Electric Field of a Charged Sphere

I can understand why the electric field inside a charged sphere is zero when there's electrostatic equilibrium.

I can understand why the electric field outside a charged sphere equals kq/r².

But I remember reading somewhere that the electric field AT the surface of a charged sphere equals the electric field at a point infinitely close to the surface of the sphere divided by 2, i.e. half the field at extreme proximity.

I can't understand why that is, even though I thought a lot about it. Do you think this statement is true, and if so, can you prove it?

fluidistic
Gold Member
I've read it about 3 days ago, yes it's true if I remember well.
I think any EM book has the proof. It's 1 page long or a bit more if I remember well...
I know the proof is in Purcell's book of electricity and magnetism.

Born2bwire
Gold Member
Unless I am misinterpreting your question, I am not aware of such a phenomenon with a charged surface. We can simply think of the relationship of Gauss' Law with a spherically symmetric charge distribution. When we take the integral to find the enclosed charge, the enclosed charge is inclusive of the Gaussian surface that is the boundary of the volume. So the right hand side of Gauss' Law will not change if we take a Gaussian surface that is on a surface of charge or if it is just above the surface. This means that the field that we find using this technique is not discontinuous on the surface like you seem to be suggesting, but then again this is probably not a correct way to think of it since if we were to go with Coulomb's law we would still encounter a singularity with any charge located at our observation point. Things can change when we have an inhomogeneous medium though. A permittivity contrast would cause discontinuous electric fields across a surface. However, I do not think that the electric field on the surface, the normal component at least, is considered to be defined due to this discontinuity.

In the end, thinking of the field from a spherical shell of charge is reminiscent of problems with residue theory in complex variables or say a principle value integration. Still I can't recall anything off hand that applies to this nor do I find anything in the standard undergraduate literature.

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fluidistic
Gold Member
The example I was talking about is in page 30 of Purcell's book (but in Spanish, in English I guess it shouldn't be far from page 30). I don't think the sphere is conductor though... I'm confused.
I think the OP has confused the force acting on each charge over the sphere which is $$F=\frac{(E_{\text{internal}}+E_{\text{external}})}{2}\sigma$$. In order to reach this result, Purcell had to assume a width of the surface of the sphere. He says that within it, the electric field can vary linearly, cubicly, etc. depending on the distribution of the charges within this small width, but the same result holds.
I don't know if something changes if we assume the sphere to be conductor. What I find strange is that the charges don't move but a net force is acting on them... I think Purcell explain this with some atomic force and so the net force on each particle is 0.

Thank you very much for your replies.

In fact, I have found in some high school books the assertion that the electric field at points at the surface equals the electric field at a point infinitely close to the surface divided by 2. I, too, have found this very unsettling, but I believe I may have thought of something which corroborates to that assertion:

When a spherical gaussian surface is very close to the surface of the charged sphere, but not AT the surface of the sphere, the electric field is, according to Gauss' Law:

Q/(epsilon 0 times area of the surface)

where Q is the charge of the sphere. But when the gaussian surface is AT the surface of the sphere, half the charge is inside the surface, and half the charge is outside. (Think of a line of atoms with orbiting electrons. The surface is leveled with the nuclei, so statistically speaking, half the electrons are above it, and half are below it.) Therefore, according to Gauss' Law, the electric field is:

Q/2(epsilon 0 times area of the surface)

Which is exactly half the electric field at extreme proximity. Does this seem sensible to you?