- #1
musemonkey
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1. Find the electric field at the tip of a cone of height and radius R with uniform surface charge density \sigma.
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. Homework Equations
The field from a hoop of radius z, charge q, at height z above the hoop center is
E_{hp} = \frac{q}{4\pi\epsilon_0} \frac{1}{2\sqrt{2}z^2}.
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
dq = 2\pi z \sigma ds = 2\pi z \sigma \sqrt{2} dz
where ds = \sqrt{2} dz is a differential arc length along the side of the cone.
Each hoop contributes to the field
dE = \frac{dq}{4\pi\epsilon_0} \frac{1}{2\sqrt{2}z^2}
= \frac{2\pi z \sigma \sqrt{2} dz}{4\pi\epsilon_0} \frac{1}{2\sqrt{2}z^2}
= \frac{\sigma}{4\epsilon_0} \frac{dz}{z},
and the field is then
\int dE = \int_0^R \frac{\sigma}{4\epsilon_0} \frac{dz}{z} = \left . \frac{\sigma}{4\epsilon_0} \ln z \right |_R^0,
which blows up at 0.
Correct? If so, what to make of it? Special surfaces can mimic point charges?
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. Homework Equations
The field from a hoop of radius z, charge q, at height z above the hoop center is
E_{hp} = \frac{q}{4\pi\epsilon_0} \frac{1}{2\sqrt{2}z^2}.
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
dq = 2\pi z \sigma ds = 2\pi z \sigma \sqrt{2} dz
where ds = \sqrt{2} dz is a differential arc length along the side of the cone.
Each hoop contributes to the field
dE = \frac{dq}{4\pi\epsilon_0} \frac{1}{2\sqrt{2}z^2}
= \frac{2\pi z \sigma \sqrt{2} dz}{4\pi\epsilon_0} \frac{1}{2\sqrt{2}z^2}
= \frac{\sigma}{4\epsilon_0} \frac{dz}{z},
and the field is then
\int dE = \int_0^R \frac{\sigma}{4\epsilon_0} \frac{dz}{z} = \left . \frac{\sigma}{4\epsilon_0} \ln z \right |_R^0,
which blows up at 0.
Correct? If so, what to make of it? Special surfaces can mimic point charges?