A proton enters a parallel-plate capacitor traveling to the right at a speed of 1.276 x 10-5 m/s. The distance between the two plates is 1.59 cm. The proton enters the capacitor halfway between the top plate and the bottom plate; that is, a distance r = 0.795 cm from each plate. The capacitor has a 2.95 x 10-4 N/C uniform electric field between the plates that points downward from the top plate to the bottom plate. Neglecting gravitational forces, what horizontal distance does the proton traverse before the proton hits the bottom plate?
The Attempt at a Solution
Since the electric field is 2.95 x 10-4 N/C I used that to get my force by multiplying the charge of an electron 1.6E-19 thus I got F = 4.72E-23 from that I used F = ma to calculated the acceleration which I got 28265 m/s^2. But now I am stuck. The question will have been easier if they gave the width of the capacitor but they gave me the distance between. How can I do this problem? Thanks!