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Electric Potential Difference on a Cone

by grindfreak
Tags: cone, difference, electric, potential
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Dec26-10, 01:54 PM
P: 39
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.
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Andrew Mason
Dec27-10, 10:24 AM
Sci Advisor
HW Helper
P: 6,679
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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