration will be -

$a = \frac{qE}{m}$

$\begin{array}{l} \frac{v d v}{d x}=\frac{q E_{0}}{m}\left(1-a x^{2}\right) \\ \Rightarrow \quad \frac{v^{2}}{2}=\frac{q E_{0}}{m}\left[x-\frac{a x^{3}}{3}\right]=0 \\ \Rightarrow \quad x=\sqrt{\frac{3}{a}} \end{array}$

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