Confusion about Circuits

Question

I have two questions:

1. When dealing with simple DC circuits, it is often said that the electrons in the wire move because the charge density of the surface charges decreases. Is that due to the fact that the amount of electric field of the battery decreases with distance?

2. I don't understand why the potential difference between any two points in a wire is equal. It would make sense to me when the decrease rate (or gradient) in surface charge density would be the same in every point. But I just don't see why this should be the case. I couldn't find any explanation...

I would be very glad, If one could help me with my confusion :)

Answer 1

If you divide the wire into small segments, as you go from the positive to the negative terminals of the source, at each segment charge concentration decreases from a maximum at the positive terminal.

Let's say at the positive terminal the charge is 100 units. And the next segment it might be 99, then 98 and so on.

This makes the E fields constant between segments, and the voltage drop (E * segment length) the same at every segment also.

If like in this example there are 100 segments, the voltage drop across each is 1%, until you lose the entire voltage back at the negative terminal of the DC source.

If there is a resistor along the way, then for the segment with the resistor, you'd have such charge concentration between its ends as to result in the higher voltage gap necessary to maintain the same current through the resistor as through all the other segments. That's why resistors capture more of the voltage. (To see this clearly, we can picture the charge as located on segments, and the resistor would span the boundary between two subsequent segments. Thus the charge gap in the segments between which the resistor is nested would determine the current through the resistor.)

In the model when everything is a short circuit except the resistor, the charge concentrations in all the segments prior to the resistor are the same as at the positive terminal of the voltage source, resulting in no E and no voltage change prior to the resistor. Then at the resistor the charge gap expands all the voltage to pass the current. After the resistor there is again no difference in charge concentration between segments and hence no potential difference (no E fields) until the negative terminal of the DC source.

It's important to note: the current doesn't change that surface charge distribution because the current entering a segment or a ring of wire equals to the current leaving the same segment or a ring of wire. The E field is the result of different surface charge concentrations between subsequent segments or rings, and these remain constant as long as the DC sources voltage is constant.

The "static" surface charge distribution is then produced quickly as DC voltage is applied, and current flow is superimposed on top of it.

To address a comment: "But you assume that the charge distribution decreases uniformly (100,99,98,... in your example). And my question is, why is that so?"?

Let's say that after the quick redistribution of static surface charge (it takes about a nano second to achieve that static redistribution after the DC source is connected), there is a greater voltage between segments 20 to 21 than between segments 19 to 20 (while these segments are of the same resistance). In that case, very briefly, more charge will flow into segment 21 (20 --> 21) then enter segment 20 (19 --> 20). In the new redistribution, subsequent rings of the same resistivity will have the same gap\difference in surface charge concentration.

Only, it's important again to note, that this static surface charge is constant after about a nano second (provided the DC voltage is constant), and only on top of it, current flows. The current is enabled by the static surface charge distribution, but doesn't change it - the same current is entering and leaving each segment.

Thus in the example of disbanded between 19 --> 20 --> 21, the movement was not the final current but only a quick redistribution of surface charges. On top of the resulting static distribution (takes a nanosecond for everyday wires) current will flow uniformly.

PS For segments of different resistance the logic is the same. A resistor requires a greater static surface charge gap between its ends, as otherwise the current entering the resistor will be more than the current leaving it, increasing the voltage gap across the resistor until equilibrium is reached. At equilibrium, the same current is entering and leaving the resistor, such that the surface charge concentration remains undisturbed by the uniform DC current flowing on top of (superimposed on top of) the static surface charge distribution.

Answer 2

We often model a wire as an object where the potential difference between any two points is zero. This is a useful simplification for many problems. If wire resistance is important, we use Ohm's law to relate current to potential difference.

We don't normally, when modeling DC circuits, consider the surface charge density, since this depends on unimportant details of how the wires are arranged geometrically and is difficult to compute. Instead, we use potentials, currents, and Ohm's law.

Answer 3

I will resort to the water analogy to make things easier to understand.

1. A river flows because there is a slight height gradient in riverbed. The water starts out at a high point and flows towards a low point. The height of the water is here the potential. In equilibrium, for a constant slope river bed, this will have constant gradient (see picture below). The velocity of the water will be proportional to the slope, again, in equilibrium. Since the slope is constant, the velocity of the water will be constant as well.

To translate this to an electric potential is easy. The height will become the potential, and the water velocity becomes the current. The electric field will be roughly constant along the wire. Note that in electrical wires there are surface effects as well, as you mentioned, so consider this diagram below as a simplification.

2. In a real wire this is simply not true. The wire has a finite resistance, so there is a finite voltage drop over any piece of wire. It is an often made simplification to consider a wire as 'ideal', which has zero resistance anywhere. The resistance in a wire is often neglible compared to other resistances. If the resistance of a particular is not neglible compared to some other component, you can simply draw the wire as a resistor in series with that component, connected by ideal wires.