Given electric field

The electric field varies with position as $E(x)=E_0(1-ax^2)$ along the $x$-direction.

A particle of charge $q$ and mass $m$ starts from rest at $x=0$.

Use work-energy theorem

The kinetic energy gained by the particle in moving from $x=0$ to some position $x$ is equal to the work done by the electric force: $K(x)-K(0)=\int_0^x F\,dx$ Since initially the particle is at rest, $K(0)=0$ and $F=qE(x)=qE_0(1-ax^2)$

Therefore, $K(x)=\int_0^x qE_0(1-ax^2)\,dx$

Integrate

$K(x)=qE_0\int_0^x (1-ax^2)\,dx$ $K(x)=qE_0\left[x-\frac{ax^3}{3}\right]_0^x$ $K(x)=qE_0\left(x-\frac{ax^3}{3}\right)$

Set kinetic energy to zero

We want the position other than $x=0$ where kinetic energy again becomes zero: $qE_0\left(x-\frac{ax^3}{3}\right)=0$

For nonzero $qE_0$, $x-\frac{ax^3}{3}=0$ $x\left(1-\frac{ax^2}{3}\right)=0$

One solution is $x=0$ (initial position), and the other is $1-\frac{ax^2}{3}=0$ $ax^2=3$ $x^2=\frac{3}{a}$ $x=\sqrt{\frac{3}{a}}$

Since the question asks for distance from the origin, we take the positive value.

Check options

• A: $a$ ✗
• B: $\sqrt{\dfrac{2}{a}}$ ✗
• C: $\sqrt{\dfrac{3}{a}}$ ✓
• D: $\sqrt{\dfrac{1}{a}}$ ✗

Final answer

The distance from the origin where the kinetic energy again becomes zero is $\boxed{\sqrt{\frac{3}{a}}}$

Comparison with stored answer

Stored correct answer: C

Our derived answer is also C, so they agree.