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Consider for instance a parallel plate, a spherical or a cylindrical capacitor. Usually we analyze it by considering the surface charge density on each plate uniform, i.e. constant along the plate.

Why do we understand that? The only explanation I have found on the web is: "Because like charges repel one another".

Ok, I understand that, because of Coulomb interaction, all charges I put on a metal surface tend to separate themself in order to minimize that force, but I do not understand why should it be a uniform distribution.

Consider for instance a situation like this (reference), in which obviously the same principle ("Because like charges repel one another") is correct:

enter image description here

You see that charge density becomes most concentrated at the location of greatest curvature, it is a known effect. Charges repel themselves, but charge density is not uniform.

Well, I do not see differences with a plate of a capacitor: in both cases a charge is put on a conductor. In case of a parallel plate capacitor, it has a rectangular shape, in the last picture, it is like a warped circle. What will happen if I build a capacitor with two parallel plates with that shape?

So, I do not understand which is the connection between the capacitor geometry and surface charge density (it seems that "nice geometry" means uniform distribution...).

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No, the charge density is not uniform in an arbitrary shaped conductor. I will show the dependence of charge density on radius of an uniform sphere, from which you can get a hint about an arbitrary shaped conductor.

As you already might know that the capacitance of a single isolated spherical conductor of outer radius $b$ and inner radius $a$, is $ C = 4 \pi \epsilon_0 (\frac{1}{a} - \frac{1}{b})^{-1}$. In the limit that $b$ tends to $ \infty $ and $a$ tends to radius $R$ , $C=4 \pi \epsilon_0 R$. Then charge density $\sigma = \frac{Q}{4 \pi R^2}$, where $Q= CV$ is total charge and $V$ is voltage applied. Putting everything, we have $\sigma \propto \frac{1}{R}$.

But, as the radius is same for sphere everywhere, charge density is uniform and the same goes for cylindrical capacitor and plane capacitor ($R= \infty$). While it is not the same for an arbitrary capacitor.

Note that the images you reproduced are of 3d objects. You can't build a plane capacitor with that shape, plane one being 2d object. Even, if you get a 2d plate of that shape, it is still the same, with $R$ still being $\infty$.

Edit 1(in response to a comment): For parallel plate capacitor, $C = \frac {A \epsilon_0}{d}$, from which you can see that $\sigma$ is uniform. Note that you don't need to assume uniform surface charge density for arriving at an expression for $C$.

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  • $\begingroup$ Your answer is totally correct. But I think the OP meant something else - whether the positive/negative charge distribution on the plates of a capacitor or uniform throughout or non-uniform. More precisely, the question is not about constancy of $\sigma$ with increase or decrease in the curvature, whereas it's about the constant of $\sigma$ within the same material under normal conditions. $\endgroup$ – Guru Vishnu May 11 at 11:33
  • $\begingroup$ Why have you said that sigma = Q/4piR^2? I'll say it is a mean value for charge density, and in general not a local expression of it $\endgroup$ – Kinka-Byo May 11 at 12:09
  • $\begingroup$ @Kinka-Byo I have used a spherically symmetric object, in which case the local expression can be easily integrated over to get total surface charge density. $\endgroup$ – Rounak May 11 at 13:50
  • $\begingroup$ What would be the local expression? $\endgroup$ – Kinka-Byo May 11 at 18:01
  • $\begingroup$ @Kinka-Byo The local expression for $\sigma$ would be fairly simple as $\frac{dq}{ds}$ or you can use $\sigma = \int E.ds * \epsilon_0$. But to actually arrive at a number is somewhat tedious. The electric field can be calculated from the electrostatic potential which in turn is a solution to Laplace’s equation (with appropriate boundary conditions). But the equation can't always be solved for arbitrary shaped conductor. $\endgroup$ – Rounak May 13 at 2:04

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