Electric Potential Difference on a Cone

  1. 1. The problem statement, all variables and given/known data
    I'm working out of Griffith's "Intro to Electrodynamics" and the problem states: "A conical surface (an empty ice-cream cone) carries a surface charge [tex]\sigma[/tex]. The height of the cone is h as is the radius of the top. Find the potential difference between points a (the vertex) and b (the center of the top).

    2. Relevant equations and Attempt at a solution
    So, since this is the chapter that I'm in, I'm going to use
    [tex]\[V(R)=\frac{\sigma }{4\pi \varepsilon _{0}}\int _{S}\frac{da{}'}{R}\][/tex].
    Now since a is at the vertex I chose
    [tex]\[\vec{a}=0\][/tex] and [tex]\[\vec{b}=h\hat{z}\][/tex].
    Thus the equation would become
    [tex]\[V(\mathbf{b})-V(\mathbf{a})=\frac{\sigma }{4\pi \varepsilon _{0}}\int _{S}\left [ \frac{{da}'}{\sqrt{(h-{z}')^2+{s}'^2}} -\frac{{da}'}{\sqrt{{z}'^2+{s}'^2}}\right ]\][/tex]
    Now da' is what I was having a little trouble attaining, so I thought the best place to start would be with the surface area of the cone:
    [tex]\[a'=\pi s\sqrt{s^2+z^2}\][/tex]
    but since the radius s is equal to the height z in our case the formula becomes
    [tex]\[a'=\pi s\sqrt{s^2+s^2}=\sqrt{2}\pi s^2\][/tex].
    Now since fractions of this area can be represented by multiplying in terms of the angle that determines the fraction of area,
    [tex]\[\frac{\theta }{2\pi }\][/tex].
    Thus [tex]\[a'=(\sqrt{2}\pi s^2)\cdot (\frac{\theta }{2\pi })=\frac{\sqrt{2}}{2}s^2\theta \][/tex]
    and if I consider the angle to be small
    [tex]\[a'=\frac{\sqrt{2}}{2}s^2d\theta \][/tex].
    Now to find the differential area I should subtract to get
    [tex]\[da'=\frac{\sqrt{2}}{2}(s+ds)^2d\theta -\frac{\sqrt{2}}{2}s^2d\theta=\frac{\sqrt{2}}{2}d\theta(s^2+sds+ds^2-s^2)=\frac{\sqrt{2}}{2}sdsd\theta\]
    since ds^2 is to small to matter.
    The main equation then becomes:
    [tex]\[V(\mathbf{b})-V(\mathbf{a})=\frac{\sigma }{4\pi \varepsilon _{0}}\int _{S}\left [ \frac{{da}'}{\sqrt{(h-{z}')^2+{s}'^2}} -\frac{{da}'}{\sqrt{{z}'^2+{s}'^2}}\right ]=\frac{\sigma }{4\pi \varepsilon _{0}}\int _{S}\left [ \frac{1}{\sqrt{(h-{s}')^2+{s}'^2}} -\frac{1}{\sqrt{{s}'^2+{s}'^2}}\right ](\frac{\sqrt{2}}{2}{s}'{ds}'{d\theta}' )\][/tex] [tex]\[=\frac{\sqrt{2}\sigma }{8\pi \varepsilon _{0}}\int_{0}^{2\pi }\int_{0}^{h}\left ( \frac{{s}'}{\sqrt{(h-{s}')^2+{s}'^2}}-\frac{\sqrt{2}}{2} \right ){ds}'{d\theta}' \][/tex]
    [tex]\[=\frac{\sqrt{2}\sigma }{4\varepsilon _{0}} [(-hln({s}'-h)+\frac{{s}'^3}{3}+{s}')|_{0}^{h}-\frac{\sqrt{2}}{2}h]\][/tex]
    but the above does not converge when evaluated so I'm at a loss. This isn't for a class or anything, I'm just self studying so answer at your convenience.
  2. jcsd
  3. Andrew Mason

    Andrew Mason 6,832
    Science Advisor
    Homework Helper

    Try slicing the cone along the vertical axis into rings of area [itex]dA = 2\pi s dz[/itex] where s = radius of the ring at height z, which is a linear function of z. So each ring carries a charge that is proportional to z. That should be easy to integrate.

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