This answer gives a great explanation of why the field inside a wire connected to a battery must be equal at all points: Why doesn't the electric field inside a wire in a circuit fall off with distance from the battery?

The answer uses the concept of surface charge buildup to show why the field has to be equal at all points and why it must be perpendicular to the wire.

However, there are a lot of ways a field could exist in a wire that is not perpendicular to the wire, but would not cause surface charge buildup:

For example these fields:

enter image description here

could exist instead of:

enter image description here

and there would still be no surface charge build up.

So my questions are:

1) Can these "exotic" fields exist"?

2) If not, why not?

3) If these exotic field exist, how would current be defined in the wire? Since the electrons are not flowing perpendicular to the wire, would current be defined as the component of the movement of the electrons perpendicular to the wire or just the entire movement of the electrons?

4) The image below is a resistor. The "lines" represent the current direction and density ("the current density streamlines") . The "gray objects" represent wires through which a voltage difference is applied. If the electric field is always parallel to the surface how can the current lines be at an angle in the first resistor?

enter image description here


Please address all my questions separately. Thank you.


  • $\begingroup$ If the electric field at the surface is not perpendicular, it will induce a current. The charges will move around until they're stationary (and the field will be perpendicular). If they don't stop moving... it's not an electrostatics problem. $\endgroup$ – chase Mar 12 '14 at 18:38
  • $\begingroup$ @Chase The point is to induce a current. The wire is connected to a battery, so a current will be induced. I never claimed that it was an electrostatics problem, its a problem about electricity and current. $\endgroup$ – user41086 Mar 12 '14 at 19:04
  • $\begingroup$ Why the down votes? $\endgroup$ – user41086 Mar 14 '14 at 20:40

Remember that all electric fields are ultimately created by electric charge. If you wanted to create those exotic field configurations, you must have a non-uniform charge buildup. If you could arrange electrons on the surface and in the volume to create that field configurations, they would quickly spread themselves throughout the conductor by their own electrostatic repulsion to create the standard uniform electric field depicted in textbooks.

Also note that the "field configurations" that you have drawn are somewhat ambiguous depictions of a vector field that is defined at each point in space. Textbooks draw straight arrows like the one you drew for the correct configurations to mean that every point inside the conductor has the same uniform electric field. In your three exotic examples, it's unclear how you define the field where you don't have arrows or where the arrows overlap.

To answer your fourth question about the wire in the resistor, the electric field is always perpendicular to the surface of a conductor. If you did have a component of the electric field parallel to the surface, that would cause charge to flow into a different configuration to cancel that parallel component. The lines do appear to be at an angle to the wire in figure (a), but if you were to zoom in on the actual field configuration (not an illustration from an artist), you would see that the field is indeed perpendicular to the surface of the conductor.

  • $\begingroup$ 1) For question number 4, you said that the field is perpendicular to the surface, and if I zoomed in enough I would see that. If the field is perpendicular to the wire shouldn't the two images look identical? Also I don't get how zooming in into a curve makes it look straight. $\endgroup$ – user41086 Mar 15 '14 at 17:12
  • $\begingroup$ 2) I still don't understand why the exotic fields don't exist. You said if the surface charge were to distribute itself to form the exotic fields, the electrostatic forces would repel each other. Why? Why wouldn't the electrons be held in place by the field of the battery? Essentially what I'm asking is if the electrons don't repel themselves when they form a distribution to create a non-exotic field why would they repel themselves when they create an exotic field? $\endgroup$ – user41086 Mar 15 '14 at 17:17
  • $\begingroup$ The two figures in Fig. 26-8 have different configurations for the wires at the ends of the resistor. In (b), the flat ends of the wire conveniently allow for straight field lines from end to end. In (a), they have to shoot out perpendicular from the wire and then turn sideways to go down the length of the resistor. Different boundary conditions lead to different solutions. $\endgroup$ – Paul Mar 16 '14 at 16:40
  • $\begingroup$ The electrons are repelling themselves when they form the standard, uniform configuration, but the uniform configuration (in the absence of any external field besides the battery) is the lowest energy state. Think of the example of three charges tied to a ring. They always repel each other, and they quickly arrange themselves so they are equally spaced from each other. A similar thing happens in a conductor. In a neutral conductor, the positive ions of the metal also attract electrons that are piled up in another part of the metal. $\endgroup$ – Paul Mar 16 '14 at 16:47
  • $\begingroup$ And how can zoom into a curve make it look straight? Take the surface of the ocean from a satellite: it is a spherical arc. Now, get closer, go on a boat. Do you see the curvature of the surface of the sea? Not unless you look at objects very far away. $\endgroup$ – Davidmh Mar 18 '14 at 13:33

Pauls' answer is correct, but I would like to back it up with a quick example. Taking your first example, we can model the field like

$$ \vec{E} = E_x \hat{x} + E_y \sin(2 \pi x) \sin(\pi y) $$

Making a vector plot of this:

Vector plot

But as Paul points out, this field corresponds to a charge distribution within the wire given by Gauss's law:

$$ \begin{aligned} \rho &\propto \nabla \cdot \vec{E} \\ &= \sin(2 \pi x) \cos(\pi y) \end{aligned} $$

If we make a density plot of this

Density plot

We can see the areas of non-zero charge density within the wire (white and dark purple). In any real conductor these charges would quickly dissipate, leaving the conductor neutral. And the only solution for the field inside a neutral conductor is uniform. This is covered in the chapter on separation of variables in Griffiths E&M book.


When you have a field variation, at some point it is $E$ and at some different point $E+e$.

A simple current field would then have $e=0$ everywhere, that is $E$ is constant. If $E$ is constant, then $\nabla\cdot E=0$.

(1) An exotic field might exist around an impurity or a mechanical damage to the conductor. Some charge might accumulate as a result of directing the streamlines into a narrower space. However, this charge would become constant.

(2) If an exotic field exists, then there is a net accumulation of charge in the conductor. While it is permitted for conductors to hold charge, they would flow to produce no net field inside the conductor. For this reason, $e=0$, and the sole field is due to a potential difference between the contacts, divided by the length.

(3) In the case of a reduction of cross-section, one would suppose Joulean heating to increase, possibly causing the wire to melt. Fuses, toasters, and other devices work on this principle.

(4) When charge is deposited at one end of the resistor, and some $E$ exists, the charge is drawn to the other end of the resistor. In the first example, charge is appearing over the full width of the resistor, and so the transverse repulsion is constant. This makes the net value of $e$ to be $0$.

In the second example, $E$ is still constant, but the appearance of a mass of charge at a point would cause a radial expansion $e$. This charge will expand until $E=e$, and thus the gradient of $e$ would be non-zero: charge would accumulate at the ends to form the streamlines. But eventually the radial expansion of the charge will disappear, and so the local value of $e$ at that segment will also be $0$. The current proceeds in a parallel manner as shown in the picture.


An electric field really cannot have a component perpendicular to the axis of the conductor. As per Ohm's law, $\vec{J} = \sigma\vec{E}$. If an electric field exists in a direction perpendicular to the axis of the conductor, it implies that there would be current flow in that direction. The perpendicular component of the field would cause the electrons to be displaced opposite to the direction of the electric field. This would cause a net positive charge on the respective atomic nucleus. This charge would then create an electric field in the opposite direction which would cancel out the external electric field. Ohm's law states all this quite simply - no current means no electric field! So all the field types you have shown in your diagram cannot exist in reality.


The defining property of an ideal conductor is that charge is flee to flow until it hits a boundary. If there is a component of the field perpendicular to the direction of the wire, such as in all of the 'exotic' fields shown in your question, then charge will flow in that direction until it cancels the field.

So, if there were some situation in which the fields you've drawn were to arise in an insulator, they still would not survive in a conductor because the surface charge would build up until it turned the 'exotic' fields into the field you've drawn at the bottom.

  • $\begingroup$ I don't understand why the charge would cancel the field. In the "exotic" fields I drew above, the charge would not buildup because as soon as they get near the boundary of the wire, they are pushed down again. $\endgroup$ – user41086 Mar 14 '14 at 0:50
  • $\begingroup$ A ideal conductor is an infinite source of charge. There is more than enough to cancel all of the fields you've drawn unless they are very strong. It sounds like you are thinking about individual particles moving along the conveyor belts of the electric fields. This is not the case, it is only the average velocity which moves along the direction of the electric field. Some of the average charge distribution will cancel out the perpendicular components of the fields. $\endgroup$ – Chris Mueller Mar 14 '14 at 1:01
  • $\begingroup$ No, I get that there are enough electrons to cancel out the fields, but theres no reason for it too. The electrons would cancel out the field, if theres a build up of them. In the scenarios I gave, there can't be a build up of electrons. For there to be a buildup of electrons at a point, the flow of the electrons going "into" that point has to not be equal to the flow going out. This isn't possible in the diagrams I drew above because the electron flow going "into" any point is the same as the electron flow going out. Did I misunderstand your comment? $\endgroup$ – user41086 Mar 14 '14 at 1:09

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