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...).


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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