Problem 11-35 Potential and kinematics - Part 7 - 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

$E = \Delta V/d = 8.9 \times 10^{-3} m/75.3\; V = 8460\; V/m \\ F = qE = (1.6 \times 10^{-19} \; C)(8460\; V/m\; \text{or} \; N/C) = 1.354 \times 10^{-15} \; N \\ a = F/m = (1.354 \times 10^{-15} \; N)/(1.67 \times 10^{-27} \; kg) = 8.1 \times 10^{11}\; m/s^2$

Correct!

$v^2 = v_0{^2} + 2ax = 0 + 2(8.1 \times 10^{11}\; m/s^2)( 8.9 \times10^{-3}\;m)^2 = 1.44 \times 10^{10} \\ v = 1.2 \times 10^5\; m/s$

Did you notice during this solution that the distance $8.9 \times 10^{-3}\; m$ appeared twice in the calculation? It was used as a divisor to get the field $E,$ and then as a multiplier to get $v^2$. In other words it was irrelevant to the calculation.

Now you should pursue the energy approach and see how much simpler it is and that the distance is indeed irrelevant.

You have completed this problem.

Now try the energy approach.