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Homework Help: Charge Density of a Conducting Disc

  1. Jul 7, 2006 #1
    Oh since this got moved: It's not for any course or class so I'm not following any book.

    I'm not sure if anyone will be able to answer this but I'll ask anyway:

    I've solved laplaces equation in cyclindrical coordinates for a disc of radius 'a' and constant potential V on the disk (disc in z=0 plane, centered on z axis) and got this as the potential everywhere

    [tex]\phi(r,z) = \frac{2V}{\pi} \arcsin{\frac{2 a}{\sqrt{z^2 + (r + a)^2} + \sqrt{z^2 + (r-a)^2}}}[/tex]

    I want to now find the charge density on the disc. From Gauss and the fact that at the surface the electric field will be perpendicular to the surface we have [tex]E_{z} = 4 \pi \sigma[/tex] where sigma is the charge density.

    So I differentiate Phi w.r.t z to get the z component of the electric field and take the limit as z->0. I get zero as my answer.

    What am I doing wrong? I can only get this to work in oblate spheroidal coordinates. They are very convenient for this problem but not so when I try to generalize to more discs of different radii and potential.
     
    Last edited: Jul 7, 2006
  2. jcsd
  3. Jul 7, 2006 #2
    Hi BP,

    Calculating Ez for r=0 could help you pinpoint the problem.
    It would also be useful to plot a numerical example.

    Michel
     
    Last edited: Jul 7, 2006
  4. Jul 7, 2006 #3
    Dear Michel,

    Thanks for the prompt reply.

    I'm still thinking about it
     
    Last edited: Jul 7, 2006
  5. Jul 7, 2006 #4

    nrqed

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    I am sure this is a really stupid question and I am missing something obvious, but if one sets z=0 in your solution, one gets a function of r. But you stated that the potential is constant on the disk. I am sure I am missing something obvious! (is your solution for r>a??)
     
  6. Jul 7, 2006 #5
    Do a taylor expansion of what is left inside the arcsin for r<a and you will end up with 1. arcsin[1]=Pi/2
     
  7. Jul 7, 2006 #6

    nrqed

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    Oh, Ok I see. Actually, that was obvious even without a Taylor expansion since, for r<a, one sees right away that the r's cancel out. Sorry!

    Would it be too much of a mess to show your result for the derivative?
     
  8. Jul 7, 2006 #7
    Well I wasn't doing anything wrong with the derivative since looking at some graphs Ez does go to zero as z->0.
     
  9. Jul 7, 2006 #8
    Here are my contour plots for the electric fields. You can imagine where the disc of radius 1 is.
    Ez:
    http://www.maths.tcd.ie/~dleen/ez.jpg
    horizontal axis is r and vertical z

    and Er:
    http://www.maths.tcd.ie/~dleen/er.jpg
    horizontal axis is r and vertical z


    and while I'm at it:
    http://www.maths.tcd.ie/~dleen/potential.jpg
    z axis is phi, x and y are r and z

    I'm a bit confused about "Calculating Ez for r=0 could help you pinpoint the problem." Why did I get non zero answer?

    I was beginning to think that I would calculate Er...
     
    Last edited: Jul 7, 2006
  10. Jul 8, 2006 #9
    BP,

    I think you are using Mathematica or some similar program.
    These programs must be handled with care when sigularities are involved.
    In your problem, the electric field has a discontinuity across the plate.
    There are many consequences like:
    the limit for z->0 does not exist, it exists on the left and the right only,
    the Taylor series does not exist​

    In this situation, it is useful to consider special cases before handling the full question.
    Here, the analysis on the axis of symetry (r=0) is easily done "manually" and you can spend a little time to understand all aspects of the problem. Eventually you can manage later to find out the charge distribution everywhere, but in a first step calculate it in the center.

    In the attachment you will find the fields on the axis of symmetry. Clearly Ez is not zero. Proceeding carefully at other places will give you the charge distribution. Note that in general f(x) diagrams are easier to read than contour plots or 3D's .

    Michel

    Attachment URL:

    https://www.physicsforums.com/attachment.php?attachmentid=7288&d=1152356117

    .
     
    Last edited: Nov 22, 2011
  11. Jul 8, 2006 #10
    I have this so far

    [tex]E_z = \frac{\frac{2 a z}{\sqrt{((r-a)^2 +z^2)((r+a)^2 +z^2)}}}{\sqrt{2(r^2 + z^2 - a^2 + \sqrt{((r-a)^2 +z^2)((r+a)^2 +z^2)})}}[/tex]

    I'm unsure how to proceed taking the limit. My first attempt gave

    [tex]E_z = \frac{\frac{2 a z}{\sqrt{((r-a)^2 +z^2)((r+a)^2 +z^2)}}}{\sqrt{2 z^2}}}
    =\frac{\sqrt{2} a}{\sqrt{(a^2 -r^2)^2}}[/tex]


    Along the symetry axis I got 1/a as my answer

    Your attachment is still pending approval so I can't see it.
     
    Last edited: Jul 8, 2006
  12. Jul 8, 2006 #11
    BP,

    There are some braket problems in your expression.

    You need to analyse the limit for z->0 carefully.
    First, factor out these few things that don't tend to zero.
    Then observe the parts that produce the indetermination:
    the 2az factor on the numerator
    the big square root on the denominator​
    Note indeed that the big square root goes to zero as z->0 .
    Be careful to choose the right sign when taking the square root of a square!
    Once you have observed the indetermination, you should be conviced that the limit is finite.
    Patience, care, and a few Taylor series for the big denominator should tell you the answer.

    Michel

    Postscriptum

    Here is the formula I got for Ez² (taking V=1 and a=1):

    16*z²/(Pi²*u²*v²*(-2 + u + v)*(2 + u + v))​
    with

    u = Sqrt((1 + r)² + z²)
    v = Sqrt((1 - r)² + z²)

    You can check if this is consistent with your formula.
     
    Last edited: Jul 8, 2006
  13. Jul 8, 2006 #12
    Yes mine is the same as yours.

    What do you mean by a few taylor series for the big denominator? Should I expand it around z=0? I haven't had any experience with this so thanks for bearing with me.
     
  14. Jul 8, 2006 #13

    nrqed

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    But I thought your problem was that you were getting Ez=0 at z=0,no?
     
  15. Jul 8, 2006 #14
    Hi BP,

    I used Mathematica v2.0 to find out Ez.

    I had to control each step by myself because Mathematica v2.0 seems very unreliable for these kind of limits. Series expansions are not more reliable.

    I worked with the expressions from my last postscriptum.
    Essentialy I had to calculate the limit of z²/(-2+u+v) , other factors make no problem.
    I expanded the divergent factor (-2+u+v) from the denominator in series of z.
    I injected that back in Ez² and this removed the indeterminacy.
    Going back to Ez and taking care for the correct sign gives a simple answer.

    Michel

    Postscriptum

    I got the following result: Ez = -2/Pi * 1/sqrt(a²-r²) .
    By the way, how did you find the formula for the potential?
     
    Last edited: Jul 8, 2006
  16. Jul 8, 2006 #15

    nrqed

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    EDIT: I see how you got your answer. But it's incorrect. Th emistake is that you neglect z^e relative to a^2 and r^2 in part of the expression (under the square root) while retaining z^2 in another part of th eexpression. This is inconsistent. I work out the correct expnasion below.


    I don't get sqrt(2 z^2) for the expansion of the
    denominator. Using (I am assuming r<a)
    [tex] {\sqrt { (a-r)^2 + z^2 }}\approx a-r + {1 \over 2 } { z^2 \over a-r} [/tex]
    [tex] {\sqrt { (a+r)^2 + z^2}} \approx a+r + {1 \over 2 } { z^2 \over a+r} [/tex]
    SO the product of the two, for small z, gives
    [tex] a^2 - r^2 + {1\over 2} { a+r \over a-r} z^2 + {1 \over 2} { a-r \over a+r } z^2 [/tex]
    Therefore, if I add r^2 -a^2 +z^2 to the above, I get
    [tex]
    z^2 + {1\over 2} { a+r \over a-r} z^2 + {1 \over 2} { a-r \over a+r }z^2 = z^2 + z^2 { a^2 +r^2 \over a^2 -r^2} [/tex]

    All this must still be multiplied by two and put under a square root, giving
    [tex]
    {\sqrt {2 z^2}} {\sqrt { { 2 a^2 \over a^2 -r^2}}}[/tex]
    Instead of your sqrt(2 z^2).

    I will post and then check that there is no typo.
    Edit: it looks right now.
     
    Last edited: Jul 8, 2006
  17. Jul 8, 2006 #16
    I solved the laplace equation in cyclindrical coordinates with the appropriate boundry condition. Weber's discontinuous integrals really helped.

    Are you using the Series function in mathematica?

    Why does the sign matter so much when taking the square root of the square in this problem? This seems like where mathematica is having the problem.
     
  18. Jul 8, 2006 #17
    I think I've got it now. Thank you very much for your help. I have learned a lot today and hopefully I'll be able to apply this to the rest of the problem.

    nrged sorry, I was still confused at that stage and I don't know why I posted that.
     
  19. Jul 9, 2006 #18

    pervect

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    Buried deep in a 56 page postscript file that can be found at
    http://www.ttc-cmc.net/~fme/captance.html [Broken]

    the above potential function and the associated charge distribution is given. I've attached this as a .png file.
     

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  20. Jul 9, 2006 #19
    BP,

    Yes I do.

    However, it fails if the function has a discontinuity or its derivative has a discontinuity and so on. All derivatives should be continuous, otherwise the series expansion needs further specifications that Mathematica does not handle (on the right or on the left).

    Still the functionality was useful for you problem. I did not use it directly on the considered function (Ez), but on elements that can be developped in series, like one the square root. Assembling the expanded parts leads to the result.

    Michel

    Postscriptum

    Here is how I did it with Mathematica.

    V = Vo 2/Pi ArcSin[2 a / (Sqrt[z*z+(a+r)*(a+r)] + Sqrt[z*z+(a-r)*(a-r)])]
    Ez = Simplify[D[V,z]]
    Ezuv = Ez //.{Sqrt[(a+r)^2+z^2]->u,Sqrt[(a-r)^2+z^2]->v}
    Ezuv = Ezuv//.{1/Sqrt[(a+r)^2+z^2]->1/u,1/Sqrt[(a-r)^2+z^2]->1/v}
    Ezuv2 = Factor[Ezuv^2]
    Ezuv2 = Ezuv2//.{u+v->upv}
    upv = Sqrt[(a+r)^2+z^2] + Sqrt[(a-r)^2+z^2]
    upv = Simplify[Normal[Series[upv,{z,0,2}]//.{Sqrt[(a-r)^2]->a-r,Sqrt[(a+r)^2]->a+r}]]
    Ezuv2 = Simplify[(Ezuv2//.{1/u^2->1/((a+r)^2+z^2),1/v^2->1/((a-r)^2+z^2)})//.{z->0}]

    Clearly I could not have the answer automaticaly.
    It had to make sure that the term I called "upv" was properly handled.
    I had also to guide Mathematica with the square root, to be sure it comes with positive sqrt.

    I have a very old version of Mathematica.
    I would like to know how you did it yourself and the tool you used (paper and pencil ?).
     
    Last edited: Jul 9, 2006
  21. Jul 9, 2006 #20
    Yes I have been mainly using pen and paper. I was getting sick of those long expressions (and they double in length for the capacitor) so I started using mathematica. However since I was not used to watching out for singularities etc I started running into problems. I've learned now though which is great.
     
  22. Jul 9, 2006 #21
    Well another question:

    I worked out the potential for a capacitor:

    [tex]\phi = \frac{2 V}{\pi}\sum_{n=0}^{\infty}\arcsin{(2a/(\sqrt{(2nL+|z-L|)^2 +(r+a)^2}+\sqrt{(2nL+|z-L|)^2 +(r-a)^2}))}
    \\ \\
    -\arcsin{(2a/(\sqrt{(2nL+|z+L|)^2 +(r+a)^2}+\sqrt{(2nL+|z+L|)^2 +(r-a)^2}))}[/tex]

    The discs are located at z=L and z=-L. The absolute value was needed to do an integral.

    I'm now trying to find the charge density again.

    Ez will not be the same above and below like in the single disc case. The charge density is now [tex]E_{z}^{+} - E_{z}^{-} = 4\pi \sigma[/tex]. E+ is Ez on top of the disc. E- on the bottom.

    First I took the limit z->L coming from infinity and going to L to get the Ez on top of the disc. This makes the absolute value signs unneccessary since z+L will always be positive and so will z-L.

    Then I took the limit z->L coming from minus infinity to get Ez on the other side of the disc at z=L. |z-L| will become -(z-L) for all values of z however I'm not so sure about |z+L|. It seems that this will change sign. Is it ok to just look at it while z>0 but z<L? This will leave |z-L| as -(z-L) and |z+L| as just z+L. Does this make sense?

    I seperated n=0 from the sum before taking the limit since we will have singularities here again and worked it out just as before.

    Also when L -> infinity I should get the same answer as before.

    I just want to make sure I'm not missing anything before I try work it out fully. Or I could be doing it completely wrong.
     
    Last edited: Jul 9, 2006
  23. Jul 14, 2006 #22
    BP,

    I had a look at this nice expression.
    I have the following remarks:

    - it is easier to develop [tex]\phi[/tex] in series directly instead of calculating Ez first

    - the term n=0 gives the same singularity as for your first example
    . this singularity is related to the ArcSin(x) function near x=1

    - the singularities in the terms n>0 are related to the Abs function
    . they are easy to calculate but do not reduce to a simple compact form

    - the end result is a full series that I don't expect to result in a compact form
    . it should be summed numerically, probably

    - if in the end calculations have to be numerical, why not doing numerics directly from [tex]\phi[/tex]
    . that's a question: do we win something by further analytic calculations

    - in the end the capacity should be calculated by integration over the disk
    . is there something useful to do analytically here ?

    By the way, could you explain shortly how you managed to get the solution for one disk, and also for two disks?

    Thanks,

    Michel
     
    Last edited: Jul 14, 2006
  24. Jul 24, 2006 #23
    BP,

    I investigated a little bit your formula and where it comes from.
    Apparently, your formula was obtained according to the following paper:


    It will cost you 30$ if you want to read it! Fortunately I could get a copy for free.
    The starting formula [2] from this paper made me suspicious: the Ez field would be discontinuous for z=L even outside the plates!!!
    But I investigated a little bit further becasue of my lack of self-confidence. Then I came upon a comment on this paper:


    The abstract reads as follows:

    A recently proposed analytic solution (Atkinson et al., 1983) to a celebrated problem in classical potential theory is shown to be incorrect.​

    Of course, I did not pay the 30$ to read in details why this formula was false. This would be really too much:

    + 30$ for a mistaken formula
    + 30$ to know why it is wrong​

    The whole thing in an expensive "refereed" publication !!!
    (note that a simple numerical test can easily show the flaw)

    Apparently, this problem can only be solved numerically. And there is no analytical formula to check it.
    Unless some of you here has a good idea to solve the problem, or another reference in the litterature.
    And maybe for free this time ...

    Michel
     
    Last edited: Jul 24, 2006
  25. Jul 26, 2006 #24
    The potential due to a circular parallel plate capacitor can be obtained
    solving a dual system of integral equation arising from the general integral representation of Laplace eq. solution in cylindrical coordinates, imposing the boundary conditions. The solution is expressed as a Tranter series of Bessel functions.
     
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