Electric Potential Difference on a Cone

[grindfreak]

Homework Statement

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 \sigma. 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. Homework Equations and Attempt at a solution
So, since this is the chapter that I'm in, I'm going to use
\[V(R)=\frac{\sigma }{4\pi \varepsilon _{0}}\int _{S}\frac{da{}'}{R}\].
Now since a is at the vertex I chose
\[\vec{a}=0\] and \[\vec{b}=h\hat{z}\].
Thus the equation would become
\[V(\mathbf{b})-V(\mathbf{a})=\frac{\sigma }{4\pi \varepsilon _{0}}\int _{S}\left [ \frac{{da}'}{\sqrt{(h-{z}')^2+{s}'^2}} -\frac{{da}'}{\sqrt{{z}'^2+{s}'^2}}\right ]\]
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:
\[a'=\pi s\sqrt{s^2+z^2}\]
but since the radius s is equal to the height z in our case the formula becomes
\[a'=\pi s\sqrt{s^2+s^2}=\sqrt{2}\pi s^2\].
Now since fractions of this area can be represented by multiplying in terms of the angle that determines the fraction of area,
\[\frac{\theta }{2\pi }\].
Thus \[a'=(\sqrt{2}\pi s^2)\cdot (\frac{\theta }{2\pi })=\frac{\sqrt{2}}{2}s^2\theta \]
and if I consider the angle to be small
\[a'=\frac{\sqrt{2}}{2}s^2d\theta \].
Now to find the differential area I should subtract to get
\[da&#039;=\frac{\sqrt{2}}{2}(s+ds)^2d\theta -\frac{\sqrt{2}}{2}s^2d\theta=\frac{\sqrt{2}}{2}d\theta(s^2+sds+ds^2-s^2)=\frac{\sqrt{2}}{2}sdsd\theta\]<br />
since ds^2 is to small to matter.
The main equation then becomes:
\[V(\mathbf{b})-V(\mathbf{a})=\frac{\sigma }{4\pi \varepsilon _{0}}\int _{S}\left [ \frac{{da}&#039;}{\sqrt{(h-{z}&#039;)^2+{s}&#039;^2}} -\frac{{da}&#039;}{\sqrt{{z}&#039;^2+{s}&#039;^2}}\right ]=\frac{\sigma }{4\pi \varepsilon _{0}}\int _{S}\left [ \frac{1}{\sqrt{(h-{s}&#039;)^2+{s}&#039;^2}} -\frac{1}{\sqrt{{s}&#039;^2+{s}&#039;^2}}\right ](\frac{\sqrt{2}}{2}{s}&#039;{ds}&#039;{d\theta}&#039; )\] \[=\frac{\sqrt{2}\sigma }{8\pi \varepsilon _{0}}\int_{0}^{2\pi }\int_{0}^{h}\left ( \frac{{s}&#039;}{\sqrt{(h-{s}&#039;)^2+{s}&#039;^2}}-\frac{\sqrt{2}}{2} \right ){ds}&#039;{d\theta}&#039; \]
\[=\frac{\sqrt{2}\sigma }{4\varepsilon _{0}} [(-hln({s}&#039;-h)+\frac{{s}&#039;^3}{3}+{s}&#039;)|_{0}^{h}-\frac{\sqrt{2}}{2}h]\]
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.

 

[Andrew Mason]

Try slicing the cone along the vertical axis into rings of area dA = 2\pi s dz 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.

AM