Problem 11-35 Potential and kinematics - Part 5B - 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
\(U = qV \\ E_{K1} + U_1 = E_{K2} + U_2\)
Correct!
In fact either (A) or (B) is correct. The zero value of the potential (energy) can be chosen anywhere. If at the negative plate (A) then it is higher \((q\Delta V)\) at the positive plate. If it is zero at the positive plate it is \(-q\Delta V\) at the negative plate.
Let us use the first choice:
\(0 + q\Delta V = ½ \;mv_2{^2} + 0 \\ v_2^2 = \frac{2q\Delta V}{m} = \frac{2(1.6 \times 10^{-19})(75.3)}{1.67 \times 10^{-27}} = 1.44 \times 10^{10} \\ v_2 = 1.2 \times 10^5 \; m/s\)
Notice that the distance \(8.9\; mm\) didn't enter into the calculation, so it is irrelevant.
You should follow through the Force/acceleration approach in order to see how much more economical and straightforward the energy approach is.
You have completed this problem.