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Homework Help: Ring of Non-uniform charge

  1. Sep 15, 2007 #1
    1. The problem statement, all variables and given/known data
    A ring of radius 'a' has a charge distribution on it that varies as [tex]\lambda (\theta) = \lambda_0 sin(\theta)[/tex], where [tex]\theta[/tex] is the angle between 'a' and the x axis.

    a) What is the direction of the electric field at the center of the ring?

    b) What is the magnitude of the field at the center of the ring?

    2. Relevant equations

    [tex]d\vec{E}= \frac{k dq}{r^2}\hat{r} [/tex]
    [tex]\lambda (\theta) = \lambda_0 sin(\theta)[/tex],

    3. The attempt at a solution

    Okay, here's what I know thus far. For a standard, uniform charge distribution, the electric field is perpendicular to the ring and follows [tex]\frac{kqz}{(z^2+a^2)^{3/2}}\hat{k}[/tex], giving us no electric field at the center of the ring, where z=0. But this is not a uniform charge distribution, so I'm not entirely sure this is going to hold true...infact, because sin has a period of 2pi, I would think that the charge on one half of the ring has an opposite charge than the other. Well, here goes nothing...

    I read through the derivation for the ring of charge that we went over in class and got to this point-

    [tex]E_z = \frac{k cos(\theta)}{r^2}\int dq [/tex] Okay, so we re-write dq as

    [tex]E_z = \frac{k cos(\theta)}{r^2}\int \lambda dl [/tex]

    Now, at this point in the derivation, we pulled lambda out because it was constant. Not so in this case. So lets leave it in there and see what we can do about that dl. Why not rewrite as [tex]dL = r d\theta[/tex]? That would make things easier with the integration. We can also pull the 'r' right out of the integral because it is going to be constant.

    Okay, so now we have...

    [tex]E_z = \frac{k cos(\theta)}{r^2}\int \lambda d\theta[/tex]
    But we know what lambda is, so..

    [tex]E_z = \frac{k cos(\theta)}{r^2}\int \lambda_0 sin(\theta)d \theta[/tex]

    Yank the [tex]\lambda_0[/tex] out of there...Why not put the limits of the integration on there too? Across the entire circle would be from 0 to 2pi, so...

    [tex]E_z = \frac{k cos(\theta)\lambda_0 }{r^2}\int_0^{2pi} sin(\theta)d \theta[/tex]

    Now, when I integrate [tex]sin \theta [/tex], I'm going to get zero. Making everything zero, no matter the position on the z axis. Making me think I made a mistake. I have a sneaking suspicion it involves my rewriting of dL as r dtheta, but to be honest, I am VERY fuzzy on how/why it happens. So hopefully someone can shed some light on this...feel free to give a crash course in integrating in non-rectangular coordinates. I have a very weak grasp on it at the moment.
  2. jcsd
  3. Sep 15, 2007 #2

    Doc Al

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    Several problems here. One, you are asked to find the field smack in the middle of the ring, not somewhere along the z-axis. Two, you should be able to find the direction of the field in the center of the ring just by symmetry--no calculations needed. Three, if you are integrating with respect to theta, you can't just pull cos(theta) out from the integrand and ignore it when integrating. Four, find the proper element of charge and its field:

    [tex]dQ = \lambda dl = \lambda R d\theta = \lambda_0 \sin\theta R d\theta[/tex]

    Find the x and y components of the field from that charge element at the center of the ring and integrate around the ring.
  4. Sep 15, 2007 #3
    Doc, thank you.

    I'll address these things one by one-

    1. I was under the impression that if i designated the center of the ring to be the origin, the point x=y=Z=0 WOULD be the dead center of the ring. So that's where that came from.

    2. I'm a little confused on this. The direction of field by symmetry alone...The only thing I can come up with is that it should be perpendicular to the ring itself, but to be honest, I can't defend that claim at all.

    3) DOH

    4) Can you give me a little more guidence as to what you mean by the x and y components from that element? I'm a little confused there.

    ...am going to go eat and ponder this.
  5. Sep 15, 2007 #4

    Doc Al

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    Yes, the dead center of the ring is at that point. (Not at some off-axis point.)

    Why would there be a perpendicular component at all? The charge and the point of interest all lie in the same plane.

    See my comment above.
  6. Sep 15, 2007 #5
    Okay, so upon further inspection, I had a few elements of this incredibly messed up.

    The theta referenced by "cos(theta)" and by the lambda expression are totally different angles. Cosine is referring to the angle between the z axis and to the circle, basically forming the side of a cone. Obviously the sin expression is referring to the angle between the radius of the ring and the z axis. I should have been much more clear there, my apologies. I will refer to the cosine angle as "phi" from here on and correct my original post. Because of that, it can stay out of the integrand.

    As far as the symmetry goes, this is all I have come up with- sine has a period of 2 pi. From 0-pi, the value is positive. From pi-2pi, the value is negative. We use the convention that the field points from positive to negative, so the field would be pointing in that direction across the center.

    As for the x and y components, the only thing I have come up with is this- I could express the X coordinate as Rsin(theta) and the Y coordinate as Rcos(theta). I don't feel that is what you meant...please bear with me here.
  7. Sep 15, 2007 #6

    Doc Al

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    I see your point about using theta to represent two different angles, but that's not really relevant to this problem. Don't waste time correcting your original post until you realize the following: Assuming that the ring lies in the x-y plane, there is no z-component to consider.

    Tell me exactly what direction the field will point at the center of the ring. (Let theta = 0 be the +x direction.)

    I'm talking about x & y components of the field at the center. Each element of charge contributes some element of electric field at the center. I'm saying to find the x-components of the field contribution and add them up to get the total x-component of field at the center; then do the same for the y-components. (Take advantage of any symmetry, of course. :wink:)
  8. Sep 15, 2007 #7
    If my assumptions about the charge distribution are true, I believe the field will be perpendicular to the x axis throughout. Because of this, there are no components of the field flowing in the x- direction.

    Will be back with an answer for the Y-components, assuming I'm not way off base on that. Initially, I see that the charge would be symmetrical about the Y-axis.
  9. Sep 15, 2007 #8

    Doc Al

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    So far, so good. Keep at it! :approve:
  10. Sep 15, 2007 #9
    Haha, well, I just spent a fair amount of time setting up an integral until I realized...wait, I'm not finding the total electric field...just the magnitude at the center of the ring...whhhooops.

    Now this is every bit as confusing to me. I can find the force between the two. I can find the dipole moment. But the magnitude of the field through the center...hm.

    I know that the charge at the "top" of the ring, where theta= pi/2, is going to be [tex]\lambda_0[/tex] . I know at the "bottom" of the ring, where theta = 3pi/2, the charge is [tex]-\lambda_0[/tex]. I know the distance between the two points is going to be twice the radius of the circle. This seems so simple...gah.
  11. Sep 15, 2007 #10
    ....Would it just be

    [tex]2\frac{k \lambda_0}{a^2} \hat{j}[/tex]?

    I found this by looking at the field from the charge at pi/2 and the field from the charge at 3pi/2 and adding them. Because one is pushing and the other is pulling, the answer is non-zero.
  12. Sep 15, 2007 #11

    Doc Al

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    Sure you're finding the total electric field at the center. You'll need to integrate.

    [tex]\lambda_0[/tex] is a charge per unit length, not a point charge. It's simple, but not that simple.

    Does that expression even have correct units?

    Do as I suggested in post #2. Set up the expression for the field components and integrate.
  13. Sep 15, 2007 #12
    Well as you pointed out, lambda is charge/length and I just treated it like a point charge...so no, it has the wrong units. I believe the phrase I am looking for is "rats."

    Electric field in the x is going to be zero.

    [tex]E_y = \int \frac{k dq}{r^2} sin \theta [/tex], but [tex]E_y = E[/tex], so I can ignore the theta in that expression...espically since it is the angle that the field is pointing at, DIFFERENT than the theta used in my charge distribution expression.

    [tex]E_y = \frac{k \lambda_0}{r^2} \int sin \theta d\theta[/tex]

    Symmetrical across the Y axis, so I'm going to integrate from pi/2 to 3pi/2 and double the value...
    [tex]E_y = \frac{2k \lambda_0}{r^2} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} sin\theta d\theta [/tex]

    Right track?
  14. Sep 16, 2007 #13

    Doc Al

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    Right. By symmetry, the x-component of the total field in the middle will be zero.

    OK. (Except for the sign: Realize that the field points away from the positive charge.)
    No. The only non-zero component of the total field in the middle will be the y-component, but you must add the y-components: [tex]E = \Sigma E_y[/tex]. Without the sin(theta), you are not adding y-components. And that theta is the same theta used in the charge distribution expression!

    Please review what I gave as the charge element in post #2. (And put back that other sin(theta)!)

    Once you get the correct integrand, that will work.
  15. Sep 16, 2007 #14
    Ah yes, I left out the R from the charge expression.

    Including the extra sin expression, I would get

    [tex]E_y = \frac{k R\lambda_0}{r^2} \int sin^2 \theta d\theta[/tex]

    Is the R from the charge expression the same as the radius of the circle? I believe it would be, but they way you wrote it as a capital R is confusing me. If it is the same, obviously that can be simplified, if not, well...then I need to figure out what radius it is referring to.
  16. Sep 16, 2007 #15

    Doc Al

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  17. Sep 16, 2007 #16
    I have a question... Luke1294: have you drawn a diagram?
  18. Sep 16, 2007 #17
    Initally ,yes. When I was going through the E_z steps. I'm not sure why I didn't step back and re-draw it after I figured out I was on the wrong track.

    Does the integrand look correct now?
  19. Sep 16, 2007 #18

    Doc Al

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  20. Sep 16, 2007 #19
    Okay, good. If not I think my head would explode. One last thing I'm not quite understanding... I am looking for the magnitude of the electric field at the center of the circle. Wouldn't this integral give me the field inside the entire ring?
  21. Sep 16, 2007 #20

    Doc Al

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    Why would you think that? We are just adding up the field contributions at the center.

    The field anywhere within the ring would be more difficult to find; it would be a function of position.
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