# 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).

**Question:**

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.

**Solution:**

Charge on a particle of mass *m* = − *q*

Velocity of the particle = *v**x*

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.