# Problem 11-35 Potential and kinematics - Part 2 - B

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 Energy approach.
The potential energy $U$ of a charge $q$ at a location where the potential is $V$ is:

(A)   $U = qV$

(B)   $U = q/V$

(C)   $U = V/q$