Coulomb's law and energy - potential energy

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Homework Help Overview

The discussion revolves around the application of Coulomb's law and the concept of potential energy in a system involving two electrons. Participants are examining the derivation of potential energy and the implications of the movement of both charges in relation to their separation distance.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are questioning the validity of the original derivation of potential energy when both electrons are in motion. There is exploration of how the distance between the electrons affects the calculations, particularly regarding the differential displacements and their implications on potential energy.

Discussion Status

The discussion is ongoing, with participants providing insights into the nuances of the calculations involved. Some guidance has been offered regarding the interpretation of variables and their meanings in the context of the problem, but no consensus has been reached on the correct approach.

Contextual Notes

Participants are grappling with the definitions of variables such as "dr" and "dx" in the context of their specific problem, indicating potential confusion or misapplication of terms. There is also mention of the initial conditions of the electrons starting from rest and moving to infinity.

laser
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Homework Statement
Two electrons. Starts from rest. Goes to infinity (PE_f = 0)
Relevant Equations
F=ma
20240318_160837.jpg


I know that the formula qqk/r applies to a system (two charges), but where is the flaw in my derivation? Thanks!
 
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Because the displacement of each electron is not dr if you move both.
 
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laser said:
Homework Statement: Two electrons. Starts from rest. Goes to infinity (PE_f = 0)
Relevant Equations: F=ma

View attachment 341989

I know that the formula qqk/r applies to a system (two charges), but where is the flaw in my derivation? Thanks!
I think this is quite instructive. First, imagine that one electron is held in place by an external force. As that electron does not move, the restraining force does no work. The second electron moves off to infinity with the calculation you have done. That electron, therefore, gets all the PE of the system. Once it's far enough away, you could release the first electron and there is no more PE to be squeezed out of the system.

That should be clear. Now, you can look more closely at your calculation and see why that does not apply if both electrons are moving. And, the reason is that the distance between the elecrons is increasing at twice the rate of your ##dx##, which relates to the motion of one electron only.
 
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PeroK said:
And, the reason is that the distance between the elecrons is increasing at twice the rate of your ##dx##, which relates to the motion of one electron only.
But does this mean that dr is actually 2dr? Would this not double the potential energy?
 
laser said:
But does this mean that dr is actually 2dr? Would this not double the potential energy?
If you label the electrons and take ##dx_1## as the differential along the path of one of the electrons, then ##dr = 2dx_1##. If you use that in your calculation, you get half the KE for the first electron. And the same KE for the other.

In general I've noticed that many students use a label such as "dr" without thinking about what it means in that specific problem. In that response, you are using ##dr## as two different things!
 
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PS if you had used ##F_1## and ##dx_1## in the first place, you might have spotted that ##dr## was something different.
 

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