I'm reading Sherwood and Chabay's brilliant textbook Matter and Interactions, in particular the section that deals with how the surface charges in an electric circuit distribute themselves to generate the electric field within the wire.

One question left unanswered, however, is why the generated electric field does not affect the surface charges themselves, only the electrons flowing through the wire.

enter image description here

In the above picture, wouldn't the charges on the rings also be affected by each other and the other ring?

  • 1
    $\begingroup$ The electric field in a wire is not generated by surface charges but by the power source that cause the current flow in the wire. The current density in the wire will then determine the local equipotential surfaces and the surface charges will be in equilibrium with that field. I am not sure what the above drawing is supposed to mean. $\endgroup$
    – CuriousOne
    Jan 4, 2016 at 7:13
  • $\begingroup$ @CuriousOne If the electric field were generated solely due to the battery, then the field would drop out proportional to the square of the distance. I don't believe this is the case, however. While the battery is the initial source of the field, feedback from this field moves surface charges into positions such that they can propagate this field at steady state. See this document that is essentially an excerpt from the textbook I mentioned. $\endgroup$
    – 1110101001
    Jan 5, 2016 at 3:08
  • $\begingroup$ Of course the far field drops with distance... why would't it? The entire setup is not even charged, as a whole, so you don't even get the field of a finite charge but you get the field of a complicated electric multipole, depending on the configuration of the source and the wires. If the book suggests otherwise then it belongs into the garbage. $\endgroup$
    – CuriousOne
    Jan 5, 2016 at 3:19
  • $\begingroup$ @CuriousOne This related physics.se question seems to suggest otherwise, that during steady state the field remains constant in magnitude throughout the wire, regardless of distance. $\endgroup$
    – 1110101001
    Jan 5, 2016 at 6:41
  • $\begingroup$ For a straight wire of constant diameter, yes. I still don't see where the surface charges come in. They simply don't play any role in this problem. You can eliminate all surface charges by coating the wire with a thin insulator and then with an outer conductive surface that will equalize all surface charges. This won't affect the conduction mechanism in the slightest. $\endgroup$
    – CuriousOne
    Jan 5, 2016 at 6:51

2 Answers 2


It is possible that the surface charges are pinned at sites on the surface, but it is also possible that they are mobile. Even if they are mobile, their contribution to the current is infinitesimal, because I = JA = sigmaEA, and the cross-sectional area associated with the surface charges is completely negligible compared to the rest of the cross-sectional area. So whether the surface charges are mobile or pinned is irrelevant.

You might find it interesting to view this VPython program in your browser (thanks to the new GlowScript version of VPython found at glowscript.org):


The surface charge distributions were calculated by a charge-field relaxation method described in the "Articles and talks" section of matterandinteractions.org. The VPython programs let you view these distributions and interactively explore the net field everywhere.

Bruce Sherwood, co-author with Ruth Chabay of the Matter & Interactions textbook.

  • 1
    $\begingroup$ I wasn't explicit about where to find the relaxation method explanation on our web site: Surface Charge Distributions in Circuits, a poster presented by Ruth Chabay and Bruce Sherwood at the winter 2016 conference of the American Association of Physics Teachers in New Orleans. $\endgroup$ Jan 4, 2016 at 3:30
  • 1
    $\begingroup$ Note too that even if the surface charges are mobile, the surface charge density at any location on the surface in the steady state does not change, so the field contributed by the surface charges doesn't change (definition of the steady state). $\endgroup$ Jan 4, 2016 at 4:22
  • $\begingroup$ Wow I didn't expect to get an answer from the author himself! I'm still sort of unclear, however, as to why the surface charge density at a given location wouldn't change if the surface charges were moving. Is it because as the charges move further down the wire they would adjust themselves to match the charge density that existed at the time before? $\endgroup$
    – 1110101001
    Jan 4, 2016 at 4:54
  • $\begingroup$ The definition of the steady state is that there is current (so not equilibrium) but the current does not change, which means that the field everywhere is constant, which means that the charge density everywhere is constant (does not change with time). Internal to the wire charges move out of a region but the same number of charges move into the region. On the surface, charge may (or may not; I don't know) move away from their present location, but in that time the same amount of charge moves into that location. $\endgroup$ Jan 4, 2016 at 5:03
  • $\begingroup$ Note that during the transient leading to the steady state the surface charge distribution DOES change with time, and the current DOES change with time. These transient movements of charge eventually lead in the case of a DC circuit to that distribution of charge that, together with the field of the battery, makes a net field everywhere that is consistent with the Kirchhoff loop and node rules. See the poster and program I mentioned. $\endgroup$ Jan 4, 2016 at 5:06

The number of electrons on the surface is EXTREMELY small compared to the number of electrons in the wire. Think of it this way: In a metal there is a "sea" of electrons that are not bound to atoms but are free to roam. If this sea expands by an infinitesimal amount, there will be a small (but essential) number of electrons on the surface, and the interior of the wire will remain very nearly neutral. The electrons that move to the surface do NOT travel a long distance to get there; there were already LOTS of free electrons right next to the surface. Sorry, but I don't understand your second set of questions. I strongly recommend that you study the paper whose link is given above.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.