# Charge on a sphere

• I

## Main Question or Discussion Point

Lets say I have a conducting neutral sphere containing a spherical hollow space. The hollow space contains a point charge at its center. This setup will result in a charge equal in magnitude with opposite sign of the point charge spreading evenly over the boundary of the hollow space and a charge equal to the point charge spreading evenly over the outside of the sphere.

Now lets say the hollow space is not a sphere but some other shape. Can it still be said that the charge on the outside of the sphere will be spread evenly?

Related Other Physics Topics News on Phys.org
BvU
Homework Helper
2019 Award
Why wouldn't it ?

Delta2
Homework Helper
Gold Member
Why wouldn't it ?
Because the symmetry breaks up due to the "irregular" or asymmetric hollow space?
Well, either the charge density in the boundary of the irregular hollow space gonna be non uniform, or the charge density at the boundary of the big sphere is gonna be non uniform or both. They cant be both uniform cause then the electric field inside the big sphere wouldn't cancel out.

BvU
Homework Helper
2019 Award
the electric field inside the big sphere
There is no electric field inside the metal, irrespective of the charge distributions on the surfaces.

 ah, we say the same thing.

Delta2
Homework Helper
Gold Member
There is no electric field inside the metal, irrespective of the charge distributions on the surfaces.

 ah, we say the same thing.
Yes I agree that there is not gonna be an electric field inside the metal, but if both of charge densities are uniform then I really cant see how the electric field inside the metal is zero. I think in the most general case, none of the charge densities will be uniform.

Doc Al
Mentor
Now lets say the hollow space is not a sphere but some other shape. Can it still be said that the charge on the outside of the sphere will be spread evenly?
Yes.

On the inner surface, the charge will distribute however it needs to cancel the field from the point charge. That means there's effectively no field from the point charge to influence the distribution of charge on the outer surface. So that charge will spread out evenly. The charge distributions on inner and outer surfaces are essentially independent.

Yes.

On the inner surface, the charge will distribute however it needs to cancel the field from the point charge. That means there's effectively no field from the point charge to influence the distribution of charge on the outer surface. So that charge will spread out evenly. The charge distributions on inner and outer surfaces are essentially independent.
Why does the charge on the inner surface have to cancel the field from the point charge? I know that the field from the point charge is cancelled in the conductor, but this means that the charge on the outer surface and the charge on the inner surface together have to cancel the field from the point charge in the conductor. Why does the charge on the inner surface do this on its own?

Doc Al
Mentor
Why does the charge on the inner surface have to cancel the field from the point charge? I know that the field from the point charge is cancelled in the conductor, but this means that the charge on the outer surface and the charge on the inner surface together have to cancel the field from the point charge in the conductor. Why does the charge on the inner surface do this on its own?
I'm not sure if I can come up with a simple explanation, but it has to do with the uniqueness theorem. (Perhaps someone can chime in.) Since it's possible to arrange the charge on the inner surface to cancel the field from the enclosed charge and to arrange the charge on the outer surface to produce no field within its boundary, it can be shown that this is the only solution.

I'm not sure if I can come up with a simple explanation, but it has to do with the uniqueness theorem. (Perhaps someone can chime in.) Since it's possible to arrange the charge on the inner surface to cancel the field from the enclosed charge and to arrange the charge on the outer surface to produce no field within its boundary, it can be shown that this is the only solution.
So you are saying that if I have any hollow 3d object in space, than for any point charge inside this object there existis a configuration of charges on the surface of the object(with the same total charge) that will produce the same field outside the object as the point charge would?

BvU
Homework Helper
2019 Award
Why does the charge on the inner surface do this on its own?
In order to achieve $|\vec E|=0$ in the conductor

In order to achieve $|\vec E|=0$ in the conductor
Yes! I have read the uniqueness theorem so I can see that if this is possible it must be the solution. My question is now why it must be possible.

Doc Al
Mentor
So you are saying that if I have any hollow 3d object in space, than for any point charge inside this object there existis a configuration of charges on the surface of the object(with the same total charge) that will produce the same field outside the object as the point charge would?
No, I don't think that's true in general. (It might be! I'll have to think on it.)

Note that if the hollow object were a conductor, then the field produced by the charges on its outer surface would not necessarily produce a field identical to that of the point charge. For example: Assume a spherical conducting shell, but have the enclosed point charge be off-center. The induced charge on the outer surface would be symmetrically arranged, thus the field would be that of a point charge at the center.

No, I don't think that's true in general. (It might be! I'll have to think on it.)

Note that if the hollow object were a conductor, then the field produced by the charges on its outer surface would not necessarily produce a field identical to that of the point charge. For example: Assume a spherical conducting shell, but have the enclosed point charge be off-center. The induced charge on the outer surface would be symmetrically arranged, thus the field would be that of a point charge at the center.
Woah, my intuition is way off...why will the induced charge on a spherical shell with an off center point charge be uniform?

Doc Al
Mentor
Woah, my intuition is way off...why will the induced charge on a spherical shell with an off center point charge be uniform?
Recall that the effect of the enclosed charge's field has been neutralized by the induced charge on the inner surface of the conductor. So as far as the charges on the outer surface is concerned, they are free to move as needed to produce zero field inside: by symmetry, that's a uniform distribution of charge.

Recall that the effect of the enclosed charge's field has been neutralized by the induced charge on the inner surface of the conductor. So as far as the charges on the outer surface is concerned, they are free to move as needed to produce zero field inside: by symmetry, that's a uniform distribution of charge.
Oh, I thought you were talking about a spherical shell, in the sense that it is infinetly thin( so there would be no inner and outter charge)

Doc Al
Mentor
Oh, I thought you were talking about a spherical shell, in the sense that it is infinetly thin( so there would be no inner and outter charge)
To realistically model a conducting shell, it's got to have some thickness, thus an inner and outer surface.

ZapperZ
Staff Emeritus
Oh, I thought you were talking about a spherical shell, in the sense that it is infinetly thin( so there would be no inner and outter charge)
But is this THE spherical shell that you used in your very first post?

The spherical shells that we deal with in General Physics classes and advanced undergraduate E&M are not "infinitely thin" if we deal with charges or induced charges on it.

Zz.

But is this THE spherical shell that you used in your very first post?

The spherical shells that we deal with in General Physics classes and advanced undergraduate E&M are not "infinitely thin" if we deal with charges or induced charges on it.

Zz.
No, in the fiest post I meant a "full" sphere with a cavity

I guess I'll have to work with "rules" until my intuition gets better(through physical intuition or a more rigorous understanding of the underlying math) even though I dont like this approach. So I will take away from this that in the case where I have a conductor with charge q1 and a cavity containing a charge q2, the charges on the inside layer will arrange to completely cancel the field outside of the cavity, and the charges on the outside layer will arrange as if the object did not contain a cavity and had net charge q1+q2. Sound about right?

Doc Al
Mentor
I guess I'll have to work with "rules" until my intuition gets better(through physical intuition or a more rigorous understanding of the underlying math) even though I dont like this approach. So I will take away from this that in the case where I have a conductor with charge q1 and a cavity containing a charge q2, the charges on the inside layer will arrange to completely cancel the field outside of the cavity, and the charges on the outside layer will arrange as if the object did not contain a cavity and had net charge q1+q2. Sound about right?
Sounds good to me.