# Charge Distribution and Stability in a Conductive Solid Sphere

My friend came to me with a simple question: What is the charge distribution on a conductive solid sphere? Of course, I answered: 'Since the solid sphere is conductive, the electric potential would be the same everywhere inside the solid sphere, hence there would be no charge inside, only on the surface of the solid sphere.'

But she pointed out if the charge were evenly distributed on the surface of the solid sphere, when we introduce a new charge in the center of the sphere, the system is in equilibrium. So one might argue this would be a unlikely yet possible distribution.

This question is discussed here.

Now I understand the center charge is in an unstable equilibrium and would not stay there. However, I was unable to resolve the following questions:

1. Showing net charge resides only on the surface minimizes potential energy of the system.
2. Showing the center is an unstable equilibrium through differential equations.

Any help would be welcomed.

In a static state, there can be no charge inside a solid conductor. This means the charge at the center will be in unstable equilibrium. An infinitesimal displacement of the charge will produce a reorientation of the surface charges to produce a force on the charge, which will quickly migrate to the surface.

• I see! Do you mean in a very short amount of time after the infinitesimal displacment occurs, the electric field inside the conductor would not be 0?(because of the dispalcemnt of the surface charge) Commented Feb 22 at 23:15
• No. The surface charge always orients it self to keep E equal zero inside. For a sphere, the final surface charge will be uniformly distributed. Commented Feb 23 at 0:50
• I think Jerold Franklins comment must be a misunderstanding. Danny wen explicitly assumes that the redistribution momentarily produces a field inside the sphere that instantly moves the charge to the surface. This must be so - if no field no force and no movement.
– Jens
Commented Feb 25 at 18:55
• Yes again. There will be an E field that moves the charge to the surface in about 10^{-11} seconds. I didn't realize he was talking about that brief instant. – Commented Feb 26 at 2:52
• Yeah I was indeed a little confused. Thanks for clearing it up. Commented Feb 26 at 21:56

Adding that there is originally a charge Q2 on the sphere, per the OP's edit: If you put a charge Q1 in the center of the conductor, either perfectly centered or just hold it in place so it can't move, other charge totaling -Q1 will flow towards it and shield it. In the immediate vicinity of charge Q1, the net charge will be zero because of this.

If the conducting sphere has charge Q2 (apart from the charge Q1 placed in the middle), then there will be charge totaling Q1 + Q2 on the surface of the sphere. It doesn't matter how you add the charge to the conducting sphere, all that matters is the total charge in the sphere.

I guess I should add, since the added charge Q1 is shielded by the charge -Q1 that flowed to it, it doesn't matter where in the sphere that charge and the shielding charge ends up. The total charge in the immediate vicinity of Q1 is 0, and the field inside the sphere is 0. So it's not an unstable equilibrium, it's just in equilibrium.

• Sorry, I was not clear in the question discription, I was discribing a system with solid sphere with charge, and we introduce another charge in the center. I edited the question and added a graph. Commented Feb 22 at 0:48
• The answer still holds. Electrostatic equilibrium will still require that $\vec{E}=0$ everywhere in the conductor, except where you are fixing the charge. If you do Gauss's law with a Gaussian surface anywhere inside the conductor you will have $\vec{E} =0 \rightarrow Q_\text{enc} =0$. So if you force a charge into the center, a "shielding" negative charge must form around it, such that the net enclosed charge is zero. Commented Feb 22 at 0:59
• @BenH We are using the fact that$\overrightarrow{E}=0$ on the Gaussian surface directly follows from electrostatic equilibrium. Which is what I'm trying to reason in the first place. The closest argument I can come up with is this: If a positive charge is unfixed in the center, it creates a positive field, that would attract electron in the conductor to cancel it out. Once an electron does, the positive charge moved to where the electron was. So it does a Brownian motion, eventually making contact with the surface of the conductor and stays there for some reason. Does this make sense? Commented Feb 22 at 4:16
• The consequence of the assumption of “electrostatic equilibrium” for a conductor is that each of the free charges within the metal have a net electric force of zero. If they did not, then they would accelerate and not be in equilibrium, which contradicts our assumption (physically we assume that charges will redistribute until they come to equilibrium). So the statement that $E=0$ inside is an immediate consequence of the assumption. It requires no further proof. Commented Feb 22 at 10:17
• Your question is not completely clear from the drawing... Do you mean the charges are both arbitrary? Or is there a relation like $Q_1$ equals $-Q_2$ (or $\frac12 Q_2$, or some other special relation?) Commented Feb 22 at 13:18

For a solid sphere, as the title now indicates, there is no equilibrium at all with a charge in the middle because a conductive solid is full of free charges and they would not be in equilibrium, neither stable or unstable. They would immediately rearrange to neutralize the centre charge.