Problem 11-35 Potential and kinematics - Part 2 - A

A proton is accelerated from rest from a positively charged plate to a parallel negatively charged plate. The separation of the plates is \(8.9\; mm.\) If the potential difference between the plates is \(75.3\; V,\) what is the speed of the proton just as it hits the negative plate?
[Ans. \(1.2 \times 10^5\; m/s\)]

Accumulated Solution

Proton diagram B

Let us then use the Force/acceleration/velocity approach.
The potential difference is given as \(\Delta V = 75.3\; V.\) What is the magnitude of the field between the plates?

(A)   \(E= d/\Delta V = 8.9 \times 10^{-3} m/75.3 V = 1.18 \times 10^{-4} \; m/V\)

(B)   \(E = \Delta V/d = 75.3 V/8.9 \times 10^{-3}\; m = 8460\; V/m\)

(C)   \(E\) is always zero between metal plates. The metal plates shield the region from electric fields.