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Electric field at tip of uniformly charged cone

  1. Nov 23, 2008 #1
    1. Find the electric field at the tip of a cone of height and radius R with uniform surface charge density [tex] \sigma [/tex].

    I get that the field diverges at the tip, which is puzzling because it's not as though there's a point charge at the tip. I thought this sort of thing can't happen when you treat charge as smeared over a surface.

    2. Relevant equations

    The field from a hoop of radius z, charge q, at height z above the hoop center is

    [tex] E_{hp} = \frac{q}{4\pi\epsilon_0} \frac{1}{2\sqrt{2}z^2} [/tex].

    3. The attempt at a solution

    I break the cone into hoops of variable radius. Because it's a right cone, the distance from the tip to each differential hoop equals the radius -- very convenient. The charge on each hoop is

    [tex] dq = 2\pi z \sigma ds = 2\pi z \sigma \sqrt{2} dz [/tex]

    where [tex] ds = \sqrt{2} dz [/tex] is a differential arc length along the side of the cone.

    Each hoop contributes to the field

    [tex] dE = \frac{dq}{4\pi\epsilon_0} \frac{1}{2\sqrt{2}z^2} [/tex]

    [tex] = \frac{2\pi z \sigma \sqrt{2} dz}{4\pi\epsilon_0} \frac{1}{2\sqrt{2}z^2} [/tex]

    [tex] = \frac{\sigma}{4\epsilon_0} \frac{dz}{z} [/tex],

    and the field is then

    [tex] \int dE = \int_0^R \frac{\sigma}{4\epsilon_0} \frac{dz}{z} = \left . \frac{\sigma}{4\epsilon_0} \ln z \right |_R^0 [/tex],

    which blows up at 0.

    Correct? If so, what to make of it? Special surfaces can mimic point charges?
  2. jcsd
  3. Nov 26, 2008 #2


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    Your derivation looks fine to me.

    From what I remember, electric fields do diverge for sharp tips. In practice, a real "tip" will be rounded off, even if at the molecular scale. The actual field depends on the radius-of-curvature for the tip.

    Still it seems a rather odd question to ask, which is essentially show that E is infinite for a particular sharp tipped object.
  4. Nov 26, 2008 #3
    I've read that high fields arise at sharp tips of charged conductors because the surface charge density is much greater there. But since this is not a conductor but a uniformly charged surface, I'm not sure the two cases are actually related.

    Your point about curvature is very interesting. I'll look into that. Thanks and thank you for checking my derivation!
  5. Apr 1, 2009 #4
    check Jackson pg 104
  6. Apr 1, 2009 #5


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    That is correct. However, a sharp tip is still a sharp tip, and the electric field will basically be pointing away from the surface, so it must spread out from that point.

    I agree, this is a strange question to ask.
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