# Gauss's Law and Conductors?

At equilibrium, the electric field at and just at the surface of the conductor must be perpendicular to the surface. If it were not, it would have a component parallel to the conductor's surface and would exert a force on the surface charge causing them to move and violating our assumption that the conductor is at equilibrium.

In the above diagram the field lines are not perpendicular to the surface, but they don't intersect any surface charges so there wouldn't be any movement and the system would be an equilibrium. So why is the book's explanation valid?

• The positive charges depicted would be pushed to the left. Feb 12, 2014 at 21:51
• @JánLalinský Why?
– dfg
Feb 12, 2014 at 21:59
• Why is the electric field pointed up and left? There must be some asymmetry in the geometry. If so, it's important in this problem to consider the regions exterior to what you've drawn. I suspect if you do this, you'll find this situation is non-statec.
– BMS
Feb 12, 2014 at 23:13
• @dfg The text doesn't mention intersection. It's not really even clear what you mean here. How does your picture disprove the book? It shows a net electric field to the left. The charges move to the left. The picture you have drawn is not in equilibrium, which is the condition under which the field is perpendicular. Feb 14, 2014 at 6:41
• @user35033 Look at the second diagram I added in. The system is a conducting plate - its in equilibrium, yet the field isn't perpendicular.
– dfg
Feb 18, 2014 at 21:29

If you break the electric field lines you've drawn into components parallel to the surface and components perpendicular to the surface, you'll find a net field pointing to the left. That means positive charges would feel a force to the left, and since this is a conductor, they're free to move, and so they move to the left. The only way for the charges to stop moving is for all of the electric field components that are horizontal/tangential/parallel to the surface to cancel out, leaving only components that are perpendicular to the surface.

Maybe the picture below can help:

It can get a little confusing, but keep in mind that a vector field has a vector associated with each point in space, and when you draw vectors that represent the field, the vectors you draw represent the field exactly at the tail of the vector (the point where the vector begins), and nowhere else, no matter how long the arrows are. So then the length is just telling you the strength of the field at the point corresponding to the tail of the vector, and the direction is telling you the direction of the vector field at the point corresponding to the tail of the vector. In other words, we don't care whether the vectors "intersect" a charge when we draw them, because the vectors are only telling us something about the point at which they originate (their tail) and nowhere else. (i.e. In my picture, I know something about the field at each charge, but nowhere else. Even though the vectors extend up and to the left, they are telling me nothing about the electric field at points above and to the left of the charges, only exactly at the charges. Thus each of these charges feels an electric field to the left, as drawn (and also a component pointed up). These components are shown in red in the bottom part of the figure. Since they (the charges) are in a conductor, they are free to move to the left as long as they feel a field pointed that way. (They won't move up though, because they would have to jump out of the conductor to do that, assuming these charges are lying on the surface of a conductor.))

I hope that helps to clear it up. To reiterate, these vectors tell us about the field at each of the charges. So each charge must be feeling a field (and therefore a force) to the left, with the vectors as drawn.

• But just because the the field lines have a parallel component, it doesn't mean that a field acts on the surface charges. It just means that a charge outside would feel a force pushing it horizontally. Unless a charge is intersected by the field, it doesn't feel a force. And the surface charges aren't intersected by the field.
– dfg
Feb 12, 2014 at 21:58
• The field has to have the same parallel component (to the surface) both outside and inside the metal. So there would be net parallel electric field inside the metal and this would move the electrons. Feb 12, 2014 at 22:05
• Along with what @Jan said, also keep in mind that the net charge on a conductor resides on the surface, so any component of the electric field that is parallel to the surface will be felt by charges that are on the surface. Feb 12, 2014 at 22:15
• @MikeBell Why would it be felt by charges on the surface? It doesn't intersect the charges on the surface! How can a charge be affected by a field its not in?
– dfg
Feb 12, 2014 at 22:21
• Since the charges are on the surface, think of them as lying in a plane. The electric field is zero inside the conductor, so the field is represented by vectors or field lines that originate in the plane, and in general have components perpendicular to the plane and components parallel to (lying in) the plane. These components in the plane would push the charges that are in that same plane. Also, keep in mind that field lines are just a helpful tool. The field is not necessarily zero somewhere just because you didn't draw a field line there (i.e. it is nonzero between the lines you drew also). Feb 12, 2014 at 23:28

Dfg, you have not drawn an equilibrium electric field and the charges will move. We can decompose the field you drew into two parts -- the perpendicular part, and the horizontal part. This horizontal part pushes the charges to the left.

You are confused about the field lines not intersecting the surface. In this case they do. Outside of equilibrium, fields can permeate conductors, and fields can have components parallel to the surface.

Perhaps it makes it clearer to think about it this way: imagine you had a conductor in equilibrium with all of the field lines perpendicular to the surface of the conductor. Then you apply an electric field to the left. Before the charges have time to move, you draw a picture of the fields and charges. That is the picture you drew.

• Fair enough. However, the positive charges are locked into a lattice structure so even if there is a parallel component to the field, the positive surface charges will be prevented from moving. You could argue that the electrons would move, but if the vector field only has parallel components on the positive ions, the electrons wouldn't move. In that case wouldn't it be possible to have a non-perpendicular electric field without movement?
– dfg
Feb 19, 2014 at 19:37
• @dfg, Okay, if you have "locked them into place" that's fine, but then you aren't talking EM anymore. In the diagram, then, you have to draw the opposing electrical force holding them into place -- you STILL get that the field is perpendicular. You are changing the question left and right, but there is no way around this! Anyway, charges are free to move in a conductor. Feb 20, 2014 at 18:27
• @dfg You appear to be on a mission to prove the book wrong. Well, the book is right, so you are going to have a tough time. Feb 20, 2014 at 18:33
• What? I am talking about EM! Positive charges are not free to move in a conductor. This isn't a hypothetical scenario, all conductors meet this criteria. I'm not trying to prove the book wrong by talking about hypotheticals, I'm simply clarifying confusions I have about all conductors.
– dfg
Feb 20, 2014 at 22:25
• @dfg You can think of the "positive" charges in a solid as "missing" electrons, in this context. And $\sigma$, the surface charge density, is the average "missing electron" density, if it's positive. But you don't need this level of sophistication, because if you just think about conductors as uncharged solid bodies that charges can move through without resistance, the physics is the same. You don't need to think about crystal lattices. And before you ask, the argument for the perpendicular field is independent of the sign of the charge. Feb 20, 2014 at 23:48