Charge on a particle of mass $m =- q$

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

Length of the plates $= L$

Magnitude of the uniform electric field between the plates $= E$

Mechanical force, $F =$ Mass $( m ) \times$ Acceleration (a) $\Rightarrow a=\frac{F}{m}$

$\Rightarrow a=\frac{q E}{m} \ldots \therefore(1)$$\text { [as electric force, } F=q E]$

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

$t=\frac{\text { Length of the plate }}{\text { Velocity of the particle }}=\frac{L}{V_{x}} \ldots(2)$

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,

$s=ut+\frac 12 at^2$

$\Rightarrow s=0+\frac{1}{2}\left(\frac{q E}{m}\right)\left(\frac{L}{V_{x}}\right)^{2}$ $[\text { From }(1) \text { and }(2)]$

$\Rightarrow s=\frac{q E L^{2}}{2 m V_{x}^{2}}$

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