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:

We have an electric field in the $$x$$-direction given to us as

$$E(x)=E_0\,(1-ax^{2}),$$

where $$E_0$$ and $$a$$ are positive constants. A particle of charge $$q$$ and mass $$m$$ is released from rest at the origin $$x=0$$. Because the particle starts from rest, its initial kinetic energy is

$$K_i = 0.$$

While the particle moves, the electric field does work on it. According to the work-energy theorem,

$$\text{Work done by all forces} = K_f - K_i.$$

Here the only force is the electric force $$\mathbf{F}=q\mathbf{E}$$, so the work done by this force as the particle moves from $$x=0$$ to some position $$x$$ is

$$W = \int_{0}^{x} F_x\,dx' = \int_{0}^{x} q\,E(x')\,dx'.$$

We want the kinetic energy at position $$x$$ to again become zero, i.e.

$$K_f = 0.$$

Substituting $$K_i=0$$ and $$K_f=0$$ into the work-energy theorem gives

$$W = 0.$$

So we must find that particular distance $$x$$ (other than $$x=0$$) for which the work done is zero:

$$\int_{0}^{x} q\,E_0\,(1-a{x'}^{2})\,dx' = 0.$$

The charge $$q$$ and the constant $$E_0$$ are common factors and can be taken outside the integral:

$$qE_0 \int_{0}^{x} (1-a{x'}^{2})\,dx' = 0.$$

Because $$qE_0$$ is non-zero, the integral itself must vanish:

$$\int_{0}^{x} (1-a{x'}^{2})\,dx' = 0.$$

Now we evaluate the integral step by step:

$$\int_{0}^{x} 1\,dx' = x,$$

$$\int_{0}^{x} a{x'}^{2}\,dx' = a\int_{0}^{x} {x'}^{2}\,dx' = a\left[\frac{{x'}^{3}}{3}\right]_{0}^{x}=a\left(\frac{x^{3}}{3}-0\right)=\frac{a x^{3}}{3}.$$

Combining these results, we have

$$x-\frac{a x^{3}}{3}=0.$$

Factor out the common term $$x$$:

$$x\left(1-\frac{a x^{2}}{3}\right)=0.$$

This product is zero when either factor is zero. The first possibility is

$$x=0,$$

which simply reproduces the starting point and is therefore not the distance we are looking for. The second possibility is

$$1-\frac{a x^{2}}{3}=0.$$

Solving this equation for $$x^{2}$$ gives

$$\frac{a x^{2}}{3}=1 \quad\Longrightarrow\quad x^{2}=\frac{3}{a}.$$

Taking the positive square root (distance is positive), we obtain

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

This is the unique position (other than the origin) where the kinetic energy again becomes zero.

Hence, the correct answer is Option C.