Density of repulsive particles on a disk? NB: the external circle (where particle or water is in it) is fixed !
It's a 2D theoretical problem (just for understand something in particular). Here, no gravity, no temperature. I put in a disk a lot of repulsive charges from center. Particles have the same charge +1. OK, if I put 40 particles in the same time, all will go at outer circle. But, if I add later 2 another particles why they will go to outer circle? I can repeat this time after time and add a lot of particles that never go to outer circle. So my questions are:
1) If I'm wrong, can you explain how new 2 particles (for example but it can be one particle) will go to outer circle (it's possible to imagine big circle).
2) If I'm wrong at (1), and if particles have a radius (not a point particle) it's not possible to put infinite particle at outer circle, so new particles need to take another layer, no?
So if I add and add particles (with radius not 0) all the sphere can be fully at 70 % for example. And my last question is:
3) How is the density? For me, the density (number of particles for one volume) increase with radius but I'm not sure.
Maybe this problem is already studied and I would appreciate some information about if you have.
Edit for first reply:
4/ Thanks for links :) For reply to your first paragraph (I'm not sure to understand): I'm agree that all particles at circumference are free to move but if I add 2 another particles, they repulsive themselves, until when ? each new particle has force from another new particle (in one direction) and force from all others particles (in the contrary direction), for me at a moment these forces are equal (absolute value) and new particle stop to move somewhere from the center to the outer circle. It's evidence if I add a single new particle in the center, why it would move at outer circle, all forces on this new particle want to let it in the center.
5/ If density change with radius, not linear. Now I add in the big disk an object compose of 2 vacuum disks, please look: . Disks have same radius but are not at the same radius inside big disk. These vacuum disks are fixed together, this object (2 vacuum disks) can turn around the center of big disk. Each vacuum disk take place of particles. This change the force on each disk. One disk has more force from another due to the lack of particles (it's not the same density). So, even this change the global density everywhere (I don't drew all forces), for me this give a torque on the object. Sure, it's not possible but I would like to understand how particles do for have a torque at 0 ? Maybe you have another link for help me ?
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With gravity it's easier to study, 2 vacuum object or one only:

It's evidence that force for sphere of gas is only radial (Archimedes) because all is symmetrical. So even the sphere of gas is attached to the sphere of water this don't change the torque. The sphere of water has less force due to the lack of matter of the sphere of gas. The pressure around sphere of water changing due to the lack of matter of the sphere of gas but the density change with the radius and change due to the presence of sphere of gas. The density is not linear and the law of attraction is not linear $k/d^2$ too, how the force of attraction and the force of pressure can be the same ?


or:


easier to calculate with:

the cosine function is nonlinear and it is use more in the calculation of one force than other, this can't be compensate by law in $k/d^2$. And the asymmetry from the circle ?

Like this it's easier to understand there is a problem with torque. Density is not the same everywhere, greater with small radius. And the position in the circle of spheres of gas give not the same height of water above them.


More easier, with walls like part of circle ! Can't give a big torque at top even there is a difference of pressure. In the contrary, the other give a torque.

 A: Interesting question, +1.
It's not clear to me why you think that (1) you can get an initial equilibrium with all charges on the circumference, or (2) once some charges are deposited on the circumference, there will be a stable equilibrium when further charges are deposited on the interior. Keep in mind that the ones on the circumference are still mobile.
You may be interested in this: Charge density of a conducting needle. David J. Griffiths and Ye Li. Am. J. Phys. 64 no. 6 (1996), p. 706, http://www.colorado.edu/physics/phys3320/phys3320_sp12/AJPPapers/AJP_E&MPapers_030612/Griffiths_ConductingNeedle.pdf
As in the Griffiths paper, your disk must have some physical shape that is modeled as a limit of some three-dimensional shape. If we model it as an ellipsoid that gets flatter and flatter, then we can use the result by Lord Kelvin, who proved that the charge density on a conducting ellipsoid is proportional to the distance from the center to the tangent plane. In your limit, the charge density vanishes except at the edge. However, Griffiths points out that (1) the result may be different depending on what model you take the limit of, and (2) the results for finite numbers of charges may not be the same as the results for a continuous distribution.
See also:
https://math.stackexchange.com/questions/112662/gaussian-curvature-of-an-ellipsoid-proportional-to-fourth-power-of-the-distance
I W McAllister 1990 J. Phys. D: Appl. Phys. 23 359 doi:10.1088/0022-3727/23/3/016 (proves that charge density is related to Gaussian curvature for 11 special surfaces)
