# If we have a net negative charge in a spherical conductor, why negative charges goes to the surface?

So imagine that we have a set of net negative charges, in physics books they said that this set of charges goes to the surface because they repel each other in such a way that this reach to an electrostatics equilibrium when this set of negative charges sits on the surface. This must create a negative potential energy in the surface of the sphere, so imagine that we have a negative charge in the middle or anywhere inside the sphere. By definition, negative charges tends to move to higher potential energy position, so why they are still going to the surface? • physics.stackexchange.com/questions/232447/…
– Buzz
Mar 4, 2021 at 21:22
• I search for similar questions, but they can't explain why an electron goes to a lower potential energy state. Mar 4, 2021 at 21:28

I was a bit skeptical of this myself when I first heard this so I made the following simulation. Every red ball is a positive charge. Every charge is confined to be inside a sphere (they can't go outside a certain radius) but apart from that they are free to move. You might think that the charges on the surface would push the newly spawned charge to the center but counterintuitively the net result is that the charge is pushed to the side.

This is a result from a theorem (I don't know what's it called) that says functions obeying the laplace equation $$\nabla^2V=0$$ can only have a maximum/minimum on the boundaries of their domain or at infinity. Since $$\vec E=-\nabla V$$ for the electric potential and $$\nabla\cdot \vec E=4\pi \rho$$ we have that $$\nabla\cdot(\nabla V)=\nabla^2V=0$$ inside a vacuum. So if you have a distribution of charges on the boundary of a sphere the potential doesn't have a minimum at the center. The best you can have is a saddle point. So if you're a particle at the center trying to find the lowest potential there is always an escape route to somewhere outside of the sphere.

The potential is hard draw in 3D but in 2D this is possible if you use the 2D potential $$V(r)\propto-\ln r$$ This looks something like this Here you see 8 charges placed in a circle and their generated 2D potential. In the middle you see a saddle point: if you place a charge exactly at the center it would remain stationary, but any deviation from that and it would accelerate towards the boundary and off to infinity. It's hard to see on this picture that it's a saddle point and not a minimum but trust me.

Here is the Mathematica code for the second plot
V[r_] := -Log[r]
R = Norm[{x - Re[#], y - Im[#]}] & /@ Exp[2 \[Pi] I Range[1, 8]/8];
Vtotal = Total@V[R];
range = 1.5;
Plot3D[Vtotal, {x, -range, range}, {y, -range, range},
PlotRange -> {-6, 2}]

• Is the theorem you’re talking about Earnshaw’s Theorem? Mar 5, 2021 at 2:26
• @Sandejo Yes, thanks. I forgot its name :) Mar 5, 2021 at 8:30
• Hey, could you share your matlab code for the 2nd simulation? Mar 11, 2021 at 0:35
• @MarceloEncisoJure I added the code. It's Mathematica but I think it could be easily converted to Matlab. I'm glad you asked for the second plot because the code for the first one is a mess and should never see daylight. Mar 11, 2021 at 11:20

"negative charges tends to move to higher potential energy position, so why they are still going to the surface?"

The surface is at no lower a potential than the rest of the sphere. We can show from the inverse square law that charges uniformly spread over the surface of a sphere produce zero resultant field at any point inside the sphere. So the potential is the same at all points inside the sphere. Any charges placed inside the sphere later will mutually repel and finish up on the surface.

This argument doesn't hold for the field inside a body bounded by a non-spherical surface. But if the body is made of a conducting material there is zero field after a very short time from placing of charges for a different reason. The charges when first introduced will indeed set up electric fields and free charges will move accordingly. But an equilibrium set-up will soon be reached, when charges have stopped moving (and reside on the surface). If there were any electric field inside the conducting body the free charges would be moving. So no (macroscopic) electric field in a conducting body in equilibrium. Nothing to stop new charges introduced from repelling each other and finishing up on the surface.

The question relates only to a charged conductor without current flow. Many students have observed the effects of static electricity generated in the classroom. A test subject stands on an insulator touching a conductive ball which becomes charged from a moving carpet belt and an electrode. Hair standing on end demonstrates that accumulating static electric charge does in fact move to points on the surface of the conductor, in this case a person. The hair remains standing until the subject dismounts the insulator and POP, the charge is discharged.

Gauss's law advises if there is net free charge in a conductor an electric field is produced within the conductor. Excess charge in a conductor is free to move and so move to the surface, spreading out as far as possible. At equillibrium q(inside) = 0 since E=0. All the charges inside the Gaussian surface sum to zero so any charge on the conductor is in fact on the surface.

A useful analogy is that the force of gravity at the center of the earth is zero since it is surrounded by mass and we feel gravity is strongest at the surface of the earth. Likewise the static charge in the center of the conductive sphere is zero.

I think you are confusing the potential with the potential energy. The potential is a function of position that tells you how much potential energy a charge would have at that position. The potential energy is a function defined on the configuration space that tells you the potential energy of a configuration of the system.

Systems tend to move to configurations with lower potential energy, regardless of the details of the system. Looking at the example of charges in an electric field, positive charges tend to move to positions with lower potential, and negative charges tend to move to positions with higher potential. However, both of these tendencies correspond to lower potential energy.

Going back to your question about why the negative charges go to the surface of a conductor, the reason is that this is not a configuration with higher potential energy, it is the configuration with the lowest potential energy. Another way to think about this is to consider charge symmetry, which tells us that the laws of physics do not change if we flip the sign of every charge, so we would expect the same situation with a negatively charged conductor as with a positively charged conductor: all the charge goes to the surface.