Question

A particle of mass m and charge (−q) enters the region between the two charged plates initially moving along x-axis with speed vx (like particle 1 in Figure). 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 qEL²/(2mvx²).

Answer 1

Charge on a particle of mass m = − q
Velocity of the particle =vx
Length of the plates = L
Magnitude of the uniform electric field between the plates = E
Mechanical force, F = Mass (m) × Acceleration (a)
⇒a=F/m
⇒a=qE/m…………….(1) [as electric force, F = qE]
Time taken by the particle to cross the field of length L is given by,

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,

Hence, vertical deflection of the particle at the far edge of the plate is qEL²/2mV_x². 

This is similar to the motion of horizontal projectiles under gravity.