Electric field inside a spherical shell

[confucius]

I've read many times now that the electric field inside a spherical shell is zero and all explanations of that fact have been based on imo false interpretation of the gauss law. The law itself says no such thing, atleast not explicitly. It merely says that the flux is zero. The only thing i can conclude from it is that the field at the very center of the shell is zero but not at any other point in the shell.
Explanations wellcome, thx.

 

[Doc Al]

I've read many times now that the electric field inside a spherical shell is zero

What kind of shell? Is it empty? Are there charges?

 

[confucius]

Yes the shell is empty, but the surface is charged.

 

[vanhees71]

In electrostatics it's a mathematical fact about harmonic functions and the uniqueness of the boundary-value problem of the Legendre/Laplace equations.

 

[Doc Al]

Yes the shell is empty, but the surface is charged.

And why do you think Gauss's law doesn't allow you to conclude that the field is zero everywhere within? Would you agree that symmetry demands that the field at any distance from the center must be the same for all points with the same radius and must be radial?

 

[confucius]

Doc Al, yes i would agree with u only if there is charge inside that is producing that field. U can use ur assumptions only if the r>R, where R is the radius of the sphere and r is the point in which u r observing ur field. As for the comment above, I must admit it is a bit 2 advanced for me. But as far as i understand i must have a continuous function of potential but that does not go in the way of el. potential varying inside of the sphere, it just says that limV when r approaches R+ is the same as when r approaches R-. I know i am probably wrong but that's why I'm here ;)

 

[Meir Achuz]

Yes the shell is empty, but the surface is charged.

If the surface charge is uniform, then the E field inside is zero.
If it is not uniform, then E inside is not zero.

 

[confucius]

Yes I am aware of the facts, but I am interested in the physics behind them...

 

[Doc Al]

Doc Al, yes i would agree with u only if there is charge inside that is producing that field. U can use ur assumptions only if the r>R, where R is the radius of the sphere and r is the point in which u r observing ur field.

No. As long as there is spherical symmetry with the charge distribution (assumed) then my assumptions about field symmetry apply for any distance from the center. Whether r < R or r > R, the assumptions hold equally well.

If you disagree, please explain why.

 

[vanhees71]

The interior is even field free for asymmetric charge distributions as long as there's no charge inside. You can check this by calculating the Green's function for the sphere with help of the mirror-charge method (see, e.g., Jackson, Classical Electrodynamics).

It is also clear that there's no field inside a conductor as long as there are no charges there, no matter of the charge distribution outside.

Of course, this is true only for the static case!

 

[Meir Achuz]

The interior is even field free for asymmetric charge distributions as long as there's no charge inside.

That is not true. Consider the case of a single point charge on the surface of the sphere.

 

[Meir Achuz]

It is also clear that there's no field inside a conductor as long as there are no charges there, no matter of the charge distribution outside.

The field inside a conductor will be zero for any shape surface.
If the conductor's surface is a sphere, then any surface charge will be uniformly distributed.

 

[vanhees71]

That is not true. Consider the case of a single point charge on the surface of the sphere.

This is physically impossible, since in this case the boundary conditions are not met, i.e., that's not a stable static condition. A current will flow until one has the chrage uniformly distributed over the whole conducting surface, and then the field inside the sphere vanishes again, and outside you have a Coulomb field as if the total charge is concentrated on the center of the sphere.

 

[Doc Al]

The interior is even field free for asymmetric charge distributions as long as there's no charge inside.

Please give an example of such a situation.

 

[confucius]

No. As long as there is spherical symmetry with the charge distribution (assumed) then my assumptions about field symmetry apply for any distance from the center. Whether r < R or r > R, the assumptions hold equally well.

Yes, I agree with u. U have been very helpful. It seems so obvious now, I don't know why has it been so hard for me to notice. I just wanted it 2 be more complicated:) Once more, big THANKS!

 

[vanhees71]

As I said, it's valid for any shape of the conductor and any charge distribution: As long as there's no charge inside and the situation is static, there's no electric field inside the conductor ("Faraday Cage").

For the sphere you can even evaluate the Green's function explicitly with help of the image-charge method. The Green's function G(\vec{x},\vec{y}) is given by the solution for the situation, where a unit charge sit's at \vec{y}. It's clear that the spherical shell must be an equipotential surface. The boundary conditions tell us that the tangential component of \vec{E} must be continuous, and thus the potential inside the whole sphere is constant (i.e., \vec{E}=0 inside the sphere). The field outside is given by the original unit charge at \vec{y} plus the field of an image charge inside, located at along the direction \hat{y} (the center of the sphere is in the origin of the coordinate frame) the sphere such that the sphere becomes an equipotential surface.

Then one has to think about the total charge on the surface of the sphere. If the sphere is grounded, this influenced charge is given by the mirror charge. If the sphere is neutral and insulated, one has to add additionally the Coulomb field of the oposite image charge in the center.

Thus, outside the sphere one has a relatively simple superposition of the field of the original unit charge and one or two image charges inside the sphere. In reality, as argued above, there are no charges inside the sphere, and thus the interior is field free.

You find the details of this calculation in any standard textbook on electrodynamics, e.g., Jackson, Classical Electrodynamics.

 

[Doc Al]

As I said, it's valid for any shape of the conductor and any charge distribution: As long as there's no charge inside and the situation is static, there's no electric field inside the conductor ("Faraday Cage").

Sorry, didn't realize you were restricting it to a conductor. Of course I agree.

But for the general case of an asymmetrical shell of charge it's not true that the field within the shell is zero. (But can be in specific cases, such as with a charged conductor.)

 

[vanhees71]

Of course, if the shell is not conducting, i.e., a dielectric medium, you have a field inside. I thought you are talking about conducting shells.