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I Coulomb's law, electrostatics?

  1. Jan 13, 2018 #1
    Equation:

    ΣF=Σk(qi)(qj)/(r^2)

    Question:

    Considering more than a couple of particles. How can a net force on a charged particle be calculated if Coulomb's law is under the restriction of static forces?

    Thanks!
     
  2. jcsd
  3. Jan 13, 2018 #2

    Doc Al

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    What does the number of particles matter?
     
  4. Jan 13, 2018 #3
    To imply a generalized approach.

    To restate, how can a net force be present under static conditions?
     
  5. Jan 13, 2018 #4

    Doc Al

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    Perhaps you are thinking of Earnshaw's theorem?
     
  6. Jan 13, 2018 #5

    Dale

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    You are correct. Statics usually assumes no net force. However:

    1) The electric force may not be the net force if there are other non-electric forces involved.

    2) The Coulomb force is not “static” but “electrostatic”, meaning that the currents are at least momentarily 0.
     
  7. Jan 13, 2018 #6
    Assuming no external forces, I suppose Coulombs Law investigates impending motion of charged particles.
     
  8. Jan 14, 2018 at 9:08 AM #7

    ZapperZ

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    I think you have misunderstood the term "static" in "electrostatics".

    It means that any "d/dt" term is zero, or that there is no time-variation in the electric field. In terms of the presence of charged particles, it means that (i) these charges do not move or change position (ii) the amount of charge of each particle does not change over time (iii) you are finding the electric field at a fixed field location. This will then allows you to find the coulomb's force at that field location if you place another charge there.

    So yes, there can be a force under static condition. This is not "static" as in "engineering statics", where the NET force on each member is zero.

    Zz.
     
  9. Jan 14, 2018 at 11:32 AM #8
    In fact, I am taking an engineering statics approach. But statics is statics.

    From my understanding of mechanics; d/dt=0 implies a steady state condition, which is not the same as d/ds=0 which implies a static condition, thus no net force. However, the integral of ds is taken over time. So, I suppose this would agree with your argument.

    Once again, I’m led to the notion that Coulombs Law defines how a charged particle impends motion on another. Rather than, explaining a static situation explicitly.

    Additionally, impending motion is covered in engineering statics with respect to dry friction. Which, nonetheless, can be referred to as Coulombs friction.
     
  10. Jan 14, 2018 at 12:19 PM #9

    Dale

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    Coulombs law certainly applies to situations that violate the no external forces assumption. I would not try to restrict it through unnecessary additional assumptions. It is valid in all “electrostatic” situations, which is a different criterion than “static”.
     
  11. Jan 14, 2018 at 12:34 PM #10

    ZapperZ

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    No, it isn't. That's like insisting "theory" is theory, and using the layman's use of it to apply everywhere. Where is it written that this is the ONLY way to use that word?

    If that is the approach you are taking, then this whole discussion is pointless, because you are applying an irrelevant set of rules to something where it wasn't meant to be used. There's nothing written anywhere that "statics is statics"? You made that up, and arbitrarily applying it. Go to your Engineering Statics professors and try to sell your idea to them and see how that goes.

    Besides, what makes you think this application is a one-way street? Why can't the "statics" as used in electrostatics be applied to Engineering Statics? After all, if you think about it, the condition used in E&M is a lot more GENERAL, and encompasses what is used in Engineering Statics. What arbitrary rules did you made up to allow for this one-way application?

    I see this topic as going nowhere fast.

    Zz.
     
  12. Jan 14, 2018 at 12:52 PM #11

    Dale

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    That is an interesting point. I had not considered it before, but that is right. It also incorporates magnetostatics and various equilibrium assumptions
     
  13. Jan 14, 2018 at 1:53 PM #12
    I think I see what your getting at, perhaps I'm misunderstanding the concept of electrostatics. In an attempt to learn, I'm using concepts familiar to me and trying to form a link to enlighten myself elsewhere. I must be misunderstanding the intent of particular examples problems introducing Coulombs Law. I shall continue to study further.

    Although, I believe I have an idea of how Coulomb formulated his Law through his torsion balance experiment, it has been eluding me how it can apply to charges in space. I now see I have been assuming too much, and the example problems I'm using aren't intended to show equilibrium, as such of Engineering Statics.

    Thanks for everyones help:)

    EDIT: Wikipedia defines electrostatics as "charges at rest." This would agree with what I'm trying to define as "static," or in equilibrium.
     
    Last edited: Jan 14, 2018 at 2:26 PM
  14. Jan 14, 2018 at 4:03 PM #13

    Dale

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    Even “charges at rest” would still allow non electric forces.
     
  15. Jan 14, 2018 at 4:09 PM #14
    Yes, I understand equilibrium. But not a net force in equilibrium.
     
  16. Jan 14, 2018 at 4:14 PM #15

    ZapperZ

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    But here again you are confusing "source" charge with "test" charge!

    If the source charge, i.e. the charge that is the source of the electric field, are "static" or not moving, then the electric field are also fixed in space at time. This is EXACTLY what I stated in one of my posts!

    However, you then claim that this means there's no net force, as in "engineering statics", where everything is not only stationary, but also has net zero force. This is not true in electrostatic, because I can put another charge (the "test" charge) in the static electric field and that charge will experience a net force! This is the coulomb force.

    The problem here is that it appears that you have on learned about electric field yet, and how coulomb's law can be obtained by the introduction of a charge into this electric field.

    Zz.
     
  17. Jan 14, 2018 at 4:16 PM #16

    Dale

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    Yes, but the total electric force may be nonzero, which is what Coulomb’s law describes.
     
  18. Jan 14, 2018 at 4:33 PM #17
    I believe to understand all that you have said, this isn’t my first time dealing with electromagnetism.

    If, instead, we drop an actual charge. Would it not either repel/attract from/to the source charge?

    If so, than all I’m claiming is impending motion of the test charge, which describes a net force.

    If not, than I’m utterly confused.

    Thanks:)
     
  19. Jan 14, 2018 at 4:34 PM #18
    This would then give rise to motion, no?

    Thanks:)
     
  20. Jan 14, 2018 at 4:58 PM #19

    Dale

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    No, not if there were other forces.

    This goes back to your extra unnecessary assumption that I pointed out earlier

    It is only a net force under an additional and unnecessary assumption
     
  21. Jan 14, 2018 at 4:59 PM #20

    ZapperZ

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    I have no idea what you said here.

    The "test" charge that I stated is the charge that you put in an electric field, and which you are finding the Coulomb's force for!

    Zz.
     
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