- #1

Opus_723

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## 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]

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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]

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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.

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