# Surface charge in a wire?

The image above shows the distribution of the surface charge in a current carrying wire. The surface charges distributes themselves to make sure the field inside the wire is always perpendicular to the surface of the wire.

Why is exactly half of the wire covered in negative surface charges and the other half with positive surface charges? I've noticed this in every surface charge distribution I've seen.

I don't understand the question and the diagram associated with it, partly because it contradicts with something I learned.

I was told that if I have a DC current in a straight current, by Ohm's law $\mathbf{J} = \sigma \mathbf{E}$, the electric field lines must be all directed along the current. So I don't see a way by which surface charge can build up on a straight conductor.

If I suddenly bend the conductor, the current leads to charge accumulation on the bent part of the conductor. However, this charge accumulation quickly causes the electric field to "turn" along with the conductor, and guides the current through the bend. Thus a steady amount of charge is accumulated on the bends of the conductor, which must be positive on one side (inside / outside) and negative on the other depending on which way the field must be bent to keep along the conductor.

Now, maybe there is a way by which your diagram makes sense, but I don't know how to relate it to the above description. In typical Physics-II courses, when you deal with capacitors, you also consider the so-called spherical capacitor, which is just a conducting sphere. The other plate is imagined to be at infinity. If the sphere must have a potential, it must have a corresponding amount of charge, and by $Q = CV$ it has some capacitance.

Now, each segment of wire (assumed to be ideal) in a circuit has a potential. Maybe you could think of it as a capacitor with the other plate at infinity and assign it a capacitance. By $Q = CV$ (where $V = 0$ at infinity), it must have a charge on it.

On the positive side of the circuit, there is a positive potential, so $Q = CV$ with $V > 0$ gives you a positive charge etc.

Here's a simple explanation: the wire is assumed initially to have a net charge of zero. That is, for every electron, there is a proton. Because charge is conserved, and you haven't provided any mechanism to add or remove charge from some external mechanism, there must be an equal number of positive charges (protons missing an electron, holes) and negative charges (extra electrons).

You can also think of this: if you understand why there are equal and opposite charges on a capacitor (points 5 and 1, in your diagram), then the same reasoning applies to the wires on opposite sides (points 4 and 2). Two wires make a capacitor just like two parallel plates, the only difference being that the capacitance is much less, since the wires have less area, and are separated by a greater distance. In fact, if you imagine yourself as a charge in this system, the plates and the wires are really all the same thing: conductor. The wires are just an extension of the plates, as far as the charges are concerned.

• Sure, but why are the charges split in the middle instead of being mixed together? – dfg Feb 11 '14 at 20:26
• @dfg I understood this image as a capacitor on the top, and an electrochemical cell (or something else that generates a voltage between its ends) on the bottom. The cell separates the charges. Is that not how you understood it? – Phil Frost Feb 11 '14 at 20:30
• Its supposed to represent a closed circuit with an electrochemical cell at the top generating a voltage :) Ignore the two "lines" at the bottom. – dfg Feb 11 '14 at 20:45
• @dfg if that's the case, this must be a physicist's joke. You can't have an ideal wire across a voltage, because an ideal wire is infinitely conductive, and so by definition, can't have a voltage across it. So that leads to the question: just what kind of wire is that, or where did you find an electrochemical cell that doesn't burst into flames when you demand that it supplies infinite power? I should probably stick to engineering. – Phil Frost Feb 11 '14 at 20:57
• I included another diagram, does this help? – dfg Feb 18 '14 at 21:12