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

Question

Philoid · Accepted Answer

Given,

Mass of the particle = m

Charge of the particle = -q

Velocity of the particle = v_x

Length of the plate = L

Electric field between the plates = E

From Newton’s second law of motion,

Where, F = Force on the particle

m = mass of the particle

a = acceleration of the particle

⇒

⇒ (Since F = qE)

Time taken by the particle to cover the distance L is given by

(Time = Displacement/Velocity)

For the movement in vertical direction, initial velocity (u) is zero.

According to Newton’s second equation of motion,

⇒

This is the vertical displacement of the particle at the far edge of the plate.

This motion is very similar to the motion of a projectile in a gravitational field. In a gravitational field, the force acting on the particle is mg and in the given case it is qE. The trajectory followed by the object will be similar in both the cases.

NOTE: A projectile is any object thrown into space by the exertion of a force. The path followed by a projectile is known as its trajectory.