Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Circular wire prob

  1. Mar 2, 2005 #1
    A uniformly charged infinitely thin circular ring of radius a has total charge Q. Place the ring in the xy-plane with its center at the origin. Use cylindrical coordinates...

    Find the potential at a point in the xy-plane a distance 2a from the orgin give your answer as a number times kQ/a
    I figured the potential to be for some point in the xy-plane,[tex] (2a, \Phi , 0)[/tex]
    [tex]k_e j a \int_0 ^{2 \pi} \frac{d \theta}{\sqrt{(2a)^2+a^2-2(2a)a cos(\Phi - \theta)}}[/tex]
    [tex]= k_e j a \int_0 ^{2 \pi} \frac{d \theta}{\sqrt{4a^2+a^2-4a^2 cos(\Phi - \theta)}}[/tex]
    where, [tex] j = \frac{Q}{2 \pi a}[/tex]
    [tex]= k_e j \int_0 ^{2 \pi} \frac{d \theta}{\sqrt{5-4cos(\Phi - \theta)}}[/tex]
    did I set up this intergral right? how do I evaluate it? hints answers it all helps
    Last edited: Mar 2, 2005
  2. jcsd
  3. Mar 3, 2005 #2
    Did I at least set it up correctly? I don't KNOW what to do! i know because the ring is symetric you can drop the phi...

    [tex]= k_e j \int_0 ^{2 \pi} \frac{d \theta}{\sqrt{5-4cos(\theta)}}[/tex]
  4. Mar 3, 2005 #3
    You want the potential in the same plane as the ring? Btw, you're forgetting a negative.

    And Mathematica likes that integral about as much as I do.

    Last edited: Mar 3, 2005
  5. Mar 3, 2005 #4
    ok check this out let me know if this a valid move...

    [tex]r = \sqrt{5-4cos(\theta)} [/tex] which is polar so...
    polar ints have the form
    [tex]\frac{1}{2} \int r^2 d \theta [/tex]


    [tex] k_e j \frac{1}{2} \int_0 ^{2 \pi} \frac{1}{5-4cos(\theta)} d \theta[/tex]
    thus making [tex] = \frac{k_e Q}{a} \frac{1}{6} [/tex]
    let me know what you think...
  6. Mar 3, 2005 #5
    anyone? come on this can't be that hard
  7. Mar 3, 2005 #6
    You can't use r2 because it's not an area integral. You've already accounted for curvature when you say [itex]ds = r d\theta[/itex].

    I don't see a problem with your setup, but your integral isn't easily computed. Mathematica says that it's

    [tex]\frac{2Q \mbox{EllipticK} [-8]}{a \pi}[/tex]

    Then again, my brain isn't in such great shape at the moment, so I could very well be wrong.

  8. Mar 4, 2005 #7
    can i get an actual value for EllipticK(-8)
    [tex] \int_0^{2\pi}\sqrt{1+8sin^2(\theta)}d\theta [/tex]
  9. Mar 4, 2005 #8
    [tex]\frac{2Q \mbox{EllipticK} [-8]}{a \pi} = \frac{0.536591 Q}{a}[/tex]

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook