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 vₓ² ). Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.

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) x Acceleration (a)

=>a = F/m

a = qE/m ……………… (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)

t=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 + at²/2

=> 0 + 1/2  (qE/m) (L/V_x)²                   [From (1) and (2)]

=> s= (qEL²)/(2mV²_x)

Hence, vertical deflection of the particle at the far edge of the plate is (qEL²)/(2mV²_x).

This is similar to the motion of horizontal projectiles under gravity.