An electron `(e,m)` is projected into a uniform electric field `E` vertically downward between parallel plates at separation `d`, with velocity `v_(0)`. The electron enters the electric field at a point midway between the plates. Find
(a) the magnitude of electric field if electron emerges from the electric field just at the edge of upper plate.
(b) the direction of the velocity of electron as it emerges from the electric field.
(c ) the equation of trajectory followed by electron.
The length of plates is `L` and consider only electric force.

Explanation

This problem is similar to when a ball is thrown from the top of a tower horizontally in uniform gravational field , this is a case of two - dimensional motion.
Let the particle reaches to point `P` after time `t`.
`x`-direction (uniform motion):
`v_(x) = v_(0)` ...(i)
`x = u_(x) t = v_(0) t` ....(ii)
`y`-direction (acceleration motion) :
`v_(y) = u_(y) + a_(y) t = (e E)/(m) t` ....(iii)
`y = u_(y) t + (1)/(2) a_(y) t^(2) = (1)/(2) (eE)/(m) t^(2)` ....(iv)
(a) When the paticle reaches to point `A`
`x = v_(0) t rArr L = v_(0) t rArr t = (L)/(v_(0))`
`y = (1)/(2) (eE)/(m) t^(2) rArr (d)/(2) = (eE)/(2m) ((L)/(v_(0)))^(2) rArr E = (mv_(0)^(2) d)/(eL^(2))`
(b) At `A` :
`tan theta = (v_(y))/(v_(x)) = (e Et //m)/(v_(0)) = (eEL)/(mv_(0)^(2))`
`theta = tan^(-1) ((e EL)/(mv_(0)^(2)))`
where `theta` : angle made by the velocity of electron with initial direction of velocity.
(c ) To find trajectory, eleiminating `t` from `(ii)` and `(iv)`
`y = (1)/(2) (eE)/(m) ((x^(2))/(v_(0)))^(2) = (eEx^(2))/(2 mv_(0))`
This is the equation of parabola , hence trajectory of electron is parabolic.