Question

A particle of charge $q$ and mass $m$ moves rectilinearly under the action of an electric field, $E=4-2x$, where $x$ is a distance from the point where the particle was initially at rest. Calculate the distance travelled by the particle till it comes to rest.

• A. $1\text{m}$
• B. $2\text{m}$
• C. $3\text{m}$
• D. $4\text{m}$

Answer 1

The correct option is D $4\text{m}$

Force acting on the charged particle,
$F=qE$
$\Rightarrow F=q(4-2x)$
Acceleration of the charged particle,
$a=\frac{F}{m}=\frac{q}{m}(4-2x)$
$\Rightarrow v\frac{dv}{dx}=\frac{q}{m}(4-2x)[a=v\frac{dv}{dx}]$
$\Rightarrow \underset{0}{\overset{v}{\int }}vdv=\underset{0}{\overset{x}{\int }}\frac{q}{m}(4-2x)dx$
$\Rightarrow \frac{v^2}{2}=\frac{q}{m}(4x-x^2)\dots \dots \left(1\right)$
When the particle comes to rest momentarily,
$v=0$
$\Rightarrow 4x-x^2=0$ [from $\left(1\right)$]
$\Rightarrow x(4-x)=0$
$\Rightarrow x=0$ or $x=4\text{m}$
Hence, option (d) is correct.