Question

A charged particle (mass $m$ and charge $q$ ) moves along $X$ axis with velocity $V_0$. When it passes through the origin it enters a region having uniform electric field $\vec{E}=-E\hat{j}$ which extends upto $x=d$. Equation of path of electron in the region $x>d$ is:

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

Answer 1

Answer: (A)

Let particle have charge $q$ and mass 'm' Solve for $(q,m)$ mathematically
$F_x=0,a_x=0,(v{)}_x=$ constant
time taken to reach at ' $P^{\prime }=\frac{d}{v_0}=t_0$ (let) $\dots$.. (1)
(Along $-y),y_0=0+\frac{1}{2}\cdot \frac{qE}{m}\cdot t_0^2\dots .(2)$
$v_x=v_0$
$v=u+at$ (along -ve 'y')
speed $v_{y0}=\frac{qE}{m}\cdot t_0$
$\tan \theta =\frac{v_y}{v_x}=\frac{qE{t}_0}{m\cdot v_0},\left(t_0=\frac{d}{v_0}\right)$
$\tan \theta =\frac{qEd}{m\cdot v_0^2}$
slope $=\frac{-qEd}{m{v}_0^2}$
Now we have to find eq ${}^n$ of straight line
whose slope is $\frac{-qEd}{m{v}_0^2}$ and it pass through
point $\to \left(d,-y_0\right)$
Because after $x>d$
No electric field $\Rightarrow F_{net}=0,\vec{v}=const.$
$y=mx+c,\{m=\frac{qEd}{m{v}_0^2},(d,-y_0)\}$
$-y_0=\frac{-qEd}{m{v}_0^2}\cdot d+c$
$\Rightarrow c=-y_0+\frac{qE{d}^2}{m{v}_0^2}$
Put the value
$y=\frac{-qEd}{m{v}_0^2}x-y_0+\frac{qE{d}^2}{m{v}_0^2}$
$y_0=\frac{1}{2}\cdot \frac{qE}{m}{\left(\frac{d}{v_0}\right)}^2=\frac{1}{2}\frac{qE{d}^2}{m{v}_0^2}$
$y=\frac{-qEdx}{m{v}_0^2}-\frac{1}{2}\frac{qE{d}^2}{m{v}_0^2}+\frac{qE{d}^2}{m{v}_0^2}$
$y=\frac{-qEd}{m{v}_0^2}x+\frac{1}{2}\frac{qE{d}^2}{m{v}_0^2}$
$y=\frac{qEd}{m{v}_0^2}\left(\frac{d}{2}-x\right)$