# Charge on Sphere: Evenly Spreading?

• Fibo112
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?Yes! If you have a hollow space, then there is a point charge at the center, and the charge on the outside of the space will be spread evenly.

#### Fibo112

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 let's 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?

Why wouldn't it ?

BvU said:
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 going to be non uniform, or the charge density at the boundary of the big sphere is going to be non uniform or both. They can't be both uniform cause then the electric field inside the big sphere wouldn't cancel out.

Delta² said:
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.

BvU said:
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 going to be an electric field inside the metal, but if both of charge densities are uniform then I really can't 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.

Fibo112 said:
Now let's 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.

Doc Al said:
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 canceled 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?

Delta2
Fibo112 said:
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 canceled 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.

Doc Al said:
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?

Fibo112 said:
Why does the charge on the inner surface do this on its own?
In order to achieve ##|\vec E|=0## in the conductor

BvU said:
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.

Fibo112 said:
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.

Doc Al said:
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?

Fibo112 said:
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.

Doc Al said:
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)

Fibo112 said:
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.

Fibo112 said:
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.

ZapperZ said:
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 don't 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?

Fibo112 said:
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 don't 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.

Fibo112