Clarifying Electric potential energy

  • #1
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I read somewhere online that according to Coulomb's law, in the context of an atom, the potential energy of an electron is proportional to the distance (squared) between the positive and negative charges, this however doesn't show up in the actual eq F = kq1q2/r^2

I thought about it in terms of grav. potential energy and it made sense that a charge separated further would have more potential energy as would an object that were lifted higher above the earth. However, what I didn't get was the attribution to Coulomb's Law that somehow shows PE is proportional to R^2 (rather than being inversely proportional).

I wanted to post this to see if I were missing something here or if what I read was not so accurate.

Thanks
 
  • #2
First of all, the energy eigenstates of a hydrogen like atom do not have a fixed specified radius, only a distribution. Of course, this distribution will have an expectation value, which is typically what would be quoted.

Regarding your question, either your reference was wrong or you interpreted it wrong. It is impossible to tell which unless you provide the reference.
 

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