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Electric Potential of Uniformly Charged Cone

  1. Dec 31, 2009 #1
    [SOLVED] Electric Potential of Uniformly Charged Cone

    I'm actually a senior in physics graduating this year, but wanted to review some E&M before grad school in the fall. I was apparently never assigned this problem during my sophomore year, but it's from Griffiths. I'm also aware that this has been asked on this forum a number of times, but my issue is getting the answer without resorting to integral tables or simply taking the instruction manual's word for what the antiderivative of the main function should be.

    1. The problem statement, all variables and given/known data

    "A conical surface (an empty ice cream cone) carries a uniform surface charge [tex]\sigma[/tex]. The height of the cone is [tex]h[/tex], as is the radius of the top. Find the potential difference between points a (the vertex) and b (the center of the top)."

    I've already determined the potential at the vertex of the cone, it's pretty straightforward; I'm just having trouble with one specific integral as needed to get the potential at the center of the top.

    2. Relevant equations

    [tex]V(b) = \frac{\sigma}{2 \sqrt{2} \epsilon_0} \int_0^{\sqrt{2}h} \frac{r}{\sqrt{h^2 + r^2 - hr \sqrt{2}}}dr[/tex]

    3. The attempt at a solution

    The most straightforward way I could think of to do this integral is to use the u substitution of [tex]u = \sqrt{h^2 + r^2 - hr \sqrt{2}}[/tex]. This then allows me to rewrite the integral as the difference of two functions of u (the remaining [tex]r[/tex] terms can be eliminated by writing [tex]r[/tex] in terms of [tex]u[/tex]. The problem, however, is that, when I do this, the limits of integration (in terms of [tex]u[/tex]) both become [tex]h[/tex], which makes me think I'm going about this the wrong way. The frustrating part here is also that, looking at the solution in the instructor's manual, I'm getting nearly the right answer, but Griffiths skips a few steps and I want to actually understand how this integral can be done from start to finish. Any advice would be greatly appreciated.
     
    Last edited: Dec 31, 2009
  2. jcsd
  3. Dec 31, 2009 #2

    diazona

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    Your range of integration includes the minimum point of the function [itex]u(r)[/itex] - it occurs at [itex]r = h/\sqrt{2}[/tex]. I think that means that for one of the limits you need to use the other "root" when you substitute, i.e.
    [tex]u = -\sqrt{h^2 + r^2 - hr\sqrt{2}}[/tex]
     
  4. Dec 31, 2009 #3
    Hm, I don't think I follow... Since the square root is defined to be positive, how do I justify putting a negative in front?
     
  5. Dec 31, 2009 #4

    diazona

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    Hmm... okay, I think I goofed. Ignore that last expression.

    Here's my reasoning, though: when you substitute u for r, you come up with (or could come up with) an expression like
    [tex]r = \frac{1}{\sqrt{2}}(h \pm \sqrt{2u^2 - h^2})[/tex]
    right? Now, if you use only the plus sign, no matter what value of u you plug in (for u real), you can never get r to equal zero. In fact, you can only have [itex]r \ge h/\sqrt{2}[/itex] using the plus sign. Conversely, if you use only the minus sign, you can never get the upper limit [itex]r = \sqrt{2}h[/itex]; you're limited to [itex]r \le h/\sqrt{2}[/itex]. So in order to really do the integral properly, you need to split it into two parts at [itex]r = h/\sqrt{2}[/itex] and use the plus sign for the upper part and the minus sign for the lower part. That's why you're getting the same limits when you plug in naively; [itex]r = 0[/itex] and [itex]r = \sqrt{2}h[/itex] both correspond to [itex]u = h[/itex], they're just on different branches of the square root function. (You're trying to integrate over a branch cut, basically.)
     
  6. Dec 31, 2009 #5
    Ah, I think I understand what you're saying... but I'm still not getting the answer I expect. I broke up the integral into two integrals (one from 0 to [tex]\frac{h}{\sqrt{2}}[/tex] and the other from [tex]\frac{h}{\sqrt{2}}[/tex] to [tex]h \sqrt{2}[/tex], but when I work this out and use the new limits of integration, I end up with a constant times (ln(1)). Since you seem to have an idea of where the math is going wrong, can you work it out on paper and let me know what's inside the logarithm term? FWIW, I should be getting [tex]\ln{(1 + \sqrt{2})}[/tex].
     
  7. Dec 31, 2009 #6

    diazona

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    Actually, why don't you show your work and the collective scrutiny of PF will track down the mistake? I'm actually having a tricky time working with the u-substitution you used... but I did plug the integral into Mathematica and confirm that [itex]\ln(1+ \sqrt{2})[/itex] should appear in the answer.
     
  8. Dec 31, 2009 #7
    Thanks so much for the quick replies. I'll convert my work to LaTeX and post it up shortly, but since you mentioned it, is there a more straightforward way to do the integral that you suggest I try? It surprises me that this has taken me so much effort considering that it's not even one of the "dotted" problems in Griffiths' book.
     
  9. Dec 31, 2009 #8

    ideasrule

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    Yes, there's a much easier way to do the integral. Try completing the square for the expression [itex]h^2 + r^2 - hr \sqrt{2}[/itex]. After doing this, you'll get an integral in the form r*dr/[(r-a)^2-a^2]. Doing the u-substitution u=r-a, and using the formula for the antiderivative of 1/sqrt(r^2-a^2), will quickly get you the answer. (If you don't want to search up the antiderivative of 1/sqrt(r^2-a^2), you can compute it yourself using trigonometric integration.)
     
  10. Dec 31, 2009 #9
    That did it! Thanks so much, guys; I'd been sitting on this problem for over a day. For anyone else who, like me, got stumped by the math of this problem, here's a recap:

    1. Complete the square of the denominator. You will have one squared term containing [tex]r[/tex] and another with just a constant.
    2. Do a u-substitution using the first term being squared. At this point, you can replace the stray [tex]r[/tex] in the numerator using this u-substitution, as well.
    3. As the integral will now be a rational function with two terms in the numerator, the integral can be broken up into two integrals. It also doesn't hurt to change the limits of integration as they will now be symmetric about zero (simplifying the final step).
    4. The first integral can be solved easily doing another algebraic substitution while the second requires a common trigonometric substitution.

    Once again, thanks to ideasrule and diazona for helping me out with this. I only wish I knew about this place when I was actually taking this class! :)
     
    Last edited: Dec 31, 2009
  11. Dec 31, 2009 #10

    diazona

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    Aww, all that work and I only get a parenthetical credit? :cry: (just kidding, it's all good)
    That's what I did, or would have done... but for continuity's sake I tried to make myself work with the other way.
     
  12. Jun 20, 2011 #11
    Hello everyone,

    I am stuck with this particular problem and I get the right answer.However, my integral is different from the one in solution manual before evaluation, I have additional terms in the ln function. Is there something wrong with the solution manual or am I doing something wrong ? I have attached the screenshot of the page containing the solution.
    Thanks in advance.
     

    Attached Files:

  13. May 18, 2012 #12
    How do you obtain the equation for V(b)? I come to the same equation with the sole difference that I have a r^2 in the numerator.
     
  14. May 18, 2012 #13
    nevermind... stupid mistake...
     
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