The length of plane-parallel electrodes is `l`, the distance between them is h. An electric field of intensity E is set up between them. An electron enters the field close to the lower plate at an initial speed `v_(0)` and at an angle `alpha` to the plates (Fig.). What should the field intensity be for the electron to pass between the electrodes without striking either of them? For what angles `alpha` is this possible?

Answer

`0 lt tan alpha lt (4h)/(l); (mv_(0)^(2) sin^(2)alpha)/(2eh) lt E lt (mv_(0)^(2)sin 2alpha)/(el)`.

Explanation

No forces act on the electron in the x-direction and it moves along this axis at a constant speed `v_(x)= v_(0) cos alpha`. In the y-direction a force `F_(y)= -eE` acts on the electron, and it moves with an acceleration `a_(y)=-eE//m`, its instantaneous velocity is
`v_(y)= v_(0) sin alpha +a_(y)t =v_(0) sin alpha u- (e Et)/(m)`
To prevent the electron moving upwards from striking the upper plate, a field has to be established that would in time `t_(1)` reduce the vertical velocity component to zero and would guarantee the condition `h_(1) lt h`. But `2a_(y)h_(1)= -v_(0y)^(2)`
therefore
`|a_(y)|=(v_(0)^(2)y)/(2h_(1)) gt (v_(0)^(2)sin^(2)alpha)/(2h)`
Hence, the first condition may be written in the form
`E gt =(mv_(0)^(2)sin^(2)alpha)/(2eh)`
The time the electron moves upwards is `t_(1)= (-v_(0) sin alpha)/(a_(y))` The time it moves downwards will be the same (prove this!). To prevent it form striking the lower plate it must be made to travel during this time a distance in the x-direction exceeding the length of the plate, i.e. `x=2v_(x)t_(1) gt `l. Hence
`(-2v_(0)^(2)sin alpha cos alpha)/(a_(y)) gt l`
and the second condition assumes the form
`E lt (2mv_(0)^(2)sin alpha cos alpha)/(el)`
Combining both solutions we obtain the result sought.