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}}.

Answers (1)

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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