Problem 11-35 Potential and kinematics - Part 5 - 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

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

No. The electric force is related directly to the field not the potential, therefore \(F = qE\) not \(q\Delta V.\)
 
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