Problem 11-35 Potential and kinematics - Part 4 - 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

Proton diagram B

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

Correct!

The force on the proton is:

(A)   \(F = qE = (1.6X \times 10^{-19}\; C)(8460\; V/m\; \text{ or} \;N/C) = 1.354 \times 10^{-15} \; N\)

(B)   \(F = q \Delta V = (1.6 \times 10^{-19} \; C)(75.3\;V) = 1.2 \times 10^{-17} \; N\)