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Is it able to make point charges stationary by electric force?

  1. Dec 11, 2014 #1
    First by "stationary" I mean "with respect to each other" so as to avoid introducing relativity problems.

    I'm wondering whether there's a way to prove or disprove that given [itex]N > 1[/itex] point charges [itex]q_1, q_2, ..., q_n[/itex], is there always a way of putting them in 3-dimensional space such that all of them remain stationary after release. Assume that when putting a new single point charge to the desired position I use mechanical force to fix the old ones at where they are.

    What I've tried:

    For [itex]N = 1[/itex] obviously I can do this but for [itex]N = 2[/itex] I can't. I first came up with the idea to investigate the case of [itex]N[/itex] charges based on the case of [itex]N-k \, (0 < k < N)[/itex] charges. However I didn't think of anything valuable in this way, it's not clear to me how the equilibrium of [itex]N-k[/itex] charges could be related to the equilibrium of [itex]N[/itex] charges.

    Any help will be appreciated :)
     
  2. jcsd
  3. Dec 11, 2014 #2
    You can't cancel out forces completely in this way. Same holds for point masses and their gravitational fields.
     
  4. Dec 11, 2014 #3

    A.T.

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  5. Dec 11, 2014 #4
    Sure. Force must be applied to maintain equillibrium state (Earnshaw's tm implication). If not applied, field in space created by free charges will drive system out of equillibrium
     
  6. Dec 11, 2014 #5

    A.T.

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    Earnshaw's theorem says that no stable stationary equilibrium is possible. It doesn't rule out an unstable stationary equilibrium, where all the E-forces still cancel, but do not restore the equilibrium when it is perturbed.
     
  7. Dec 11, 2014 #6
    Obviously OP isn't interested in unstable equillibrium case.
     
  8. Dec 11, 2014 #7

    A.T.

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    That isn't obvious to me. But I was actually responding to your claim:
    As I understand the Earnshaw's theorem, it doesn't rule out that the E-forces cancel out on all point charges, in an unstable or even neutral static equilibrium.
     
  9. Dec 11, 2014 #8
    It is enough that E-force doesn't cancel out on one point charge to have instability in system.
     
  10. Dec 11, 2014 #9

    A.T.

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    But Earnshaw's theorem allows static configurations where the E-forces cancel out on all point charges.
     
  11. Dec 11, 2014 #10
    "Earnshaw's theorem states that a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges."

    I really don't know what is not clear here.
     
  12. Dec 11, 2014 #11

    A.T.

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    That quote is perfectly clear, but it doesn't support your claim:
    Take two point charges of 4C and place a third point charge of -1C midway between them. The forces on all charges cancel.
     
  13. Dec 11, 2014 #12
    Then, I'm sure you'll be able to explain OP how to stabilize system of any given electrical charges by introducing new charges (at different positions):


    Good luck ;)
     
  14. Dec 11, 2014 #13
    Thank you so much for the answers, in fact I didn't know "Earnshaw's theorem" before viewing your answers and I spent some time learning it :(

    To be more clear I was NOT asking for stable stationary equilibrium, what I'm curious about is whether I can have -- in your terms -- an unstable stationary equilibrium in which all electric forces on every single point charge cancelled out. I DIDN'T expect the configuration to be interfered, thus I DIDN'T expect the charges to resume equilibrium after interference.

    I'm sorry for not making the term "given charges" clear enough. Actually I assumed that the values of charges are not specified, only the number N (N > 1) is specified and you're free to pick different values of charges and put them to an unstable stationary equilibria if possible.

    So let me describe my problem again in this way: is there a non-empty set [itex]S[/itex] of natural numbers satisfying [itex]S=\{n \, | \, \text{n is a natual number & you can pick n charges to form an unstable stationary equilibrium.} \}[/itex]. If there is, how to find all elements of [itex]S[/itex]? As far as I know, [itex]2[/itex] is not an element of [itex]S[/itex], [itex]1[/itex] and [itex]3[/itex] are in. I'm trying [itex]4[/itex] but obviously it's impossible to try all natural number :L
     
  15. Dec 12, 2014 #14

    Nathanael

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    For N=4 you could place 3 identical charges on the vertices of an equilateral triangle with a fourth opposite charge in the center (I didn't bother calculating the magnitude of this central charge but it wouldn't be the same as the vertex-charges).

    This should also work for N=5 using a "3D equilateral triangular pyramid" or whatever it is called... I was going to explain why this should work, but I stopped because words are quite clumsy with geometric ideas. Hopefully with a little thought you will either agree or show me to be wrong.

    Perhaps there are some geometric generalizations you can make.

    I don't have much to contribute to the solution, I just want to encourage geometric approaches to the problem.
    P.S. Thanks for the interesting problem :oldsmile:
     
    Last edited: Dec 12, 2014
  16. Dec 12, 2014 #15

    Nathanael

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    It would also work with two equilateral triangles of equal length and a shared center (which are relatively rotated by 180 degrees) by putting identical charges on the vertices and one oppositely charged particle in the shared center (N=7).

    Similarly it should work for two of the "3D equilateral triangular pyramid" if you rotate one of them by 180 degrees about the shared center in the direction such that there is no overlapping charges. (N=9)
     
    Last edited: Dec 12, 2014
  17. Dec 12, 2014 #16

    Nathanael

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    I edited my last two posts a few times because this problem has made me go a little crazy o0) (I thought I was wrong then I realized I wasn't, but anyway....)

    I believe that the solution is all natural numbers except N=2... And the funny thing is (if I'm not wrong) you only need 2 dimensions.

    As far as I can tell, every regular (equilateral) polygon with identical charges placed on the vertices and an opposite charge in the center would be a solution (you would have to find the right value for the central charge depending on the number of sides of the polygon, but that's another problem).
    This, along with A.T.'s solution for N=3 (and the obvious "solution" for N=1) gives you a solution for all N≠2 (assuming I'm not wrong, but I leave that for you to prove).

    Now how can we put that third dimension to good use... o0)
     
    Last edited: Dec 12, 2014
  18. Dec 13, 2014 #17
    Hey @Nathanael you've come to a nice solution! I'm a little late to check this thread so I missed the good moment right when you posted it :w (bcz in weekdays I'm busy being a software engineer)

    I think you're totally right about the 2D solution and that's great! So sad that I was not aware of such simple fact :L

    Now talking about the use of the 3rd dimension, I'm not sure if there's anything interesting related to it. Since there're only 5 platonic solids (4, 6, 8, 12, 20) I guess the trick of "perfect" symmetry can only be applied to them. Thus if the purely 2D configurations are precluded, then the best I can achieve is to put N = {5, 7, 9, 13, 21}in.

    By far my original problem is solved and thank you again @Nathanael for the solution!

    Still expecting non-trivial solutions in 3D ;)
     
  19. Dec 13, 2014 #18

    Nathanael

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    All the solutions I've been thinking about have a number of identical charges (which I call the "vertex charges") with a single opposite charge in the middle (which I call the "central charge").

    For there to be a magnitude of the central charge such that all charges are in equilibrium, I see only two requirements:
    (Please excuse my imprecise wording :redface:)

    1.) The system is symmetrical from the point of view of all vertex charges.
    (Perhaps there's a better way to say this? Maybe, "The perspective from any vertex charge must be identical to that of all the other vertex charges.")
    (Maybe I could say, "the relative positions of the charges must be independent of which vertex charge's frame you are viewing from.")

    2.) The net force on any single vertex charge from all other vertex charges must be in line with the central charge.
    (Let me try to say it another way: "If you take the central charge as the origin, the net force on any vertex charge (excluding the force from the central charge) must point in the direction of the radius vector of that charge.")

    I have a feeling that the second requirement is a necessary consequence of the first requirement, (and thus redundant) but I'm uncertain.



    All of the platonic solids seem to satisfy these conditions, but I believe you can overlap and rotate them to get more configurations. For example, if you take two tetrahedrons of equal size with a shared center and rotate one of them by 180 degrees (but not in the wrong direction) then the conditions are also apparently satisfied. (Unfortunately, this gives you N=9, which you could simply achieve with a cube.)

    Actually,
    now that I think about it, I think the only one you can do this with is the tetrahedron. This is because it is the only platonic solid with the same number of faces as edges (and thus rotating it like I described gives you one vertex popping out of the center of each face, and the required symmetry is maintained).

    I'm thinking perhaps you can overlap an octahedron with a cube to get another solution. My thinking is that you would have to choose the right sized octahedron so that all of the vertices (of both shapes) lie on the surface of a sphere. (In other words, all the vertices would have to be equidistant from the shared center.) I'm slightly confident this will work. (This gives you N=15, which you can't achieve with any single platonic solid.)
    [[[P.S. I'm not quite sure this satisfies my first requirement, but it might]]]

    By logical analogy, perhaps you could overlap an icosahedron with a dodecahedron (with a shared center) so that all vertices lie on the surface of a sphere (which has the same center as the two shapes). The reason I say, "by logical analogy" is because it's now beyond my ability to visualize. (The analogy is that the dodecahedron has as many vertices as the faces on an icosahedron.)

    Perhaps there are more complex combinations you could make, perhaps even some generalized way of combining them.

    Do you think there are any 3D solutions that aren't built from platonic solids? Platonic solids have the required symmetry, but perhaps there are other shapes which will also work?

    It is now past 2 a.m. and I'm visualizing overlapping rotated cubes... o_O (Which, contrary to my initial thoughts, don't work.) I think I'll go to bed now... (and probably dream about crazy shapes o_O hahah)
     
    Last edited: Dec 13, 2014
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