# Which charges pushes exactly a charge inside a conductor to the surface and why does it prefer curved surface?

I know this has a lot of questions about it, however my question is different: why does a charged particle inside a conductor get pushed toward the surface?

The usual explanation is that in a conductor there can't be an electric field or else the charges would reorganize inside that conductor and remove the produced electric field. Usually the explanation is that considering gauss law inside a conductor there can't be any net charge because there can't be a field. However this seems very unintuitive.

My question is: what is the intution behind charge being attracted by the surface in a conductor, even though the surface is actually repelling that charge?

Imagine a conductor with a random positive charge in it like so:

How would you determine that? I guess the charges on one side repel the particle more than the other, however I can't really formalize this.

Then my second question is, why does it get pulled towards the curved part of the surface of the conductor? Again this is usually explained with two spheres connected by a cable and showing the dependence of the surface charge density on the radius of the sphere. However my question is exactly how the coulomb force produces this charge rearrangement inside the conductor?

To understand this concept it is absolutely essential that you not think of the surface charges as some pre-existing fixed charge distribution pushing on an individual interior mobile charge. All of the charges are mobile and they are all pushing on each other.

The key to remember about their joint motion is that the forces are repulsive, so they are all constantly pushed away from each other. In other words, they are trying to expand. The most expanded that they can get is to all go to the surface.

Once on the surface they are still mobile. They can slide along the surface in response to forces in the plane of the surface. As one surface charge pushes on a neighbor, the component of the force parallel to the surface is reduced as the surface curvature is increased. Thus on high-curvature regions the repulsive interactions are weaker so more charges accumulate there.

• Why is the distance between the charges maximized when at the surface? Is there a simple way of showing this? i.e. $d/dr \sum q_i q_j/r = 0$ Commented Apr 24, 2022 at 16:52
• There may be a way to show it mathematically, but intuitively they are all pushing each other outward and the surface is the furthest outward they can go.
– Dale
Commented Apr 25, 2022 at 11:23
• Wouldn't be a more symmetric configuration across a volume be possible as well? I mean obviously not or else this wouldn't happen, but that could intuitively still be possible right? Commented Apr 26, 2022 at 10:36

For your first question, there are a couple of steps to go through. Using local Ohm’s law, you have $$\vec j=\sigma \vec E$$ ($$\sigma$$ the conductivity), so plugging it into the charge conservation and using Gauss’ law: $$\frac{\partial \rho}{\partial t}+\frac{\sigma \rho}{\epsilon_0}=0$$ so in the stationary solution is trivial, with a relaxation time $$\frac{\epsilon_0}{\sigma}$$ that vanishes in a perfect conductor.

Physically, charged clumps generate electric field lines going toward them (if negative) or away from them (if positive). The current follows these lines which has the effect of depleting the clumps and leveling the distribution. The fact that the charge ends up at the boundary is therefore the simple consequence that it is the only place it can go to.

For your second question, its a bit tricky as it does not only depend on the conductor’s geometry. Indeed, outside charges can displace the surface charges, and the general behavior is non local.

Hope this helps and tell me if you find some mistakes.