A particle of charge q and mass m is subjected to an electric field  $E=E_0(1-a{x}^2)$  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

Here, Initial Kinetic energy ,

$K_I=0(atx=0)$

Final Kinetic energy 

$K_f=0[atx=x_0(say\left)\right]$

So, charge in Kinetic energy 

$\Delta k=k_f-k_I=0$

From work energy theorem Work done=change in Kinetic energy

i.e  $W=\Delta k\therefore W=0$

$\Rightarrow \int Fdx=0\Rightarrow \int qEdx=0(\therefore F=Eq)$

$\Rightarrow \int q\left[E_0(1-a{x}^2)\right]dx=0[\because E=E_0(1-a{x}^2]$

$\Rightarrow q{E}_0\underset{0}{\overset{x_0}{\int }}(1-a{x}^2)dx=0$

$\Rightarrow q{E}_0(x-\frac{a{x}^3}{3}{]}_0^{x_0}=0$

$\Rightarrow \left[\{x_0-\frac{a{x_0}^3}{3}\}-(0-0)\right]=0$

$\Rightarrow x_0-\frac{a{x_0}^3}{3}=0$$\Rightarrow x_0=\frac{a{x_0}^3}{3}\Rightarrow 1=\frac{a{x_0}^2}{3}$

$\Rightarrow {x_0}^2=\frac{3}{a}\Rightarrow x_0=\sqrt{\frac{3}{a}}$