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  • 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.

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 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 \frac{qEL^{2}}{2mv_{x}^{2}}.

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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 electric field will experience a constant force (Since the electric field is constant.)

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

\therefore a = -qE/m (-ve sign implies here in the 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 deflects in the downward 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 while in this case, it is qE. The trajectory will be the same in both cases.

Posted by

HARSH KANKARIA

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