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 *qEL*^{2}/ (2*m`v_s^2`*).

#### 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=F/m`

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,

`s=ut+1/2at^2`

`s=0+1/2("qE"/m)(L/v_x)^2`

`s=(qEL^2)/(2mV_x^2)` ...(3)

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.

#### Notes

*Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.*