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Conical Surface Potential Difference

  1. Mar 27, 2016 #1
    Hey guys!
    The question is related to problem 2.26 from Electrodynamics by Griffiths (3ed).

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

    A conical surface (an empty ice-cream cone) carries a uniform surface charge σ. The height of the cone is h, as the radius of the top. Find the potential difference between points a (the vertex) and b (the center of the top).

    2. Relevant equations
    Here I will call the potential V.

    First of all, I assumed that at the vertex: V(a) = 0. (I can do that because I'm interested in V(b) - V(a), am I right?)

    Then I calculated V(b). So:
    V(a) - V(b) = - V(b) = -σh/(2ε) * ln (1 + (21/2/2))

    But the book's solution didn't consider V(a) = 0, and found:
    V(a) - V(b) = σh/(2ε) [1 - ln (1 + (21/2/2))]

    Finally, my questions are:
    Why is my assumption wrong?
    How to calculate it assuming V(b) = 0?
     
  2. jcsd
  3. Mar 27, 2016 #2

    TSny

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    In deriving your expression for V(b), where did you choose V to be zero?
     
  4. Mar 27, 2016 #3
    At the point a = (0,0,0)
     
  5. Mar 27, 2016 #4

    TSny

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    Can you show an outline of your derivation of your expression for V(b)?
     
  6. Mar 27, 2016 #5
    Yes, of course.
    r is the vector along the central axis and r' is the vector along the conical surface.

    $$ V(b) = \frac{1}{4\pi\epsilon_0}\int_0^b{\frac{\sigma\cdot da'}{|\textbf{r}-\textbf{r}'|}} = \frac{\sigma}{4\pi\epsilon_0}\sqrt{2}\pi\int_0^{\sqrt{2}h}\frac{r'dr'}{\sqrt{r'^2 + h^2 -\sqrt{2}hr'}} = \frac{\sigma h}{2\epsilon_0}\ln (1+\sqrt{2}) $$
     
    Last edited: Mar 27, 2016
  7. Mar 27, 2016 #6

    TSny

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    The integrand represents the potential at b due to a small element of charge ##\sigma da##. It has the form ##V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}## for a point charge. Note that this expression assumes a particular place where V = 0. Where is that place?
     
  8. Mar 27, 2016 #7
    For a point charge, V = 0 at the infinity.
    You helped me a lot to identify my own misunderstanding. In the previous exercise that I solved the potential wouldn't go to 0 at the infinity (it was due to a infinity distribution).

    Thank you a lot!
     
  9. Mar 27, 2016 #8

    TSny

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    OK.

    I'm not quite getting the result that you quoted as the given solution. I get that the argument of the log should be ##1 + \sqrt{2}## rather than ##1 + \frac{\sqrt{2}}{2}## . But I could be messing up somewhere.
     
  10. Mar 27, 2016 #9
    No no, you are right. My mistake. I already edited there.

    Thanks again.
     
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