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In a static situation we defined voltage as energy/unit charge, or j/c. As the distance between the charged particles increased, the voltage decreased. Now why do we not apply this in a simple DC circuit? When the electrons move inside the wire, they are changing their distance from the cathode. By our definition of voltage, the voltage throughout the wire should vary along the length of the wire, resistor or not. However, it only changes when there is a resistor. Can anyone explain this? Thanks!

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This is because a wire is much more complicated than a bunch of charges.

What happens is that there is a tiny potential gradient along a wire, caused by surface charges on the wire. The plates of a battery also case a small potential gradient due to fringe fields. However, a battery has zero net charge, so charges which aren't too close to the battery don't really get affected by it. Remember, the electrons are getting farther from the cathode, but they get farther from the anode as well. Also, the plates of a battery aren't charged. Replace the battery with a capacitor, and you still have the argument that "you go away from the anode as well as the cathode, so there is not much change in energy".

In the end, any energy difference while moving along a wire is caused by surface charges. However, these only have a gradient if the wire has resistance. If not, then these are uniformly spread. While analyzing circuits we usually assume that wires are resistanceless -- a reasonable assumption. Even if we didn't assume that, the energy/potential difference across a wire would be too negligible to matter.

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Thanks very much, that made it a lot more intuitive. – Ovi Apr 5 '13 at 0:42
Wait a second, the fact that as an electron moves it gets farther away from the cathode and the anode doesn't support your point that there is not much change in potential energy. Potential or potential energy (I forgot which, but it doesn't matter) is defined by the force on the charge times the distance. If the battery has a net charge of 0, the approximate net force is also 0, so by that logic the charge should have 0 potential at a reasonable distance from the battery. – Ovi Apr 9 '13 at 0:24
@Ovi: Approximate. Fringe fields exist on the battery, and they create a tiny potential gradient.. This won't happen with an idea wire as the surface charges balance out, though. Remember, we're talking about potential difference here. On one side of the battery, we have negative potential, on the other side it is positive. It's fine if it becomes zero somewhere. – Manishearth Apr 9 '13 at 6:25
(basically, the surface charge gradient comes from the fringe field) – Manishearth Apr 9 '13 at 6:30
I know we're talking about the difference, and that difference is defined by the force times the distance between the test charge and the battery in this case. Since the cathode is relatively close to the anode, you are going to have one repulsive and one attractive force of approximately equal magnitude and in approximately opposite direction. That means that the net force is approximately 0, and since potential difference=F*r, the voltage should be approximately 0 going by that logic. – Ovi Apr 9 '13 at 19:11

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