$A$ particle of mass $m$ and charge $(-q)$ enters the region between two charged plates,initially moving along the $x$-axis with speed $v_{x}$ (like particle $1$ in the Figure). The length of the plate is $L$ and a uniform electric field $E$ is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is $q E L^{2} / (2 m v_{x}^{2})$. Compare this motion with the motion of a projectile in a gravitational field.

(N/A) Given:
Mass of the particle = $m$
Charge of the particle = $-q$
Initial velocity along $x$-axis = $v_{x}$
Length of the plates = $L$
Uniform electric field = $E$
$1$. Force and Acceleration:
The electric force acting on the particle is $F = qE$. Since the particle is negatively charged,the force acts in the direction opposite to the electric field. The acceleration $a$ of the particle in the vertical direction ($y$-axis) is given by Newton's second law:
$a = \frac{F}{m} = \frac{qE}{m}$
$2$. Time of Flight:
The time $t$ taken by the particle to cross the plates of length $L$ with a constant horizontal velocity $v_{x}$ is:
$t = \frac{L}{v_{x}}$
$3$. Vertical Deflection:
In the vertical direction,the initial velocity $u_{y} = 0$. Using the kinematic equation $s = u_{y}t + \frac{1}{2}at^{2}$:
$s = 0 \cdot t + \frac{1}{2} \left( \frac{qE}{m} \right) \left( \frac{L}{v_{x}} \right)^{2}$
$s = \frac{qEL^{2}}{2mv_{x}^{2}}$
$4$. Comparison with Projectile Motion:
This motion is analogous to the motion of a projectile launched horizontally in a uniform gravitational field $g$. In projectile motion,the vertical displacement is $y = \frac{1}{2}gt^{2} = \frac{1}{2}g(x/v_{x})^{2} = \frac{gx^{2}}{2v_{x}^{2}}$. Here,the role of gravitational acceleration $g$ is played by the electric acceleration $a = qE/m$.