Problem 11-35 Potential and kinematics - Part 4B - C
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
\(U = qV\)
\(U_1 = U_2\)
This says that at the end of the process the potential energy is unchanged. Has the proton been accelerating?