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An electric field of intensity E is set ...

An electric field of intensity E is set up between two parallel plates of length L. An electron beam enters the field at an angle `alpha gt 0` to the plates and leaves it at an angle `beta lt 0` (Fig.). Find the initial velocity of the electrons. The force of gravity is to be neglected.

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The correct Answer is:
`v_(0)= sqrt((eEL cos beta)/(m cos alpha sin (alpha-beta)))`
Since there are no forces in the direction of the x-axis acting on the electron (we neglect the force of gravity), `v_(x)=` const, and the electron transit time in the field is `t=L//v_(x)`. The force acting on the electron in the direction of the y-axis is `F_(y) = - eE`. This force imparts to the electron an acceleration `a_(y)= -eE//m`. The projection of the velocity on the y-axis varies with time according to the equation `v_(y)= v_(0y)+a_(y)t`. At the point of the electron.s exist from the field we have
`v_(x) tan beta = v_(x) tan alpha - (e EL)/(mv_(x))`, from which `v_(x)^(2)(tan alpha-tan beta)=(e EL)/(m)`
Since `v_(x)=v_(0) cos alpha`, we obtain
`v_(0)^(2)cos^(2)alpha (tan alpha-tan beta)= eEL//m`
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