# (Challenging) Conceptual Question on Induced Charge on Conductor

Here is the question that we have been debating for the past day or so:

Take a conducting cubic box and center a conducting sphere with charge +q inside of it. Will the induced charge density on the sphere be uniform or not?

My gut instinct is no. The cube and sphere have different symmetries, so if the density were to be uniform, that would have to be true in general for spheres inside of conducting cavities.

On the other hand, if you imagine replacing the sphere with a point charge, the induced charge on the cube will be identical in that situation. It seems like you could make the argument that this is just an "image charge"-like configuration that matches our boundary conditions so we are done. However, I think that this actually does not quite work since we are either fixing the potential on the sphere (erroneously) or changing our charge density in the region of interest.

So I am leaning towards non-uniform, with that reasoning. Any thoughts on this? It's a surprisingly difficult conceptual question, and I would love a fully satisfactory answer...

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mfb
Mentor
On the other hand, if you imagine replacing the sphere with a point charge, the induced charge on the cube will be identical in that situation.
Why?

The potential difference between sphere and cube is constant, but the distances between the conducting parts is not. Therefore, the electric field strength and the charge density are not constant. In the extreme example of a sphere which nearly touches the box, nearly all your charge will accumulate close to those "nearly touching"-points.

Why?

The potential difference between sphere and cube is constant, but the distances between the conducting parts is not. Therefore, the electric field strength and the charge density are not constant. In the extreme example of a sphere which nearly touches the box, nearly all your charge will accumulate close to those "nearly touching"-points.

My thoughts were: the E-field produced by a uniformly charged sphere will be the same as the E-field produced by a point charge, by Gauss's Law. As long as the sphere remains uniformly charged, this is true, isn't it? If it were a uniformly charged insulating sphere, I think this would be the end of the story.

However, given that it is a conducting sphere, it's charges will re-arrange based on the "applied" field, which really is from the induced charges. I would imagine that for the "first instant" though, that you could basically say the problem is analogous to the point charge. Instantly after, though, things will get messed up.

I don't know how to wrap my head around this or rectify any of this with your example. For your extreme case, how do we know that that is lower energy than leaving the inner sphere uniformly charged and moving around charges on the conducting box?

mfb
Mentor
My thoughts were: the E-field produced by a uniformly charged sphere will be the same as the E-field produced by a point charge, by Gauss's Law. As long as the sphere remains uniformly charged, this is true, isn't it? If it were a uniformly charged insulating sphere, I think this would be the end of the story.
Sure.

What is a "first instant"?

I don't know how to wrap my head around this or rectify any of this with your example. For your extreme case, how do we know that that is lower energy than leaving the inner sphere uniformly charged and moving around charges on the conducting box?
Both happens.
It is easier to look at the conditions an equilibrium has to satisfy. If you go from the sphere to the box on an arbitrary path, you have to see the same potential difference as both sphere and box are conducting. As an example, let your box have a side length of 1 m and the sphere have a diameter of 0.998 m. We compare the two paths "across a 1mm gap" and "from one corner to the closest point of the sphere (~70cm)", both in straight lines so by symmetry we follow field lines. No matter how exactly the field looks along those lines, the electric field strength at the surface of the sphere will be much larger at the shorter path. That means a larger charge density there.

Sure.

What is a "first instant"?

Both happens.
It is easier to look at the conditions an equilibrium has to satisfy. If you go from the sphere to the box on an arbitrary path, you have to see the same potential difference as both sphere and box are conducting. As an example, let your box have a side length of 1 m and the sphere have a diameter of 0.998 m. We compare the two paths "across a 1mm gap" and "from one corner to the closest point of the sphere (~70cm)", both in straight lines so by symmetry we follow field lines. No matter how exactly the field looks along those lines, the electric field strength at the surface of the sphere will be much larger at the shorter path. That means a larger charge density there.

Love it. Thanks for the help. Just to re-iterate to make sure that I understand:

1. Both the sphere and the cube are equipotential surfaces.

2. Therefore, ΔV is constant between any point on the the sphere and any point on the surface.

3. We know from electrostatic boundary conditions that the field will point directly from the point on the sphere closest to the wall, to the center of the wall, along those six directions.

4. Since ΔV=-∫E⋅dl, and that is the shortest possible path, E must be largest near those six points, and therefore, the induced charge density must be greatest at those six points.

5. All of this is true regardless of how charge distributes itself on the cube, since we never used that. However we would expect the charge density on the cube to essentially "match" the charge density of the sphere. (i.e. at those six points on the cube, we would expect the charge density on the cube to be greatest).

Sound right?

mfb
Mentor
Right.