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Homework Help: Determining direction and magnitude of electric field in a ring.

  1. May 31, 2010 #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]. 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?

    The textbook includes a picture of a ring/circle with radius a, angle [tex]\theta[/tex], and charge distribution [tex]\lambda[/tex]. It's pretty basic and straightforward.

    2. Relevant equations

    Only one that I can think of:
    [tex]E_{z} = \frac{kQz}{(z^{2} + a^{2})^{3/2}}[/tex]


    3. The attempt at a solution

    I really don't know where to start. I mean I don't have any value for anything (other than if I were to just manipulate my given function and put it in the equation in "Relevant equations"), and z isn't even given in my function. I'm not understanding the picture (sorry I couldn't post the picture on here); it shows basically a circle with its radius forming an angle between the radius and the x axis. I really can't relate that to how this question is worded.

    Since I haven't really attempted anything, I'm not looking for someone to do my homework, but rather, for some hints and suggestions.

    Thank you all in advance!
     
  2. jcsd
  3. May 31, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Some hints. On your diagram, sketch the charge at various points along the circle. You should be able to answer (a) without any calculation. Use symmetry.

    Instead of trying to find a canned formula, you'll have to come up with your own. Consider a small element of the charged ring. What field does it produce at the center? Again, use symmetry.
     
  4. May 31, 2010 #3
    Alright, well after some thought, here's what I've got.

    [tex]E = \int dE

    = \int \frac{kdq}{a^{2}}
    = \frac{k}{a^{2}}\int dq[/tex]

    after a little manipulation and substituting:

    [tex]\frac{k}{a^{2}}\int a\lambda^{2}d\theta
    = \frac{k}{a}\int \lambda^{2}_{0}sin^{2}\theta d\theta[/tex]

    [tex]\frac{k}{a^{2}}\int a\lambda^{2}d\theta
    = \frac{k}{a}\int \lambda^{2}_{0}sin^{2}\theta d\theta[/tex]

    bring lambda-not out of the integrand, and we get:

    [tex]= \frac{k \lambda^{2}_{0}}{a}\int sin^{2}\theta d\theta[/tex]

    but after I noticed another thread on this same question here at PF, I noticed the answer shouldn't have theta-not squared, it should just be theta-not...what went wrong there? Other than that, are my steps correct and the resulting formula for E correct?
     
  5. May 31, 2010 #4
    or, i think I might have put in the squared by mistake...instead would it just be:

    [tex]
    = \frac{k \lambda^_{0}}{a}\int sin\theta d\theta
    [/tex]
     
  6. May 31, 2010 #5

    Doc Al

    User Avatar

    Staff: Mentor

    What did you get for (a)?
     
  7. May 31, 2010 #6
    well, the field will point outwards, from the ring, and thus in the middle, there is zero field. So the direction points outwards from the ring.
     
  8. May 31, 2010 #7

    Doc Al

    User Avatar

    Staff: Mentor

    I don't understand how you arrived at this conclusion. Take several small segments of the ring and see which way their field points at the center of the ring. (Getting part a straight will help with part b.)
     
  9. May 31, 2010 #8
    I think the good Doc wants you to look at the symmetry between the various E-field vectors produced by various charges around the circle. Symmetry can help you cancel some stuff and make the problem easier.
     
  10. May 31, 2010 #9
    Argh...I'm so confused...

    I mean it just seems like on any given point on the ring, the point opposite of it would have the same magnitude. Like charges repel, so the fields would all point outwards from the ring. That's what I thought...

    I think i'm gonna read my textbook and go over this stuff again. After that, I'll come back on here and go from there.
     
  11. May 31, 2010 #10
    The same magnitude of E-field contribution, Yes it would and would also be pointed in the opposite direction...

    And you are not looking at the Force of one charge on a charge opposite of it in the circle. You are looking at the contribution of a bunch of point charges that are arranged in circle to an E-field at a certain point in space. And that point in space happens to be right in the middle of the ring of charge.
     
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