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)` (as in the fig.). The length of plate is L and an 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)//(2mV_(x)^(2))`.
Compare this motion with motion of a projectille in gravitational field discussed in section 4.10 of `1^(st)` Year Textbook of Physics.

Explanation

Mass of particle `= m`
Charge of particle `=-q`
Speed of particle `=V_(x)`
Length of plates `=L`
Electric field between the plates `=E` (from positive plate to negative plate).
Let the deflection in the path of the charge -q is Y, because the force acting in + Y axis direction. The direction of force is from negative plate to positive plate because the charge in 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 `y=0+1/2 xx(+(qE)/m)L^(2)/V_(x)^(2)`
`Y=(qEL^(2))/(2mV_(x)^(2))`
In the case of projectile motion `y=1/2 g t^(2)`. Thus, it is exactly similar to the projectile motion in the gravitational field.