Question

An electron is projected with an initial speed v0 = 5.05

Answer 1

Let the distance between the plates = d
Let the length of the plate = L
The electron enters the field, experiences a uniform acceleration in the vertical direction
= (e*E/m) (where e and m are the charge & mass of an electron and E is the field strength)....(1)

Since the field is purely vertical, there is no force in the horizontal direction and so the horizontal component of the velocity remains constant while traversing the field = u and therefore the time of travel = t = (L/u)......................................

In order for the electron to barely miss the upper plate, the distance travelled in the vertical direction = (d/2) = (1/2)*f*t^2................................
where t = time and f = acceleration. (d/2since the electron enters midway between the plates)

the figure in the link mentions that d = 1cm; L = 2cm........................................

Substituting the value of t obtained from (2) into (3) and then applying (4) you get

L/4 = (1/2)*E*(e/m)*[(L/u)^2]....................

After simplification, this yields E = (u^2/2)*(m/e)*(1/L) = (3.55*E+3) N/C (newtons per coulomb).