# Q 1.33: A particle of mass m and charge (–q) enters the region between the two charged plates initially moving along x-axis with speed vx .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 $\frac{qEL^{2}}{2mv_{x}^{2}}$.

Let s be the vertical deflection, t be the time taken by the particle to travel between the plates

$\therefore$ $s = ut + \frac{1}{2}at^2$

Here , u =0 , since initially there was no vertical component of velocity.

The particle in the elecric field will experience a constant force (Since, Electric field is constant.)

F = ma = -qE  (Using Newton's Second Law, F = ma)

$\therefore$ a = -qE/m (-ve sign implies here in downward direction)

Again, t = Distance covered/ Speed  = $L/ v_{x}$

(In x-direction, since there is no force, hence component of velocity in x-direction remains constant = $v_{x}$.

And, the distance covered in x-direction = Length of the plate = L)

Putting these values in our deflection equation,

$\\ s = ut + \frac{1}{2}at^2 = (0)(L/v_{x}) + \frac{1}{2}(-qE/m)(L/v_{x})^2 \\ \implies s =\frac{-qEL^2}{2mv_{x}^2}$   (S is -ve, which implies it deflect in downwards direction.)

$\therefore$ The vertical deflection of the particle at the far edge of the plate is $\frac{qEL^{2}}{2mv_{x}^{2}}$.

This motion is similar to the motion of a projectile in a gravitational field, which is also a constant force. The force acting on the particle in the gravitational field is mg whiles in this case, it is qE. The trajectory will be the same in both cases.

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