Question

A charged particle $\left(\text{mass}m\text{and}\text{charge}q\right)$ moves along $x-\text{axis}$ with velocity $v_0$. When it passes through the origin, it enters a region having uniform electric field $E=-E\hat{j}$ which extends up to $x=d$. The equation of path of electron in the region $x>d$ is :

• A. $y=\frac{qEd}{m{v}_0^2}x$
• B. $y=\frac{qE{d}^2}{m{v}_0^2}x$
• C. $y=\frac{qEd}{m{v}_0^2}(\frac{d}{2}-x)$
• D. $y=\frac{qEd}{m{v}_0^2}(x-d)$

Answer 1

The correct option is C $y=\frac{qEd}{m{v}_0^2}(\frac{d}{2}-x)$

$F_x=0,a_x=0,(v{)}_x=\text{constant}=v_0$

Time taken to reach point $P$
$t_0=\frac{d}{v_0}$

$\text{Along}y-axis,$

$\text{At origin},u_y=0,a_y=\frac{qE}{m}$,

$\Rightarrow y_0=0+\frac{1}{2}.\frac{qE}{m}.t_0^2$

$\text{At point}P,$

$v_y=\frac{qE}{m}{t}_0$

$\Rightarrow \tan \theta =\frac{v_y}{v_x}=\frac{qE{t}_0}{m.v_0}$

$\text{As},t_0=\frac{d}{v_0}$

$\tan \theta =\frac{qEd}{m.v_0^2},\text{slope}=\frac{-qEd}{m.v_0^2}$

In the region $x>d$, There is no electric field, therefore, the path will be straight line.

$y=Mx+c$

Where, $M=\frac{-qEd}{m.v_0^2}$

The cordinates of point $P$ is $(d,-y_0)$

$\Rightarrow -y_0=\frac{-qEd}{m{v}_0^2}d+c$

$\Rightarrow c=-y_0+\frac{qE{d}^2}{m{v}_0^2}$

$\Rightarrow y=\frac{-qEd}{m{v}_0^2}x-y_0+\frac{qE{d}^2}{m{v}_0^2}$

$\text{As}{y}_0=\frac{1}{2}.\frac{qE}{m}.t_0^2=\frac{1}{2}.\frac{qE{d}^2}{m{v}_0^2}$

$\Rightarrow y=\frac{-qEdx}{m{v}_0^2}-\frac{1}{2}\frac{qE{d}^2}{m{v}_0^2}+\frac{qE{d}^2}{m{v}_0^2}$

$\Rightarrow y=\frac{-qEdx}{m{v}_0^2}+\frac{1}{2}\frac{qE{d}^2}{m{v}_0^2}$

$\Rightarrow y=\frac{qEd}{m{v}_0^2}(\frac{d}{2}-x)$

Hence, option $\left(B\right)$ is correct.