Question

A charged particle having charge $q$ and mass $m$ moves rectilinearly under the action of an electric field $E(x)=(4-3x)$ $N/C$ where $x$ is the distance in metre from the point where the particle was initially at rest. The distance travelled by the particle till it comes instantaneously to rest again for the first time, is

• A. $\frac{2}{3}m$
• B. $\frac{4}{3}m$
• C. $2m$
• D. $\frac{8}{3}m$

Answer 1

Answer: (D)

$F=qE(x)$
$F=q(4-3x)$
$a=\frac{q}{m}\left(4-3x\right)$
$\frac{dv}{dx}\frac{dx}{dt}=\frac{q}{m}\left(4-3x\right)$
$\underset{0}{\overset{v}{\int }}vdv=\frac{q}{m}\underset{0}{\overset{x}{\int }}\left(4dx-3xdx\right)$
(The initial and final value of velocity is zero)
$\frac{v^2}{2}=\frac{q}{m}\left[4x-\frac{3x^2}{2}\right]$
$v^2=\frac{2q}{m}\left[4x-\frac{3x^2}{2}\right]$
As the particle comes to rest
$4x-\frac{3x^2}{2}=0$
$\frac{3}{2}x=4$
$x=\frac{8}{3}m$