# Linear Charge Distribution on a Needle?

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1. Jun 19, 2015

### TheDemx27

http://www.colorado.edu/physics/phy...MPapers_030612/Griffiths_ConductingNeedle.pdf

I was reading this paper, and was confused by a result in section 2-A. (Heck they even mention they weren't expecting it themselves). The purpose of the paper is to find the linear charge density for a wire and they use several models, the first of which treats the wire like an ellipsoid. They end up with an expression that doesn't depend on $x$, $y$, or $z$, but only on $Q$ and $a$:

$\lambda(x)=\frac{Q}{2a}$​

Meaning that the charge density is constant as you move along the x-axis.
Its pretty crazy how nicely things simplify in the paper.

Certainly there must be something wrong with this ellipsoid model, since we know that the charge is supposed to collect at the ends of an object. (right?) I mean, that's how static wicks on planes operate. Not only that, but every other model used in the paper produces a charge distribution you would expect: Higher charge density near the ends of the object.

I'm pretty sure that using ellipsoids like they did isn't a good way to model this judging by the discrepancy in the results. This is also coupled with the fact that it is counter intuitive for me.

Is this model really correct?
Are ellipsoids really mathematically special objects that have linear charge distributions?

Thanks in advance to anyone who can clear things up.

2. Jun 19, 2015

### Staff: Mentor

I can imagine that different shapes lead to different answers. A cylinder clearly leads to a non-uniform distribution at least at the ends, but the ellipsoid has a different tilt for its surface (=electric fields on the surface are a bit tilted outwards), so more charge closer to the center (compared to the cylinder) sounds reasonable. Interesting to see a constant result.

Hmm, checking the ellipsoid probably requires reference 2.

Edit: A constant charge density plus two extra charges at the end does not give a uniform potential in between. The line has to be more complicated.

Computing power increased by orders of magnitude in the last 20 years, it should be possible to run much better simulations.

Last edited: Jun 19, 2015
3. Jun 19, 2015

### marcusl

Ellipsoids are indeed special. Polarization in a cylinder in a uniform E field has no analytic solution, likewise magnetization of a ferrous rod in uniform B. Demagnetizing fields and all sorts of approximations were invented (in days before computers) to handle it. A ferrous ellipsoid in uniform B has an exact and simple solution, however--B and M inside are perfectly uniform! It's not so surprising that rho is uniform as well.

4. Jun 20, 2015

### TheDemx27

Thats good to know!

But back to the paper, is the ellipsoid model really applicable in this context? i.e. does it really make sense to treat the ellipsoid as a one dimensional line charge like they do by taking the $\lim_{b, c \to 0}(\frac{Q}{2a})$? Sure it makes sense from a purely mathematical point of view, but the results are completely different, and its not even really a 3 dimensional figure anymore.

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