Question

A particle of charge q and mass m is subjected to an electric field $E=E_0\left(1-a{x}^2\right)$ 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. $\sqrt{\frac{2}{a}}$
• B. $a$
• C. $\sqrt{\frac{3}{a}}$
• D. $\sqrt{\frac{1}{a}}$

Answer 1

The correct option is C

$\sqrt{\frac{3}{a}}$

Step1: Given data.

Particle of charge $=q$

Mass of the particle$=m$

Electric field intensity in $x-$direction $E=E_0\left(1-a{x}^2\right)$

Particle at rest at on $x=0$

Step2: Finding the initial position the kinetic energy of the particle becomes zero.

We know that from work-energy theorem.

$W=∆KE$

Where $W$ work done and $∆KE$ change in kinetic energy.

We have the kinetic energy of the particle becomes zero then,

$W=0$ ……..$\left(i\right)$

Also,

$W={\int }_0^{x}Fdx$

Where $F$ is force and $dx$ is displacement

$W={\int }_0^{x}qEdx$ $\left[\because F=qE\right]$

$\Rightarrow$$W={\int }_0^{x}q{E}_0\left(1-a{x}^2\right)dx$

$\Rightarrow$$W={\int }_0^{x}q{E}_0\left(1-a{x}^2\right)dx=0$ $\left[\because fromequation\left(i\right)\right]$

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

$\Rightarrow$$q{E}_0\left[x-\frac{a{x}^3}{3}\right]=0$

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

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

Hence, option C is correct.