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Homework Help: The potential at points around a cone

  1. Jan 13, 2010 #1
    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 h, as is the radius of the top. Find the potential at the centre of the top, taking infinity as reference point." - Griffiths

    My result for the potential differs from the answer's. Can someone please check my solution

    2. Relevant equations

    [tex]V = \iint_S \frac{1}{4\pi \epsilon_0}\frac{\sigma}{x} dS[/tex], where [tex]x[/tex] is the distance from the source to the point.

    3. The attempt at a solution

    My diagram is in the attached picture.

    First off, I used cylindrical coordinates with the equation z = r, z > 0 to graph the cone
    I found [tex]dS = \sqrt{2}zdA[/tex]

    The distance from any point on the cone to the origin is [tex]\sqrt{2} z[/tex], so using the cosine law,
    [tex]x = \sqrt{2z^2-2hz+h^2}[/tex]

    So we have

    [tex]V = \int_0^h\int_0^{2\pi} \frac{1}{4\pi \epsilon_0} \frac{\sigma}{\sqrt{2z^2-2hz+h^2}} \sqrt{2}z d\theta dz=\frac{\sigma h}{4\epsilon}\ln(2\sqrt{2}+3)[/tex]

    Answer: [tex]V = \frac{\sigma h}{2\epsilon}\ln(1+\sqrt{2})[/tex]


    Attached Files:

    • cone.jpg
      File size:
      14.3 KB
  2. jcsd
  3. Jan 13, 2010 #2


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    Homework Helper

    Hi Identity! :smile:

    erm :redface: … 2ln(1 + √2) = ln(3 + 2√2) :wink:
  4. Jan 13, 2010 #3
    :eek: oh my!
    thankyou tiny-tim
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