Charge on a particle of mass m =-q

Velocity of the particle = ${\mathrm{v}}_{\mathrm{x}}$

Length of the plates = L

The magnitude of the uniform electric field between the plates = E

Mechanical force, F = Mass (m) x Acceleration (a)

$\begin{array}{lr}\Rightarrow \mathrm{a}=\frac{\mathrm{r}}{\mathrm{m}}& \\ \Rightarrow \mathrm{a}=\frac{\mathrm{qE}}{\mathrm{m}}\dots \dots \dots \dots \dots .\left(1\right) \text{ [as electric force, }\mathrm{F}=\mathrm{qE}]& \end{array}$

Time taken by the particle to cross the field of length L is given by,

$t=\frac{\text{ Length of the plate }}{\text{ Velocity of the particle }}=\frac{L}{V_x}\dots \dots \dots \dots \dots \dots .\left(2\right)$

In the vertical direction, initial velocity, u = 0
According to the third equation of motion, vertical deflection s of the particle can be obtained as

$s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow s=0+\frac{\mathit{1}}{\mathit{2}}\left(\frac{\mathit{q}\mathit{E}}{\mathit{m}}\right)(\frac{\mathit{L}}{{\mathit{V}}_{\mathit{x}}}{\mathit{)}}^{\mathit{2}}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }[\mathrm{From} 1 \mathrm{and} 2]$
$\Rightarrow s=\frac{qE{L}^2}{2m{V}_x^2}$

Hence, vertical deflection of the particle at the far edge of the plate is $\frac{qE{L}^2}{2m{V}_x^2}$. This is similar to the motion of horizontal projectiles under gravity.