Problem 11-35 Potential and kinematics - Part 5 - 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

\(E = \Delta V/d = 8.9 \times 10^{-3} m/75.3\; V = 8460\; V/m \\ F = qE = (1.6 \times 10^{-19} \; C)(8460\; V/m\; \text{or} \; N/C) = 1.354 \times 10^{-15} \; N\)

Correct!

The acceleration of the proton is:

(A)   \(a = F/m = (1.354 \times 10^{-15} \; N)/(1.67 \times 10^{-27} \; kg) = 8.1 \times 10^{11}\; m/s^2\)

(B)   \(a = F/m = (1.354 \times 10^{-15} \; N)/(9.1 \times 10^{-31} \; kg) = 1.5 \times 10^{15} \; m/s^2\)