Q 1.33: A particle of mass m and charge (–q) enters the region between the two charged plates initially moving along x-axis with speed vx .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 .
Let s be the vertical deflection, t be the time taken by the particle to travel between the plates
Here , u =0 , since initially there was no vertical component of velocity.
The particle in the elecric field will experience a constant force (Since, Electric field is constant.)
F = ma = -qE (Using Newton's Second Law, F = ma)
a = -qE/m (-ve sign implies here in downward direction)
Again, t = Distance covered/ Speed =
(In x-direction, since there is no force, hence component of velocity in x-direction remains constant = .
And, the distance covered in x-direction = Length of the plate = L)
Putting these values in our deflection equation,
(S is -ve, which implies it deflect in downwards direction.)
The vertical deflection of the particle at the far edge of the plate is .
This motion is similar to the motion of a projectile in a gravitational field, which is also a constant force. The force acting on the particle in the gravitational field is mg whiles in this case, it is qE. The trajectory will be the same in both cases.