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 \(\overrightarrow{ E }=- E \hat{ j }\) which extends upto \(x = d\). Equation of path of electron in the region \(x > d\) is 

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

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

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

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

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

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