What is the Magnitude of the Electric Field to Just Miss the Upper Plate?

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To determine the magnitude of the electric field that allows an electron to just miss the upper plate of parallel plates, the initial speed of the electron is given as 2.00×10^6 m/s, with the plates measuring 2 cm in length and 1 cm apart. The uniform electric field is directed vertically downward, while the field outside the plates is zero. The challenge lies in calculating the acceleration due to the electric field and the time taken for the electron to travel through the plates. It is confirmed that the velocity in the horizontal direction remains constant during this motion. The solution requires applying kinematic equations to find the necessary electric field strength.
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Homework Statement


An electron is projected with an initial speed v_0 = 2.00×106 {\rm m/s} into the uniform field between the parallel plates in the figure (Intro 1 figure) . Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates.
If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
(length of plates = 2cm, distance between plates=1cm)

Homework Equations


F=eQ
y=.5at^2+V(t)+y(initial)



The Attempt at a Solution


Ok, so first I tried to force on the charge from the electric field, but I don't know the electric field. So then I tried the position equation to find Acceleration but I don't know the time it took to travel through the plates, I would be able to find that but I am not sure if the velocity in the x direction stays the same? (the direction parallel to the plates)
 
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As you have said, the velocity in the x direction remains the same.
 
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