Let's say you have a spherical charge distribution of radius R. This distribution has some charge density as a function of radius. I know that I can determine the electric field outside of the charge density by forming a spherical gaussian surface around the charge distribution and apply gauss' law. But what if I want to find the field directly on the surface. That requires that my surface intersects some of the charge of the distribution. In that case, when I apply Guass' Law. How do I determine the charge enclosed by the sphere. Do I discount the charge which exists on the surface or is that counted as enclosed by the sphere?

# [Physics] Charges lying on a Gaussian Surface

gauss-law

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## Best Answer

As I understand your question, you are trying to determine the electric field via Gauss law. In doing so you assume that there is a spherical symmetry in your electric field, otherwise you will only get some integral characteristics the field, which is not what you want. The symmetry allows you to write the surface integral as a product of the electric field and the surface area. So, symmetry is important.

Then you see that your charge should be uniformly distributed over concentric spheres. So, we are left with two cases -- there are spheres that carry nonzero electric charge (so we have a flat charge density instead of point-like one), or there are no such spheres and we have a continious charge density in the volume.

In the latter case, there is no problem, because there is actually zero electric charge ON your surface, and hence no ambiguity.

So, lets say there is a flat density like a charged sphere. Then answer is you dont care what is the electric field on the surface, because it gives not measurable effect. You just cant measure it, and test charges wont fill it, because in an instant they will leave this surface under the electric force. Well, this not all the truth, because you can ask about the force experienced by a small piece of your sphere, or what will a charge left on the surface do. You can find the answer in the end of my post, and now I want to add some physical sence here.

Actually you never have uniform distributions. On some smaller scale you always have electrons, positirons and so on (well, in QM this is probably subtle). However, what you care about is some mean value of your field over a small macroscopic volume. This effectively smoothes your charge distribution, and you can regard it as symmetric and continious in space, so there are no infinetely thin charged spheres and point charges, and the problem dont arise. Unless you go to a smaller scale.

On the smaller scale there are pointlike charges. Generally there is then no spherical symmertry and your way of getting the field value fails. But the Gauss theorem lives. You ask what if we take the surface to go through the charge. Then, first of all, you should answer how to understand the flux of electric field through this surface -- it is not well-defined mathematically. You have to choose some prescrition of how to assign a number to the surface integral, and this prescription will give you the answer how to count this charge -- inside, outside, half of it or whatever. There is no nice way to do it (though the half is nice). Here you just have to watch yourself and ask sensible questions.

Now, returning to charged sphere, or a general smooth surface. If you take a charged plane, then there is good reason (symmetry) to say that the field inside is zero. Futhermore, you know the value of the outside field. Now, say, you want to find the exterior force acting on some small piece of charged sphere. To find it, you need to cut this piece out and ask what is the field now? (Now it is well-defined.) It is hard to tell, but you know that it is continious, and if you add your piece, which can be viewed as a charged plane, in the very vicinity of some point inside of it, the field becomes $0$ inside the sphere, and $E$ outside. Recalling the electric field of charged plane, you recover that the

(well-defined) field, acting on your small piece is $E/2$. It gives you the measurable force. If you believe that the field is zero inside a flat plane, and that a charge put there is going to stay on its place, it is now natural to assign the mean arithmetical value of outer and inner field to points of the surface.external