Coulomb's law and energy - potential energy

[laser]

Homework Statement

Two electrons. Starts from rest. Goes to infinity (PE_f = 0)

Relevant Equations

F=ma

I know that the formula qqk/r applies to a system (two charges), but where is the flaw in my derivation? Thanks!

 

[Orodruin]

Because the displacement of each electron is not dr if you move both.

 

[PeroK]

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.

 

[laser]

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?

 

[PeroK]

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!

 

[PeroK]

PS if you had used $F_1$ and $dx_1$ in the first place, you might have spotted that $dr$ was something different.