# Electric Potential of Hollow Cone on z-axis

• Opus_723
In summary, the conversation discusses finding the potential of a conical surface with a uniform surface charge density at a point on the z-axis. The solution involves breaking the cone into rings and using integration to find the potential. The final answer is found to be dependent on the height and radius of the cone, with the voltage decaying as the height of the cone becomes large compared to its radius. The general form for the potential in terms of the cone's slope is also given.
Opus_723

## 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 $\sigma$. What is the potential of this cone at a point z on the z axis (i.e. at the point $\vec{r}$=z$\hat{z}$)?

## 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 = $\sigma$da = $\sigma$*2$\pi$*z*$\frac{2}{\sqrt{2}}$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 $\sqrt{(H-z)^2+z^2}$

So the integral we need is

$\frac{1}{4\pi\epsilon}$$\int^{h}_{0}$$\frac{\sigma*2\pi*z\frac{2}{\sqrt{2}}dz}{\sqrt{(H-z)^2+z^2}}$

-------------------------------------

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:

$\frac{\sigma}{2\epsilon}$$\int^{h}_{0}$$\frac{zdz}{\sqrt{(z-\frac{H}{2})^2+\frac{H^2}{4}}}$

Next, I tried a hyperbolic trig substitution. Namely, z-$\frac{H}{2}$= $\frac{H}{2}$sinh$\theta$, with dz = $\frac{H}{2}$cosh$\theta$

Substituting in, I got:

$\frac{\sigma}{2\epsilon}$$\int$$\frac{\frac{H^2}{4}(sinh\theta+1)cosh\theta}{\sqrt{\frac{H^2}{4}sinh^2\theta+\frac{H^2}{4}}}$

Which simplifies to:

$\frac{\sigma}{2\epsilon}$$\int$$\frac{\frac{H}{2}(sinh\theta+1)cosh\theta}{coshθ}$

And further to:

$\frac{H\sigma}{4\epsilon}$$\int$$sinh\theta+1$

And, finally taking the integral:

$\frac{H\sigma}{4\epsilon}$($cosh\theta+\theta$)

Going back to my original substitution, $\theta$= arcsinh($\frac{2z}{H}$-1)

So I sub back into get:

$\frac{H\sigma}{4\epsilon}$$(cosh(arcsinh(\frac{2z}{H}-1))+arcsinh(\frac{2z}{H}-1))$

And then I evaluated from 0 to h (little h being the height of the cone)

$\frac{H\sigma}{4\epsilon}$$[cosh(arcsinh(\frac{2h}{H}-1))+arcsinh(\frac{2h}{H}-1)-cosh(arcsinh(-1))-arcsinh(-1)]$

Or

$\frac{H\sigma}{4\epsilon}$$[cosh(arcsinh(\frac{2h}{H}-1))+arcsinh(\frac{2h}{H}-1)-\sqrt{2}-ln(\sqrt{2}-1)]$

-------------------------------------------

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:
Well, I figured out what was happening with mathematica, and the new plot looks perfect. Still, I would really appreciate a check on this work. Also, I would like to be able to see the limit on this algebraically, but every way I try to manipulate it it looks indeterminate.

Early on, you lost reference to h, the height of the cone. The potential should decay as H becomes large compared to h.
Also, you've understated the area of the ring by a factor. The surface of the ring slopes.

I believe h came back in when I evaluated the antiderivative at the limits of the integral. I dropped it for awhile so that it wouldn't be confusing with the change of variable. Is that not correct?

You're obviously right about the slope of the cone though, that would introduce another constant factor of $\frac{2}{\sqrt{2}}$. Thank you.

Opus_723 said:
I believe h came back in when I evaluated the antiderivative at the limits of the integral. I dropped it for awhile so that it wouldn't be confusing with the change of variable.
Yes, sorry, I missed that. So if you look at your penultimate line and let H→∞ you'll see it does tend to 0. Writing δ=2h/H, the leading term after cancellation from the two cosh terms is -δ/√2, and that from the two angle terms is +δ/√2. These cancel to leave only terms O(δ2). The factor outside the integral combines with this to give O(δ).

This may be an old question of no interest but in case anyone is interested l think l have an answer for the more general situation where the cone is of height h and radius a (so the slope doesn't necessarily equal 1) as well as the particular solution given above with correct factors of 2.

General form:
define slope to be m $\equiv$h/a
then voltage at some height d is $\frac{σdm}{2ε(1+m)^{1.5}}$(m arcsinh ( $\frac{m^2+1}{dm}$ h-m) - m arcsinh(-m)+cosh(arcsinh($\frac{m^2+1}{dm}$ h -m))-cosh(arcsinh(-m)))

If h=a (m=1):
then voltage at some height d is $\frac{σd}{ε(2)^{2.5}}$ (arcsinh ( 2h/d-1) - arcsinh(-1)+cosh(arcsinh(2h/d-1))-cosh(arcsinh(-1)))

## What is the Electric Potential of a Hollow Cone on the z-axis?

The Electric Potential of a Hollow Cone on the z-axis is the measure of the electric potential at any point on the z-axis due to a hollow cone-shaped object with a uniform charge distribution.

## What is the equation for calculating the Electric Potential of a Hollow Cone on the z-axis?

The equation for calculating the Electric Potential of a Hollow Cone on the z-axis is V = kQ * (1/R1 - 1/R2), where k is the Coulomb's constant, Q is the total charge of the cone, R1 is the distance from the apex of the cone to the point of interest, and R2 is the distance from the base of the cone to the point of interest.

## How does the Electric Potential of a Hollow Cone on the z-axis change with distance?

The Electric Potential of a Hollow Cone on the z-axis decreases as the distance from the cone increases. This is because the electric field created by the cone weakens with distance, resulting in a lower electric potential.

## What is the relationship between the Electric Potential of a Hollow Cone on the z-axis and the height of the cone?

The Electric Potential of a Hollow Cone on the z-axis is directly proportional to the height of the cone. This means that as the height of the cone increases, the electric potential also increases.

## What is the significance of the Electric Potential of a Hollow Cone on the z-axis?

The Electric Potential of a Hollow Cone on the z-axis is an important concept in understanding the behavior of electric fields and charges. It can help scientists and engineers in designing and analyzing electrical systems and devices.

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