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).
A particle of mass $m$ and charge $(-q)$ enters the region between the two charged plates initially moving along $x$-axis with speed $v x$ (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 $q E L^{2} /\left(2 m v_{x}^{2}\right)$.
Compare this motion with motion of a projectile in gravitational field discussed in Section $4.10$ of Class $X I$ Textbook of Physics.
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)
$a=\frac{F}{m}$
However, electric force, $F=q E$
Therefore, acceleration, $a=\frac{q E}{m}$ ...(1)
Time taken by the particle to cross the field of length L is given by,
$t=\frac{\text { Length of the plate }}{\text { Velocity of the particle }}=\frac{L}{v_{x}}$ ...(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,
$s=u t+\frac{1}{2} a t^{2}$
$s=0+\frac{1}{2}\left(\frac{q E}{m}\right)\left(\frac{L}{v_{x}}\right)^{2}$
$s=\frac{q E L^{2}}{2 m V_{x}^{2}}$ ...(3)
Hence, vertical deflection of the particle at the far edge of the plate is
$q E L^{2} /\left(2 m v_{x}^{2}\right)$. This is similar to the motion of horizontal projectiles under gravity.