A uniform E exiest between two metal plates one negative and other positive. The plate length is l and the seperation of the plates is d.
(i) An electron and a proton start fron the negative plate and positive plane, respactively, and go the opposite plates. which one of them wins this race?
(ii) An electron and a proton start moving parallel to the plates towards the other end from the midpoint of the separation of plates at one end of the plates. Which of two will have greater deviation when they come out of the plates if they start with the
(a) same initial velocity
(b) same initial kinetic energy, and (c ) same initial momentum.

Explanation

(i) `a_(e)=(eE)/(m_(e)), a_(p)=(eE)/(m_(p)), d=1/2at^(2)`
`t=sqrt((2d)/(a))` or `t=sqrt((2md)/(eE))`
Therefore, we have
`(t_(e))/(t_(p))=sqrt((m_(e))/(m_(p)))`
As `m_(e)ltm_(p)`, so `t_(e)ltt_(p)`. Hence, electron will take less time i.e. the electrons wins the race

(ii) Time to cross the plates is `t=1//u`. Deviation is
`y=(1)/(2)at^(2)=(1)/(2)(eE)/(m)((1)/(u))^(2)`
or, `(y_(e))/(y_(p))=(m_(p))/(m_(e))((u_(p))/(u_(e)))^(2)`
(a) if `u_(p) =u_(e)` then
`(y_(e))/(y_(p))=(m_(p))/(m_(e))`
As `m_(p)gtm_(e)` so `y_(e)gty_(p)` Hence, the deviation of hte electron will be more.

b. From Eq. (i)
`(y_(e))/(y_(p))=((m_(p)u_(p)^(2))/(m_(e)u_(e)^(2)))=1` (as given)
Hence, the deviation of both the electron and the proton will be the same.
(c) From eq (i), `(y_(e))/(y_(p))=((m_(p)u_(p))/(m_(e)u_(e)))^(2) (m_(e))/(m_(p))=(m_(e))/(m_(p))`
As `m_(e)ltm_(p)`, so `y_(e)lty_(p)`. Hence, the deviation of proton will be more.