Question

A particle with specific charge $\frac{q}{m}$ moves rectilinearly due to an electric field $E=E_0-ax,$ where $a$ is a positive constant and $x$ is the distance from the point where the particle was initially at rest. The distance covered by the particle till the moment it comes to standstill is $\frac{n{E}_0}{a}$.The value of $n$ is

Answer 1

Under the influence of electric field
$F_e=qE=q(E_0-ax)$
$m\frac{dv}{dt}=q(E_0-ax)$
$⟹v\frac{dv}{dx}=\frac{q}{m}(E_0-ax)$
$⟹\int vdv=\int \frac{q}{m}(E_0-ax)dx$
$⟹\frac{v^2}{2}=\frac{q}{m}(E_{0}x-a\frac{x^2}{2})+C$

Using initial condition,
At $x=0$, $v=0$ and $C=0$
$v^2=\frac{2q}{m}[E_{0}x-a\frac{x^2}{2}]$

When the particle comes to rest (i.e $V$ = 0)
$0=E_{0}x-\frac{a{x}^2}{2}$

$⟹x=\frac{2E_0}{a}$

$\therefore x=\frac{2E_0}{a}$ is the distance covered by the particle till the moment it comes to rest.