- #1
Opus_723
- 178
- 3
Homework Statement
This isn't a homework problem, but I got it off of an upperclassman's homework and decided to give it a shot.
Consider a conical surface (like an empty ice-cream cone) with a height and top radius which are both h pointed up so that its axis lies along the z-axis and its tip at the origin. The cone has a uniform surface charge density of [itex]\sigma[/itex]. What is the potential of this cone at a point z on the z axis (i.e. at the point [itex]\vec{r}[/itex]=z[itex]\hat{z}[/itex])?
The Attempt at a Solution
For less confusion, I call the height, or z-coordinate, of the point we are interested in H, so that we can integrate over z.
Splitting the cone up into rings, we get
dq = [itex]\sigma[/itex]da = [itex]\sigma[/itex]*2[itex]\pi[/itex]*z*[itex]\frac{2}{\sqrt{2}}[/itex]dz (Since the radius is always equal to the height above the origin)
(Thanks to haruspex for pointing out the cosine factor)
The distance from H to the ring at height z is [itex]\sqrt{(H-z)^2+z^2}[/itex]
So the integral we need is
[itex]\frac{1}{4\pi\epsilon}[/itex][itex]\int^{h}_{0}[/itex][itex]\frac{\sigma*2\pi*z\frac{2}{\sqrt{2}}dz}{\sqrt{(H-z)^2+z^2}}[/itex]
-------------------------------------
Now, assuming I set it up right, the only problem is taking the integral. First, I completed the square and pulled out constants to get:
[itex]\frac{\sigma}{2\epsilon}[/itex][itex]\int^{h}_{0}[/itex][itex]\frac{zdz}{\sqrt{(z-\frac{H}{2})^2+\frac{H^2}{4}}}[/itex]
Next, I tried a hyperbolic trig substitution. Namely, z-[itex]\frac{H}{2}[/itex]= [itex]\frac{H}{2}[/itex]sinh[itex]\theta[/itex], with dz = [itex]\frac{H}{2}[/itex]cosh[itex]\theta[/itex]
Substituting in, I got:
[itex]\frac{\sigma}{2\epsilon}[/itex][itex]\int[/itex][itex]\frac{\frac{H^2}{4}(sinh\theta+1)cosh\theta}{\sqrt{\frac{H^2}{4}sinh^2\theta+\frac{H^2}{4}}}[/itex]
Which simplifies to:
[itex]\frac{\sigma}{2\epsilon}[/itex][itex]\int[/itex][itex]\frac{\frac{H}{2}(sinh\theta+1)cosh\theta}{coshθ}[/itex]
And further to:
[itex]\frac{H\sigma}{4\epsilon}[/itex][itex]\int[/itex][itex]sinh\theta+1[/itex]
And, finally taking the integral:
[itex]\frac{H\sigma}{4\epsilon}[/itex]([itex]cosh\theta+\theta[/itex])
Going back to my original substitution, [itex]\theta[/itex]= arcsinh([itex]\frac{2z}{H}[/itex]-1)
So I sub back into get:
[itex]\frac{H\sigma}{4\epsilon}[/itex][itex](cosh(arcsinh(\frac{2z}{H}-1))+arcsinh(\frac{2z}{H}-1))[/itex]
And then I evaluated from 0 to h (little h being the height of the cone)
[itex]\frac{H\sigma}{4\epsilon}[/itex][itex][cosh(arcsinh(\frac{2h}{H}-1))+arcsinh(\frac{2h}{H}-1)-cosh(arcsinh(-1))-arcsinh(-1)][/itex]
Or
[itex]\frac{H\sigma}{4\epsilon}[/itex][itex][cosh(arcsinh(\frac{2h}{H}-1))+arcsinh(\frac{2h}{H}-1)-\sqrt{2}-ln(\sqrt{2}-1)][/itex]
-------------------------------------------
Which, ideally, should be my answer. But what's got me bothered is I'm not sure that this goes to zero as H gets large. If someone could help me figure that out, I'd feel a lot better about this solution. I tried plotting in on Mathematica, but it sort of looks different depending on how far I take it out, so I'm not sure what to make of it.
Thanks for your time, I know this is a long one.
Last edited: