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) \(U = V/q\)