Uniformly Charged Cone - Potential Difference

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  • #1
nd
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We have a right circular cone of base radius a and height a with a uniform surface charge sigma. I want to determine the potential difference between the apex of the cone and the center of the base (this cone doesn't have any charge on the base).
My plan of attack for the problem was to determine the potential at the apex and center of the base, and then subtract the two, since finding the E field doesn't seem too appealing (although by the symmetry, I can determine the direction of the E field along a path through the axis).
What I did was come up with a vvf to parameterize the cone. Then I need to find a formula for r in dV = dq / (4pi ep r). Then I integrate dV over the cone. The integral for the apex is simple and comes out nicely, but the other one is not so simple. Can anyone suggest something to make this process easier on me?
 

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  • #2
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Hello nd...

I am just curious mate...how did you find the apex integral?

Cheers
Vivek
 
  • #3
OlderDan
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If you align the axis of the cone with the z axis, with the center of the circle at the origin, you can think of the cone as rings of thickness dz and radius R, where [tex]R^2 = x^2+y^2[/tex]. From your geometry, [tex]z = a - R[/tex]. I assume your result for the apex has as a component the on-axis potential for a charged ring, though you might not have thought about it that way. Every point on the ring is the same distance from the point of interest. All you need to do is integrate over z, adding the contributions from the rings. The distance from the center of the circle to a point on the ring is

[tex]r = \sqrt{z^2-R^2} = \sqrt{z^2-(a - z)^2}[/tex]

You can simplify the radicand and be on your way.
 
  • #4
nd
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I take [tex] \textbf{r}(t,u)=t(\cos u,\sin u,1)[/tex] to be a parametrization of the cone. The norm of the fundamental vector product is (this can be easily checked)
[tex]\| \frac{\partial \textbf{r} }{\partial u} \times \frac{\partial \textbf{r} }{\partial t} \| = \sqrt{2} t [/tex]
As you can tell by my parametrization, the apex is at the origin. So, we have that [tex] dV = \frac{dq}{4 \pi \epsilon_0 \sqrt{x^2 + y^2 + z^2}}=\frac{\sigma dA}{4 \pi \epsilon_0 \sqrt{2}t} = \frac{\sigma du dt}{4\pi \epsilon_0} [/tex].
It follows by integration over the cone that [tex]V = \frac{\sigma}{4\pi \epsilon_0} 2\pi a = \frac {\sigma a}{2 \epsilon_0}[/tex].
This is how I found V for the apex. Is this right?

As for the advice from OlderDan, is the approximation by rings not affected by the fact that the differential 'ring' is not actually an annulus, but is at a slant?
 
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  • #5
OlderDan
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nd said:
As for the advice from OlderDan, is the approximation by rings not affected by the fact that the differential 'ring' is not actually an annulus, but is at a slant?
I'll answer this first and then look at your calculation of the potential. I'm not seeing your parameterization clearly at this point, and it seems more complex than simply integrating along the face of the cone, but it could be just fine. Certainly the slant has an effect when you figure out how much charge is contained in each ring. You need to express the surface area associated with each ring in terms of R and dz, and that would be 2*pi*R*dz/cos(pi/4) I believe. I need to draw a diagram and look at it.

OK so far. I agree with your result

[tex]V = \frac {\sigma a}{2 \epsilon_0}[/tex].

I got there using the idea of "rings" integrating along the face of the cone from the apex to the base.
--------------------------------------------------------------

Now I'm not doing so good. I made a sign mistake in my earlier post. Looks like

[tex]r = \sqrt{z^2-R^2} = \sqrt{z^2-(a - z)^2}[/tex]

should have been

[tex]r = \sqrt{z^2+R^2} = \sqrt{z^2+(a - z)^2}[/tex]

and now I'm looking at a nasty integral. I'm sure that comes as no surprise to you.
 
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  • #6
OlderDan
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I believe the integral that needs to be done is

[tex]\int_{a}^{0} \frac{a - z}{\sqrt{z^2+(a - z)^2}}dz [/tex]

and I believe the result is

[tex] \frac{a}{2\sqrt{2}}ln\left[\frac{\sqrt{2}-1}{\sqrt{2}+1}\right] [/tex]

I did not actually do the integral. I let the computer do it. If you agree that it is the integral needed check this site and make sure that I have reduced the result correctly.

http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=advanced#reply [Broken]
 
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