# 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

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.