Let particle have charge $q$ and mass ${ }^{\prime} m ^{\prime}$

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 )$$\ldots(1)$

(Along $-y), y_{0}=0+\frac{1}{2} \cdot \frac{q E}{m} \cdot t_{0}^{2}….(2)$

$v_{x}=v_{0}$

$v=u+a t$ (along -ve $'y'$)

speed $v _{ y 0}=\frac{ q E }{ m } \cdot t _{0}$

$\tan \theta=\frac{ v _{ y }}{ v _{ x }}=\frac{ qEt _{0}}{ m \cdot v _{ o }},\left( t _{ o }=\frac{ d }{ v _{ o }}\right)$

$\tan \theta=\frac{q Ed }{ m \cdot v _{0}^{2}}$

$\operatorname{slope}=\frac{-q \operatorname{Ed}}{m v_{0}^{2}}$

Now we have to find eq $^{n}$ of straight line whose slope is $\frac{-q Ed }{ mv _{0}^{2}}$ and it pass through

point $\rightarrow\left( d ,- y _{0}\right)$

Because after $x > d$

No electric field $\Rightarrow F _{\text {net }}=0, \overrightarrow{ v }=$ const.

$y=m x+c,\left\{\begin{array}{l}m=\frac{q E d}{m v_{0}^{2}} \\ \left(d,-y_{0}\right)\end{array}\right\}$

$-y_{0}=\frac{-q E d}{m v_{0}^{2}} \cdot d+c \Rightarrow c=-y_{0}+\frac{q E d^{2}}{m v_{0}^{2}}$

Put the value

$y=\frac{-q E d}{m v_{0}^{2}} x-y_{0}+\frac{q E d^{2}}{m v_{0}^{2}}$

$y _{0}=\frac{1}{2} \cdot \frac{ qE }{ m }\left(\frac{ d }{ v _{0}}\right)^{2}=\frac{1}{2} \frac{ q Ed ^{2}}{ mv _{0}^{2}}$

$y =\frac{- qEdx }{ mv _{0}^{2}}-\frac{1}{2} \frac{ qEd ^{2}}{ mv _{0}^{2}}+\frac{ qEd ^{2}}{ mv _{0}^{2}}$

$y=\frac{-q E d}{m v_{0}^{2}} x+\frac{1}{2} \frac{q E d^{2}}{m v_{0}^{2}}$

$y=\frac{q E d}{m v_{0}^{2}}\left(\frac{d}{2}-x\right)$