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Statement of the problem:

Consider an electron and a proton that are initially at rest separated $a$ meters. Do not take into account the movement of the proton, because its mass is much greater than the electron's.
1. What is the minimum kinetic energy at which the electron must be "launched" so that the electron gets to be $b$ meters away from the proton?
2. What is the corresponding minimum speed for that situation?
3. What distance away from the proton will the electron reach when it has double the initial kinetic energy?

(The original problem says 2.00 nm instead of $a$ meters, and 12.0 nm instead of $b$ meters)

My attempt:

Question 1 and 2
We say $q$ represents the elementary charge, i.e. the charge of the proton (positive) and of the electron (negative). Then I can assume that when the electron is $b$ meters away, the KE is zero. Then the difference in potential energy would be the same as the lost KE. $$\Delta U=\frac{1}{4\pi \varepsilon_o} \left ( \frac{q^2}{a} - \frac{q^2}{b} \right ) =K= mc^2 \left ( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1 \right ) \approx \frac{1}{2}m v^2$$ Doing some algebra we can get $v$, which is what they are asking for.


My questions:

4. Is it possible to have an electron and a proton at rest?
I imagine we would need another force that is "pulling" the electron away from the proton, and that that force cancels out with the electrostatic force from the proton. But, what would be a realistic example of a source of that force? And, how would that source alter the entire system?

5. How can we launch an electron that is near a proton?
Question 3 suggests that the speed of the electron suddenly goes from zero to some positive value. But, doesn't that ("launching the electron") imply infinite acceleration? If that is indeed possible, how can we realistically do it?


EDIT: The original problem was written in a confusing way. Thanks to the help of @user7777777 I was able to interpret it the right way and fix the wording.

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closed as off-topic by Kyle Kanos, stafusa, user191954, John Rennie, glS Sep 24 '18 at 18:45

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You are absolutely correct for Questions 1 and 2. For Question 3, we can use the result from Question 2, but replacing $\frac{1}{2} m {v_1}^2$ with $2 (\frac{1}{2} m {v_1}^2)$, and comparing the two results in terms of $b$.

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  • $\begingroup$ Thanks for your answer. In question 1, why the KE has to be equal to the potential energy? I understand that one can eventually transform into the other, but at the beginning I see no reason for them being equal. $\endgroup$ – evaristegd Sep 24 '18 at 12:00
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    $\begingroup$ @evaristegd: From a classical standpoint, it is entirely possible, and I don't think the question is asking for something more complicated than that. The question is also not clear about when the kinetic energy is being computed. $\endgroup$ – user7777777 Sep 24 '18 at 13:04
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    $\begingroup$ Thank you again. I agree with you. I initially thought the question was asking for the KE when they were at rest, but that is trivial. And then question 3 is about the case when it has double the KE, which would make the question even more trivial. $\endgroup$ – evaristegd Sep 24 '18 at 13:19
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    $\begingroup$ @evaristegd: No problem. I believe that Question 3 is asking you to compare the distances reached when you launch with some kinetic energy versus twice that energy. This, obviously, depends on the initial distance to the proton as well. $\endgroup$ – user7777777 Sep 24 '18 at 13:27
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    $\begingroup$ The problem is actually a translation from a foreign language and I just noticed that what you mentioned makes considerably more sense in that language. Thanks to your last comment I think I now fully understand the problem and the solution as well. I will edit the question to make the clarification. $\endgroup$ – evaristegd Sep 24 '18 at 13:48

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