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As the title says. For example, if I take a fat metal wire with non-negligible thickness, and then wind it into a spiral, but such that none of the parts touch one another (there are no topological 'holes'), then if I add a charge to the wire, would it equally distribute over the wire? Or would the charges from neighboring 'loops' of the spiral interact with one another somehow, similar to a capacitor almost?

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  • $\begingroup$ Do you mean that you place some charge $q$ ,not a unit charge, on the wire? $\endgroup$
    – Wang Yun
    Dec 8, 2015 at 10:57
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    $\begingroup$ The electric field of - for example a charged needle - is on the tip of a needle stronger as in the other areas $\endgroup$ Dec 8, 2015 at 11:42

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It is the electrostatic conditions that are more fundamental than determining the charge distributions, namely, that the field has to be inside a conductor, otherwise, the charges will be flowing and there won't be any electrostatics to study.

So, if you leave the conducting wire alone, it will come to electrostatic conditions and whatever will be the charge distribution that could match that would be the one that'll be present.

It just turns out that when you study simple conductor systems like one spherical conductor, it is so that the electrostatic condition is achieved when there is uniform distribution.

So, finally the answer to your question is that yes other loops will definitely interfere.

Also, these are all ideal cases, every conductor's molecules and atoms and atoms have some bonding energy which the electric field has to always overcome, so no perfect conductor will have absolutely 0 elecrtic field inside, but yes, to a great extent they can be approximated. So, you can define a conductor by a material whose binding energy of electrons is less(depending on your tolerance of error) and can approximate it using ideal conductor which are an extremal mathematical model following Maxwell's equations.

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I think that the either ends of your wire no matter how you fold them have higher amount of surface density of charge than the middle, however, the amount of charge is different as follow.

Imagine your wire to be approximated by the figure below enter image description here

The first three spheres have the same radius and the rest have different ones. Same potential dictates that:

$$\frac{Q_1}{r_1}=\frac{Q_2}{r_1}=\frac{Q_3}{r_1}=\frac{Q_4}{r_2}=\frac{Q_5}{r_3}=\frac{Q_6}{r_4}=\frac{Q_7}{r_5}$$

Hence you see that the charge on the first three spheres are the same and it changes as you move to the last sphere. This as a result gives a higher surface charge density on the smaller sphere.

If you need a perfectly uniform distribution you need a very long wire.

About the curves in your wire I should say they change the distribution because geometry of is very important.

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  • $\begingroup$ How can you say that $\frac{Q_1}{r_1} = \frac{Q_2}{r_2}$ etc. since the potentials of each of the spheres will be affected by all the other spheres close to it and that must also be taken under consideration. I agree that their potentials will be equal, but their potentials won't be $\frac{Q_1}{r_2}$ etc, they'll be something else. $\endgroup$ Apr 4, 2021 at 12:41

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