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Electric Potential of Hollow Cone on z-axis

  1. Jan 30, 2013 #1
    1. The problem statement, all variables and given/known data

    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])?


    3. 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 in to 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: Jan 30, 2013
  2. jcsd
  3. Jan 30, 2013 #2
    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.
     
  4. Jan 30, 2013 #3

    haruspex

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    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.
     
  5. Jan 30, 2013 #4
    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 [itex]\frac{2}{\sqrt{2}}[/itex]. Thank you.
     
  6. Jan 30, 2013 #5

    haruspex

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    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).
     
  7. Feb 14, 2014 #6
    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 [itex]\equiv[/itex]h/a
    then voltage at some height d is [itex]\frac{σdm}{2ε(1+m)^{1.5}}[/itex](m arcsinh ( [itex]\frac{m^2+1}{dm}[/itex] h-m) - m arcsinh(-m)+cosh(arcsinh([itex]\frac{m^2+1}{dm}[/itex] h -m))-cosh(arcsinh(-m)))

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