# Electric field and potential exactly on the surface of a spherical conductor

[EDITED] I understand how Electric Field and Potential ( E and V ) can be calculated inside and outside a spherical conductor at electrical equilibrium using Gauss law, but I have a doubt when trying to find E exactly on the surface where charges are distributed: in this case, Gauss law seems to fight with Coulomb law. Given a test charge of 1 Coulomb very little like a point (a "test" charge) we try to calculate the force on it when moving from T to T'. Assuming E as the force exerted by charges on the surface on the test charge of 1 Coulomb, according to Coulomb force it should be undefined ( E is inversely proportional to the distance by the effective charges on the surface and the test charge, but the distance is zero ). There is a unit of charge (an atom without an electron) very close to the test charge. I know I have to imagine all the charges on the center of the sphere, but in "real life" charge is composed by units (e.g electrons) distributed on the surface. Why the test charge does not feel the force exerted by those units when approaching to the surface going to "infinite"? Further, considering a patch of charge on the surface (dA), why it "goes" to zero when the test charge goes to the surface ( making zero also $$(\sigma\,\mathrm dA)$$ )?

• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
Commented May 19, 2023 at 12:03

You must have been looking for a simple answer, but you have worded it in a way that summoned horrible monsters. But first, a technicality,

but if E is also the force "felt" by a charge of 1 Coulomb very little like a point (a "test" charge)

1 Coulomb is a tremendously large amount of charge. I mean, $$10^{19}$$ electrons worth, is large enough that you are already pushing on Avogadro's number. A test charge should not be that big.

Anyway, let us get back to your problem. Your problem stems from dealing with point charges and spread-out charges at the same time. Yes, per Coulomb's law, there is a distance-of-zero based singularity as the test charge passes through the surface charges. But this distance-of-zero singularity goes as $$\left|\vec F\right|=\frac1{4\pi\epsilon_0}\frac{q(\sigma\,\mathrm dA)}{r^2}$$ and so it goes to zero with $$\mathrm dA$$

So, something else must help us decide the $$\frac00$$ that is happening, and the conceptual understanding we need to add is that we want the forces behave exactly as the spherical field should do. This way, we pick that there will be no weirdness as the test charge passes through the sphere.

Now we deal with the monster. If you read JD Jackson, the $$2^\text{nd}$$ chapter starts with precisely the thing you are dealing with. The monster lies in the fact that the metal sphere will react to the presence of the test charge, and generate an image charge ($$3^\text{rd}$$ chapter) in response.

The horrific result is that, even if your metal sphere has the same sign of charge as the test charge, say, both positive, the initially far away charge will feel the repulsion, but at a close enough distance, the image charge will dominate in interaction with the test charge rather than the main charge. As such, for all test charges, there exists a distance near enough to the surface of the sphere that the test charge will be attracted to the charged metal sphere.

Because of that, the real definition of the electric field using a test charge is $$\left|\vec E\right|=\lim_{q\to0}\frac{\left|\vec F\right|}{q}$$

Needless to say, the fact that this behaviour is the one that is detected in experiments, means that we will never ever detect the repulsion of like charges at the surface of a metal sphere if you actually try to do that experiment.

• Thank you for your response. I wrote "1 Coulomb" just to try to be understood. About yor words "this distance-of-zero singularity goes as $$\left|\vec F\right|=\frac1{4\pi\epsilon_0}\frac{q(\sigma\,\mathrm dA)}{r^2}$$ and so it goes to zero with $\mathrm dA$ So, something else must help us decide the $\frac00$ that is happening,..." , I telly you that actually I wondered sumething about that limit, but why do I have to consider just $$(\sigma\,\mathrm dA)$$ and just the small $$\mathrm dA$$ going to $0$ and and not the rest of the charges on the sphere (they are still there :) ) ? Commented May 19, 2023 at 9:51
• The rest of the sphere's worth of charges are not at distance zero from the incoming test charge Commented May 19, 2023 at 9:56
• Thank you for you response, very precise. Commented May 22, 2023 at 8:34

E is inversely proportional to the distance by the effective charges on the surface and the test charge, but the distance is zero

This is true when approaching a point charge, where a non-zero charge is concentrated in a single geometrical point.

But the charge on the sphere's surface is not a point charge, it is distributed. It a sheet charge density, spread evenly over the surface (at least in classical electrostatics where we ignore the quantization of charge). The total charge on a patch of the surface goes to zero as the size of the patch goes to zero.

Therefore the surface charge "at distance 0" from the test charge is actually 0, and you have to do some calculus (or at least, some reasoning from Gauss's Law) to work out the total field due to the distributed surface charge.

• <<in classical electrostatics where we ignore the quantization of charge>> So don't I have to consider the spanned charge composition "electron by electron"? <<The total charge on a patch of the surface goes to zero as the size of the patch goes to zero. Therefore the surface charge "at distance 0" from the test charge is actually 0>> Why does the size of the patch "go to zero" when approaching the surface? I cannot see a relation between the distance going to zero and the patch dA to choose. Commented May 20, 2023 at 13:47
• On your first question: The point of my statement that you quoted is exactly that in classical electrostatics we don't consider charge electron by electron, because classical physics ignores the quantization of charge. That is, it ignores the fact that electrons exist at all, and assumes that charge can be spread out in space perfectly evenly, rather than concentrated in tiny lumps, aka quanta, aka electrons. Commented May 21, 2023 at 14:07
• On your second question, we don't assume the patch size goes to zero because the particle is approaching the surface. I'm just saying that if you consider a region of size zero, then the charge of that region is also zero. The singularity you're asking about (the E field going to infinity) only happens if there is some charge at distance zero from your test charge. But the total charge at distance zero (that is, the charge contained in a patch of the surface around your test charge with size zero) is zero. Therefore, there is no singularity. Commented May 21, 2023 at 14:09
• Thank you for your responses. Yes, you were clear. I have an other idea: a test charge inside an Electric Field is defined "so weak to not alter the electric conditions in the space", so is a "model charge". When distance goes to zero, also the "model" test charge goes to zero, to not alterate the charge on the sphere. In this way: q goes to zero, dA goes to zero, distance goes to zero ... so that we have a 0/0 condition (the limit goes to an indeterminate form). We have to consider another law. Thank you, guys! Commented May 21, 2023 at 23:47