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 Fig. 1.33). 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/ (2m`v_s^2`).
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)
However, electric force, F=qE
Therefore, acceleration, `a=(qE)/m` ...(1)
Time taken by the particle to cross the field of length L is given by,
t=`"lenght of the plate"/"Velocity of the particle"=L/v_x` ...(2)
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
`(qEL^2)/(2mV_s^2)`. This is similar to the motion of horizontal projectiles under gravity.
Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.