# Charges inside a Conductor

I have a question regarding the existence of charges inside a conductor. I am reading both Intro. to Electrodynamics by Griffiths's and Feynman's lectures. They both derive using Gauss's law that since the electric field inside the conductor is zero, the charge density inside the conductor must be zero. However, they both interpret the results differently.

Griffiths claims that $$\rho$$ is zero but "there is still charge around, exactly as much plus as minus, so the net charge density in the interior is zero". Meanwhile Feynman claims that "the charge density in the interior of the conductor must be zero... there can be no charges in a conductor". Which is correct?

When using the integral version of Gauss's law, I conclude that this should really imply that there is no charge at all inside the conductor, which would be stronger than simply having no net charge density. But I am confused what Griffiths means when he says that "there is still charge around".

• There are good textbooks and not so good ones. A good textbook would probably say something along the lines of "positive and negative charges inside an uncharged conductor without current flow are balancing each other out" rather than "there are no charges inside a conductor", which is absolutely false. Commented Oct 7, 2022 at 9:26
• @FlatterMann As far as I know, these are standard undergraduate textbooks for electrodynamics, but if you have textbooks you would recommend please do. Why would it be false that there are no charges inside a conductor? If the electric field inside the conductor is zero there can't be any charges by Gauss's law right? Or am I missing something crucial... Commented Oct 8, 2022 at 6:07
• Both claims are equivalent, what Feynman meant was that there is no net charge, afterall a conductor is made of atoms which are made of charges, net charge being zero. For more on this please look at the difference between macroscopic and microscopic electric fields. Commented Oct 8, 2022 at 8:53
• @GedankenExperimentalist If you have to do Feynman's work for him by interpreting his undergrad textbook like as if it's a religious text, then he simply didn't do a good job writing it. Commented Oct 8, 2022 at 11:38
• @AndyRoo Did you have an introductory course on solid state physics, yet? Did you study the band model of metals and semiconductors? Metals have a high charge density. That's what makes them good conductors in the first place. It just happens that, on net, these charges cancel out in the static, field free case. This is high school physics, by the way. That the inside of a conducting shell is only field free if there are no free charges in the inside volume was one of my high school physics test questions. Commented Oct 8, 2022 at 11:45

There is one important case where (volumetric) charge density inside a conductor is exactly zero: perfect conductor with infinite conductivity.

In any other case, there's a so-called skin effect:

• Charges are pushed to the boundaries of the conductor, so volumetric density falls off exponentially as you go deeper into the conductor;
• Electric (and magnetic) field also decreases rapidly as you go deeper.

The "skin" of the conductor is this region near the surface where charges, currents and fields are non-negligible. It depends on the frequency of the fields, as the formula for skin thickness shows: $$\delta=\sqrt{\frac{1}{\mu_0\gamma\pi\nu}}$$ with $$\gamma$$ the conductivity and $$\nu$$ the frequency of the signal. This explains several things:

• It confirms that, for a perfect conductor ($$\gamma=\infty$$), $$\delta=0$$ so the signal can't penetrate the conductor at all.
• For a non-perfect conductor, a cable that carry high-frequency signals is usually thin because $$\delta$$ is small, so there won't be any signal travelling near the center of the cable. On the other hand, a cable for low-frequency signals can be thick.
• For a static field ($$\nu=0$$), skin is infinitely deep so the field can penetrate inside the conductor, but once the conductor returns to equilibrium the field has to be zero (otherwise there would be a net force on charges, so they'd be set in motion).

On other words, the quotes given in the question have a limited validity. Since they're located in the intro section of their respective books, it's likely that they're about a static field with equilibrium assumed.

• Thank you for your response, and yes the quotations concern electrostatic fields only, with equilibrium assumed. Could you clarify how your explanation of skin thickness relates to the charge distribution inside the conductors? I guess I am asking from a theoretical perspective: how can charges exist inside the conductor if there is no electric field inside of it (in electrostatics)? Can't we just take smaller and smaller Gaussian surfaces and use Gauss's law to show that there can't be any charges inside - let alone any charge density? Commented Oct 7, 2022 at 9:24
• @AndyRoo As for your question, if you're limiting the discussion to static fields in electrostatic equilibrium, then yes, skin effect disappears. Also, local electroneutrality for the conductor implies zero charge density after a very short time (so it's already settled when equilibrium is assumed). Commented Oct 7, 2022 at 9:41

It depends on the scale you are looking at. On a large enough scale that individual atoms can be ignored, then any volume of space has net zero charge, and so there not even any localised volumes of non-zero charge.

However, if you are thinking of the conductor as being made of protons, electrons and neutrons, then there are clearly charged particles around, and therefore there are charges around. The classic demonstration is Geiger and Marsden's experiment, which showed that when alpha particles impact gold foil some of them bounce out. If there were no charges within the foil they would all go straight through.

I believe I have found an answer. The ambiguity at hand is between what Griffiths and Feynman consider the "interior" of the conductor. Consider a conductor with a cavity.

Feynman considers the "interior" of the conductor to be the material of the conductor between the inner and outer surfaces. It is true that within this region (and similarly for conductors without cavities) that by Gauss's law there can be $$\textbf{no}$$ charge at all - not just that the total charge is $$0$$, though that obviously follows as well. The derivation is clear, just take smaller and smaller spherical Gaussian surfaces $$S$$ that lie completely within the conductor's "interior". Everywhere on these surfaces the electric field is $$0$$, so there is no flux through $$S$$, and therefore the total charge within $$S$$ is $$0$$. But we can use the symmetry of the spherical surfaces to show further that in fact there can be no charges at all in $$S$$. In the limit of smaller and smaller $$S$$, we see that there can be no charges in a conductor (in electrostatic equilibrium).

Griffith's, on the other hand, considers the "interior" of the conductor to be the region entirely within the outer surface. While it is still true that in the region between inner and outer surfaces (as in Feynman's case) there is $$\textbf{no}$$ charge, we can no longer conclude this about the region within the cavity itself (nor on the inner surface either). Consider a Gaussian surface $$S'$$ that lies entirely in Feynman's interior and wraps around the cavity. As before, everywhere on $$S'$$ the electric field is $$0$$, and so there is no flux through $$S'$$ and the total charge within $$S'$$ is $$0$$ by Gauss's law. However, in this case, we cannot assume that $$S'$$ is spherical in nature, and thus we cannot use symmetry to claim that there is no charge within $$S'$$ as we did before.

This is the conclusion I drew (with some additional context) to see why Griffiths would claim that there is charge within the conductor whereas Feynman would claim there isn't. Hope this helps anyone else with similar confusion for this particular minutiae.