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Potential on the rim of a uniformly charged disk

  1. Aug 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the potential on the rim of a uniformly charged disk (radius ##R##, charge density ##\sigma##). [Hint: First show that ##V=k(\sigma R/\pi\epsilon_0)##, for some dimensionless number ##k##, which you can express as an integral. Then evaluate ##k## analytically, if you can, or by computer.]


    2. Relevant equations
    [tex]V = \frac{1}{4\pi\epsilon_0} \int \frac{\sigma da}{|\vec{r}-\vec{r'}|}[/tex]
    where ##\vec{r}## is the position vector to the point at which the potential is being calculated and ##\vec{r'}## is the position vector of the charge.

    3. The attempt at a solution
    Well, my first instinct was to jump in and do it brute-force with the above integral but I got stuck with a complicated integral with which even the computer struggled. After that I tried something simple - choose a convenient point on the rim rather than use a general one. Here I got a 0 for ##(R,0)## which doesn't seem right. Now, I didn't use the hint the problem gave me - I can't figure out from where am I supposed to get the ##R## in that equation (the other variables already appear in the formula above). I guess I could try to look for mistakes in the brute-force calculation but I am interested in a more elegant way of doing it which the hint seems to suggest.
    Any thoughts on how to use the provided hint to solve this problem?

    Thanks in advance!
     
  2. jcsd
  3. Aug 27, 2014 #2

    Zondrina

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    Hmm the way I see it (assuming that 'rim' means not above the surface) you need to find some way to relate the distance ##r## between the point ##P## and ##dq##.

    $$V = \int dV = k \int \frac{dq}{r} = k \int \frac{\sigma dA}{r}$$

    The question seems a bit odd though. The potential at any point on or inside the disk should be the same, so I don't see why you couldn't just use the potential of a charged disk:

    $$V(z) = \frac{\sigma}{2 \epsilon_0} (\sqrt{z^2 + R^2} - |z|)$$

    With ##z = 0##.
     
  4. Aug 27, 2014 #3
    Thank you for the reply. Firstly, I don't see why the potential on the disk is the same at every point. Can you explain how to prove it? Concerning the equation, it's the equation of the potential at a distance ##z## above the center of the disk, while the problem asks for the potential on the rim. Then again, if the potential is indeed the same at every point substituting ##z=0## in your equation should yield the correct answer. So, why would the potential be constant over the disk?
     
  5. Aug 27, 2014 #4

    Zondrina

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    Ah yes I forgot that only applies to a uniform spherical conductor. The surface charge will be the same on both sides of the disk, but on the outer rim of the disk it could differ.

    Though they mention the disk is uniformly charged, I'm not sure the same logic applies.

    The question is implying to find the potential at a point on the upper rim of the disk or the lower rim. These points could lie on the border of the top or bottom surface.
     
  6. Aug 27, 2014 #5

    Zondrina

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    Okay I think I have all the geometry for this figured out it was much more complicated than I initially anticipated. To find the potential on the rim:

    $$V = \int dV = k \int \frac{dq}{r} = k \int \frac{\sigma dA}{r} = \sigma k \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{2R \cos(\theta)} \frac{r}{r} dr d\theta$$
     
  7. Aug 27, 2014 #6
    Thank you! Could you elaborate on the limits in the last integral please? In particular the limits on ##\theta##.
     
  8. Aug 27, 2014 #7

    Zondrina

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    $$\sigma k \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{2R \cos(\theta)} dr d\theta = 2R \sigma k \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \cos(\theta) d\theta = 4R \sigma k$$

    With ##k = \frac{1}{4 \pi \epsilon_0}##:

    $$4R \sigma k = \frac{R \sigma}{\pi \epsilon_0}$$

    which is what we were trying to find. The limits allow the coverage of the whole surface of the disk for an arbitrary wedge ##W## of length ##R## and width ##d \theta##.
     
  9. Aug 27, 2014 #8
    Thank you, I get it now. Nice geometric trick.
     
  10. Aug 27, 2014 #9

    Zondrina

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    ImageUploadedByPhysics Forums1409168240.110459.jpg

    I'll upload my doodle to help you see whats going on a bit more. You could have chosen any point on the perpendicular line through the center.
     
  11. Aug 27, 2014 #10
    Thank you again. My confusion with the limits of ##\theta## was that I thought that for an arbitrary point on the rim it will not be from ##-\pi/2## to ##\pi/2## but now I see that ##\theta## is measured from the diameter and the latter can be rotated to fit any arbitrary point on the rim.
     
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