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. 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 qEL2/(2m vx2).
(compare this motion with motion of a projectile in graviataional field as discussed in class 11 Physics. )
Solution
The particle is moving along x-axis in a uniformly charged electric field E between two oppositely charged metallic plates of length L. The motion of a charged particle in an electric field is analogous to the motion of a projectile in the gravitational field. The only difference is that here the constant electric field is in upward direction and is limited to the region between the plates of length L. Since x-component of the electric force is zero therefore, acceleration along x-axis is zero. So, the velocity vx along x-axis is constant.

If x is the horizontal distance covered in time t, then
Force acting along y-axis, F = qE
Acceleration along y-axis,
where, m is the mass of charged particle (electron).
If y is the vertical distance covered by the particle in time t, then
So, within the electric field, the particle follows a parabolic path.
Let y, be the vertical deflection suffered by the particle inside the electric field.
When x = L, then

If x is the horizontal distance covered in time t, then
Force acting along y-axis, F = qE
Acceleration along y-axis,
where, m is the mass of charged particle (electron).
If y is the vertical distance covered by the particle in time t, then
So, within the electric field, the particle follows a parabolic path.
Let y, be the vertical deflection suffered by the particle inside the electric field.
When x = L, then