A particle of mass m and charge (-q) enters the region between the two charged plates initially moving along X-axis with speed `v_(x)` (like particle in figure). The length of 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 `qEL^(2)//(2" mv"_(x)^(2))`.
Compare this motion with motion of a projectile in gravitational field.
In this question, we have to discuss te motion of a particle under the effect of electrostatic field as the electric field is a uniform, so the force on the particle will be constant hence so the force on the particle will be constant, hence we can use equation will be constant, hence we can use equation of a uniformly accelerated motion.

Explanation

Mass of particle =m, charge on particle `=-q`
speed of particle `=v_(x)`, length of plate = L
Electric field between the plates E
(from positive plate to negative plate)
Let the deflection in the path of charge `-q` be Y because the force is acting in positive Y-axis direction.
The direction of force is from negative plate to positive plate because the charge is negative in nature.
Let us discuss the motion in Y-axis direction Initial velocity, `u=0`
Acceleration, `a=(F)/(m)=(=qE)/(m)`
Deflection, `y=?`
`:." Time"=("Distance")/("Velocity")=(L)/(v_(x))`
Using second equation of motion, `s=ut+1/2"at"^(2)`
Putting the values, we get
`y=0+1/2xx(+(qE)/(m))xx(L^(2))/(v_(x)^(2))=(qEL^(2))/(2mv_(x)^(2))`
In the case of a projectile projected horizontaly, `y=(1)/(2)"gt"^(2)`. Thus, is is exactly similar to the projectile motion in the gravitational field.