A proton enters a parallel-plate capacitor traveling to the right at a speed of

Question

A proton enters a parallel-plate capacitor traveling to the right at a speed of 1.279 10-5 m/s, as shown in the figure. The distance between the two plates is 1.66 cm. The proton enters the capacitor halfway between the top plate and the bottom plate; that is, a distance r = 0.830 cm from each plate, as shown in the figure. The capacitor has a 2.80 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?

Explanation / Answer

The field exerts a force on the proton, which accelerates down according to its mass; that takes some time "t," during which it traverses some distance "x."

acceleration a = F / m = qE / m = 1.602e-19C * 2.8e-4N/C / 1.672e-27kg
a = 26827.75 m/s²

r = ½at²
0.00830 m = ½ * 26827.75m/s² * t²
t = 7.86e-4 s

x = v*t = 1.279e-5m/s 7.86e-4s = 10.05e-9 m answer

BTW, that seems awfully slow for a proton.

Hope this helps!

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