1.33 A particle of mass m and charge (–q) enters the region between the two charged plates initially moving along the x-axis with speed vx (like particle 1 in Fig. 1.33). The length of the plate is Land 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/(2mvx2).

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


 

Charge on a particle of mass m =-q

Velocity of the particle = vx

Length of the plates = L

The magnitude of the uniform electric field between the plates = E

Mechanical force, F = Mass (m) x Acceleration (a)

a=rma=qEm.(1)    [as electric force, F=qE]

Time taken by the particle to cross the field of length L is given by,

t= Length of the plate  Velocity of the particle =LVx.(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+12at2
s=0+12(qEm)(LVx)2             [From 1 and 2]
s=qEL22mVx2

Hence, vertical deflection of the particle at the far edge of the plate is qEL22mVx2. This is similar to the motion of horizontal projectiles under gravity.