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The Quadrapole Ring

  1. Jun 15, 2008 #1
    1. The problem statement, all variables and given/known data The linear charge density for a ring of radius R which lies in the xy-plane (center at the origin) is given by:

    [tex]\eta(\phi) = \left\{
    \begin{array}{c l}
    +\lambda & \mbox{if } 0<\phi<\frac{\pi}{2} \\
    - \lambda & \mbox{if } \frac{\pi}{2}<\phi<\pi \\
    +\lambda & \mbox{if } \pi < \phi < \frac{3\pi}{2} \\
    -\lambda & \mbox{if } \frac{3\pi}{2} < \phi < 2\pi
    \end{array}
    \right.[/tex]

    where [tex]\phi[/tex] is the usual azimuthal angle and [tex]\lambda > 0[/tex] is a constant of appropriate dimensions. Discuss the electrostatics of this system, in particular:

    a) Show a figure with the electric field lines in the xy-plane;
    b) Calculate the electric field along the x-axis for both [tex]|x| < R[/tex] and [tex]|x| > R[/tex]
    c) an expression for the far-field; i.e., when [tex]|x| >> R[/tex]

    2. Relevant equations
    Electric field: [tex]E = \frac{kQ}{r^2}[/tex]
    Electric field along the axis of a ring of radius R with uniform charge: [tex] E = \frac{kQx}{(x^2+R^2)^{\frac{3}{2}}}[/tex], where x is the distance along the axis from the center of the ring.

    3. The attempt at a solution
    To tell you the truth, I am not really sure where to start. I drew a picture of the quadrupole, but what is confusing me is that the example we did in class involved the electric field along the axis of the ring. Here, the ring lies in the xy-plane, which is confusing me. The axis in this case would be the z-axis, not the x-axis.

    Nevertheless, I first tried inside the ring along the x-axis. I chose an arbitrary point (not on the origin). Then I wanted to find the distances from each of the segments of the ring. But, that become difficult because I couldn't draw any triangles (the boundary is a circle) like we did with the example in class. I tried the same on the outside of the circle, but I ran into the same problem. That is, I cannot figure out how to draw the shapes needed since the ring lies in the xy-plane.

    What I want to do is first find the equation for the electric field and then use that to draw my electric field lines. If you could provide any help to get me started on finding the equations, that would be great. Thank you.
     
  2. jcsd
  3. Jun 16, 2008 #2

    dynamicsolo

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    Homework Helper

    Alas, this will be of no help here, since we are not looking at the field along the symmetry axis of the ring, but on an axis in the plane of the ring. (This is as you surmised; it is confirmed by the fact that the problem asks for the field for |x| < R and |x| > R . So there is no escape: this problem is a bit nasty, as it is already hard enough to solve when the ring has a single distributed charge...)

    You can simplify the effort a bit by picking a point on the positive x-axis within the ring. (First choice? What is the field at the center of the ring?) For your point, draw a line at any angle to the x-axis that passes through both positive quadrants and your point. "Guesstimate" the field strengths and directions at your point from the bits of charge at the intersections of your line with those quadrants (that is, the field from one quadrant is stronger than the other and the two point in opposite directions). Now draw a line making the same angle to the x-axis, but which goes through the two negative quadrants and your point and make similar estimates of the fields from the charges on those quadrants.

    What do you notice? The x-components of the fields from the positive charges are cancelled by those from the negative charges. So the total field along the x-axis points in the y-direction (in fact, the negative y-direction).

    If you make a similar picture for a point on the negative x-axis within the ring, basically the same thing happens, but you'll find the total electric field now points in the positive y-direction. So there is an antisymmetry in the electric field about the y-axis within the ring. If you now try this out for the total field outside the ring, you'll find that the roles of the fields from the charges on the individual quadrants has changed a bit, but the same cancellation principle applies and the same field antisymmetry is present.

    That cuts the work down a bit, but you're just getting into the thicket. Using analytic geometry, you'll need to work out the distances of the intersection points of one of your lines with, say, the positive quadrants. You'll also need to set up integrals for the appropriate field components from the infinitesimal bits of charge along those quadrants, once you have the distance relationships. (I did say this wasn't easy... It's getting late here, so I'll look more at this later.)

    As for the last part, (c), we can guess what the proportionality ought to look like, so we will be able to check the result from part (b) against it once we have that worked out (how much time do you have to finish this?). The far-field behavior for a monopole (point charge) is proportional to 1/(r^2) and that for an electric dipole is proportional to 1/(r^3). Guess what it is for a quadrupole? (The m-poles go by factors of 2: the next one is an octupole...)

    You really don't want to do this that way. The problem is already tough to work out when you are looking at points on the x-axis. When you choose general points on the plane, you sacrifice any symmetry in working out the field and really complicate matters.

    Start with figuring out how the field from each quadrant sector behaves at points along the center, along the x-axis, and along the y-axis. You will at least get component cancellations in those locations.

    If you know consider a general point inside or outside the ring, judge roughly how the field from the center of each ring quadrant contributes to the total field. (They just want a qualitative map, not a detail vector computation -- people use computers for that...) Look at what happens for the other three points symmetric about the axes to your chosen point; you will find that you can estimate the field map in one quadrant of the plane and flip the total field vector appropriately in the other quadrants. So pick several points in the one quadrant and make the total field estimate for each; after maybe a few, you'll get a feel for how the electric field behaves in one quadrant, and thus all quadrants, of the xy-plane.
     
    Last edited: Jun 16, 2008
  4. Jun 17, 2008 #3

    dynamicsolo

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    I've had a chance to finish working this through, so I can tell you a bit more about setting this up.

    Pick a point on the x-axis and look at the triangle connecting the center of the ring, your point, and an infinitesimal bit of charge dq on the ring. You will need the Law of Cosines and Law of Sines to solve for useful angles. If you look at the corresponding bit of charge -dq for the point on the ring which is symmetrically-placed about the x-axis, you will find that the infinitesimal field from it cancels the x-component of the field from the charge dq. So you will only need to find the y-component of the field from dq and double it. (This also means you will only need to integrate over half of the ring.) The triangle solution you work out will give you the needed angle for finding the y-component. You will want to base the integration on the azimuthal angle phi around the ring measured from the origin.

    It sounds like setting up the integral will be terrible, but it turns out that when you put together the field magnitude from dq at your chosen point and the factor for the y-component, you will have an integral that can be carried out by simple substitution (not even trig. substitution)! Moreover, the geometry for setting up this integral works for all charges along the ring and for finding the field both inside and outside the ring. The one break in the integration limits is at phi = pi/2 , where the sign of the charge switches; otherwise, it's the same integrand everywhere. (I find that you should end up with three terms.)

    The field function along the x-axis will have the correct behavior at the origin (E = 0 at x = 0) and at infinity, and blows up at x = R as it should. In the negative x-direction, the field direction is simply flipped, as discussed above. Because of the (anti)symmetry of the charge distribution, it also turns out that you have now found the field magnitude along the y-axis as well, with the directions being (-i) on the positive x-axis, (-j) on the positive y-axis, etc.

    Part (c) is a festival of approximation technique (Maclaurin series down to x^2 terms will suffice) and does indeed show that the far-field behavior of the field along the principal axes falls off as 1/(x^4) [the field is miniscule for |x| > 5R].
     
    Last edited: Jun 17, 2008
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