Find the magnitude of the electric field between the two plates (assume field between plates is uniform)?

Question

an electron enters the lower left side of a set of parallel plates and exits at the upper right side. The intial speed is 7.0 * 10^6 m/s.
The plates are 21 cm long, and are separted by 10 cm.
so basically it starts here
and ends here
start
and the y distance is 10 cm. , while the x distnace is 21 cm. the top plate is postiviely charge,d while bottom plate is negatively charged.

Answer 1

The electron will be accelerated toward the top plate by the force from the electric field E. The force in that field is q*E. The acceleration a = F/m = (q/m)*E The distance traveled by a particle under acceleration a is s=.5*a*t^2, so the time it takes to go from one plate to the other is √[2*h/a], or t = √2*h*(m/q)/E, where h is the plate separation. The horizontal velocity of the electron is not affected, and remains at v0. The particle travels the x-distance w in t = w/v0; equate these times to get:

√[2*h*(m/q)/E] = w/v0 and solve for E.

E = (2*h*m*v0^2)/q*w^2

q = 1.6*10^-19 coul, m = 9.1*10^-31 kg

E = 12.6 volt/cm, 1260 volt/meter