Question

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,

• A. $a$
• B. $\sqrt{\frac{2}{a}}$
• C. $\sqrt{\frac{3}{a}}$
• D. $\sqrt{\frac{1}{a}}$

Answer 1

The correct option is C $\sqrt{\frac{3}{a}}$

Given,
Electric field, $E=E_0(1-a{x}^2)$

$\therefore F=qE=q{E}_0(1-a{x}^2)$

Also, $F=ma=m\frac{dv}{dt}$

Also, $a=\frac{dv}{dt}=\frac{dv}{dx}\times \frac{dx}{dt}=v\frac{dv}{dx}$

$\therefore mv\frac{dv}{dx}=q{E}_0(1-a{x}^2)$

$\Rightarrow vdv=\frac{q{E}_0(1-a{x}^2)dx}{m}$

Integrating both sides, we get,

$\Rightarrow {\int }_0^{1}vdv={\int }_0^x\frac{q{E}_0(1-a{x}^2)dx}{m}$

$\Rightarrow \frac{v^2}{2}=\frac{q{E}_0}{m}(x-\frac{a{x}^3}{3})$

When, $v=0,\frac{q{E}_0}{m}(x-\frac{a{x}^3}{3})=0$

After solving, we get, $x=\sqrt{\frac{3}{a}}$

Hence, $\left(C\right)$ is the correct answer.