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Potential Theory Problem

  1. Apr 4, 2008 #1
    Given n (finite) point charges in the xy-plane, is it possible to have a curve (in the plane) along which the electrostatic force vanishes (F=0)?

    I know that it's possible to have a curve through space along which the force vanishes when all of the charges are in the plane. For instance, place an even number of point charges of equal magnitude but alternating sign equidistantly on a circle. Then the line orthogonal to the plane of the charges passing through the center of the circle will have net force of 0 at every point.

    Is it possible to generate such a curve in the plane, when all of the points are located in the plane?

    I have already proven that if such a curve were to exist, the potential along this curve must also be zero (I know this result is true if all of the charges have the same sign...I think it's true in every case). Also, I know that such a curve cannot exist if all of the charges have the same sign. I just don't know if this result extends to any configuration of positive and negative charges.
    Last edited: Apr 4, 2008
  2. jcsd
  3. Apr 4, 2008 #2
    I think it is possible to make curves of zero potential in the plane. First let me take the case of 4 charges, 2 +ve, 2 -ve having the same magnitude. Now place them in a square formation with the center of the square lying at the origin of a 2-dimensional Cartesian system, with charges of same sign lying diagonally across. The potential along the x- and y-axes would be zero.

    The general case is a bit more complicated; let us assume that the x-axis is an equipotential and so is a line lying at a certain angle R to it. However there is a condition we put on R viz. that it should be a divisor of 180 deg. Now we place one of our positive charges between the x-axis and the line inclined at R and assume that both these lines are parts of 2 semi-infinite grounded (potential=0) planes, which are going through our 2-dim coordinate system.

    Then we use the method of image charges to get the image charge configuration that will mantain zero potentials at these 2 lines. Next we replace the image charges by real ones giving us the 2 lines as equipotentials. Basically I have turned the method of image charges upside-down yielding the positions of real charges. I am attaching an image (no pun intended) for the case with 8-charges (line at 45 is equipotential). But I am not sure that the force along these lines would be zero.

    Hope this helps..

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    • octa.bmp
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    Last edited: Apr 4, 2008
  4. Apr 4, 2008 #3
    First, I should clarify something from my original post: if there is a curve that the force vanishes on, and it extends through the whole plane, then the potential is identically zero on that curve. If the curve does not extend through the whole plane (from - oo to oo), then along that curve we can only say that the potential is a constant.

    I've attached a 3dplot (generated in Maple) of the potential field generated by the configuration described above. For computational purposes, I placed the +1 charges at (-1,1) and (1,-1), and the -1 charges at (1,1) and (-1,-1). Keep in mind that the force generated by this potential is given by the gradient of the field. Clearly, this shows that the x and y axes are not lines along which the force vanishes. A simple check of this algebraically shows this to be true (pick any point on either axis and calculate the force there). The only (obvious) place that the force vanishes is at the origin...there may be others, but not infinitely many along a curve. This example shows that (in R^3) the z-axis is a line upon which the entire force vanishes, and is a specific case of the example I mentioned in my original post.

    If the potential is actually a constant along a curve, then the electrostatic force is zero on that same curve (the force is the gradient of the potential: constant potential -> no slope -> zero gradient ->zero force). However, your example assumes that equipotential lines can exist in the plane. My questions is: is there a proof that equipotential lines can exist in the plane when all of the point charges generating the potential are in the plane?

    Attached Files:

  5. Apr 5, 2008 #4
    I have two points to clarify :-

    1. This assumption (equipotentials with V=0 exist in the plane) was just made to get the right configuration of charges, one can check mathematically that given the particular configuration of charges (arrived at by using the method I described) the potential will be zero along the "assumed" equipotential. So given n charges all having the same magnitude of charge (half of which are +ve and half -ve), I think that we can always create a configuration which will have curves with zero potentials. I do not have any proof for the case of a given arbitrary configuration of charges, it is not even obvious whether there will be a curve with zero potential in R^3.

    2. Unfortunately, I did not pay much attention to the force along the curves with zero potential, although I did mention I was not convinced. It is obvious in the case of 4 charges that force is not zero along the x- and y- axes which have zero potential. Now let us examine your proof. It says : "If the potential is actually a constant along a curve, then the electrostatic force is zero on that same curve (the force is the gradient of the potential: constant potential -> no slope -> zero gradient ->zero force)." Consider the case of 2 equal and opposite charges : the line which bisects the line segment joining the 2 charges is an equipotential with V=0, it extends from -oo to +oo and yet the force along it is not zero, as it should have been using your logic

    The problem lies with the interpretation of gradient as the change in potential along a curve. The rate of change in potential along a curve is given by the scalar product of the gradient and a unit tangent along the curve, and not by the gradient itself. This means that potential may be constant along a path (equipotentials) and yet the gradient of potential may not be zero (happens when gradient vector is always normal to the curve). Therefore although we can show the existence of curves of zero potential in a plane given n charges which can be arranged according to our liking, these curves (even if they extend from -oo to +oo) do not necessarily have zero field along them.
  6. Apr 5, 2008 #5
    As I said in my last post, the potential does not have to be zero if the curve does not extend through the whole plane. However, the existence of an equipotential curve is not the ultimate goal of this problem: it is to find a curve along which the force vanishes. Also, the example I gave in my original post generates a curve in R^3 with both zero potential along a curve and zero force on that curve (namely, the z-axis)

    You are correct in that I oversimplified the problem. I had in mind that constant potential was both a necessary and sufficient condition to have zero force, but instead, as these examples show, it is only a necessary condition. Zero force -> constant potential, but not the other way around.

    This is correct. Essentially, what must be found is a saddle curve in the potential. All of the examples given fail to to this in R^2. This is easy to do in R^3. Can it be done in R^2?
  7. Apr 5, 2008 #6
    thanks for the interesting question. it seems to me, we can think about this in an analogy: we are given a planar land and two point-wise tools. One tool patches up sand at one point and extends the slope to infinity, another tool scoops out sand and does the reverse. Now we are asked whether we can use the tools to dig-and-patch the land and eventually get a patch with no slope in all direction.

    IMO, it is not possible if (1) the tool is quantized to integer amount of sand only or (2) a line with charge is not allowed, i.e. only allow pointwise operation. This looks like lemma to me.:rolleyes:. One exception is, you put 1+ and 1- at exact same location and the whole plane becomes zero potential.

    If real charge is allowed, then its like we can always tune the amount of sand to obtain a slopeless patch. If a line patch is allowed, then any parallel-line patches can contain 1 line with zero slope in all directions. :tongue: But i cannot think of why in 3D it is possible. The example you have pointed out shows that it is easy in 3D. Perhaps it has something to do with geometry and dimension?
    Last edited: Apr 5, 2008
  8. Apr 5, 2008 #7
    This part I can answer, although I'll have to study the rest of your response in greater detail later.

    Everything works in R^3 because the potential is a harmonic function (regardless of what combination of positive or negative charges). In the example of equal but alternating even number of charges placed equidistantly on a circle centered at the origin, the z-axis is, although it is impossible to picture, a saddle curve of the potential.

    By the harmonic properties of the potential function, we get that the potential cannot have a local min or max, and therefore any place that the force vanishes must be a saddle point of the potential...or a saddle curve. This property is not violated in this example. The limit as x or y goes to oo of the potential is 0. If there existed a curve along which the force was zero, then clearly the potential must be constant along such a curve. Take an (x,y) far enough away from the charges, but sufficiently close to the curve. By the limiting argument and the harmonic properties, the potential must be zero on that curve, or the curve would be a local max or min, which would imply that the potential must be a constant function.

    This argument only works if the curve extends infinitely through the space.

    In the case of R^2, we lose all harmonic properties unless all of the charges are the same sign. If they are all positive, then the potential is sub-harmonic; all negative then super-harmonic. The same argument above applies to R^2 if the curve extends through the entire plane and all of the charges have the same sign. Otherwise, we lose all harmonic properties and are stuck with the simple fact that the potential must be a constant, but we can't prove that it's identically zero (unless the curve extends to infinity...I think an argument could still be made then that the potential along the curve must be zero).
  9. Apr 7, 2008 #8
    thanks for the explaination. i have not notice about the saddle point thingy, which is very much related to the maximum principle of harmonic functions. After some wikis i got the idea , but it seems that the information in wiki can help much with your original problem. perhaps the assumptions behind the problem need to be clarified first...
  10. Apr 7, 2008 #9
    The only assumption in this problem is that I have n point charges somewhere in the xy-plane. The point charges can be any combination of positive or negative, and there is no restriction on their magnitude (obviously cannot equal 0....). The k-th point charge (q_k) is located at (x_k,y_k)

    Define P=\sum_{k=1}^n{\frac{q_k}{\sqrt{(x-x_k)^2+(y-y_k)^2}} as the potential function (which is straight from the physics textbook). Then for the force we have:

    F=-grad{P}=(F_x)\hat{i} + (F_y)\hat{j}, which in terms of components is

    F_x=\sum_{k=1}^n{\frac{q_k(x-x_k)}{({(x-x_k)^2+(y-y_k)^2})^{3/2}} and

    To have the force equal 0 at some point (x,y), then we must have F_x=0 and F_y=0. To have F=0 along a curve, then we must have F_x=0 and F_y=0 for all (x,y) on that curve, which also implies that P=constant on that same curve.

    So the question is, are there infinitely many solutions (i.e. does a curve exist) to the system


    If so, then one of the implications of this (from Bezout's theorem) is that F_x and F_y have a common factor. You'll notice that termwise they are different, but this does not prove they have no common factor (unless an algebraist can establish this for me).

    Another result, obtained through implicit differentiation, is that:

    \frac{1}{2}\sum_{k=1}^n{\frac{q_k(x-x_k)^2}{[(x-x_k)^2+(y-y_k)^2]^{5/2}}}\leq \sum_{k=1}^n{\frac{q_k(y-y_k)^2}{[(x-x_k)^2+(y-y_k)^2]^{5/2}}} \leq 2\sum_{k=1}^n{\frac{q_k(x-x_k)^2}{[(x-x_k)^2+(y-y_k)^2]^{5/2}}}.

    Although this gives an interesting upper and lower bound on the sum of the y-coordinates in terms of a sum of the x-coordinates, I'm not sure if it helps. However, this is a necessary condition to have F=0 on a curve.
    Last edited: Apr 7, 2008
  11. Apr 14, 2008 #10
    I've attached my previous post as a pdf for those who don't have LaTeX.

    I've been shown that it is impossible to have the potential become constant on a curve of finite length (this follows from a result of Puiseaux), unless the curve is a loop. However, a loop will violate the harmonic properties, so any curve going through the plane must have infinite length.

    A physics prof told me that Earnshaw's theorem could be applied here to show that such a curve cannot exist. I don't see the connection though. From my understanding, Earnshaw’s theorem simply says that harmonic functions cannot have local max’s or mins, which we already know. In R^2, this doesn't apply. Perhaps someone can clarify.

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  12. Apr 14, 2008 #11


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