Why doesn't the length of a wire directly affect potential difference in a circuit?

Question

I am currently in a Physics II course and we learned about the potential difference between two points in space that are a certain distance away from a point charge. If the distance is directly correlated to the potential difference for the equation V = E*dr, how come this does not apply to a wire in a circuit?

To better explain the question, try to picture this scenario:

Imagine an electron between one negatively and one positively charged particle. Both of these particles have an electric field that applies a force to the electron in between. There's a potential difference between the position that the electron is currently in and a position closer to the positively charged particle; Now let's say I replace this path between the particles with a wire, shouldn't the potential difference be the same and depend upon the length of the wire?

Answer 1

It does, unless the wire is superconducting.

A wire of silver, copper, aluminum, gold, etc., has finite non-zero resistance. At a fixed temperature, it will behave like an ohmic resistance and there will be a small potential difference from one end to the other depending on the length of the path:$$R=\frac{\rho L}{A},$$ where $\rho$ is the temperature-dependent resistivity of the material, $L$ is the length of the wire, and $A$ is the cross-sectional area.

In analyzing simple electrical circuits in introductory physics courses, we ignore those voltages across wires because they are very small compared to those in other parts of the circuit. (That's a first-order modeling assumption.)

In high-end applications like NASA spacecraft in the 1960s, silver was used for wiring because it had lower resistivity than copper. They wanted to minimize voltage changes in the wiring because they needed to minimize electrical power consumption.

The effective resistivity of air or other gases is much higher than for conducting metals, so charge distributions very quickly get minimalized and the electric field inside a current carrying wire is very small.

That doesn't happen in air until you have a huge electric field and lightning occurs.

Bottom line: The potential difference between a charge and a point in space depends on the dielectric properties of the material between the charge and the point in addition to the charge value and distance.

Answer 2

The integral of - e.dl is the potential or ie. -* the amount of work done by a charge in moving through a distance of infinity to a point r,where dl is the path element corresponding to the line connecting infinity to your point r . if i were to calculate the potential at a point r+1 this would be lower than at point r as it moves through a shorter distance from infinity so i change my relative end position the charged reaches relative to a charge q. this is different then extending the wire. as by entending the wire you do not change the final starting point of a charge and end point. We know that the line integral from the positive terminal to the negative terminal (pd across the circuit) is PATH INDEPENDANT therefore increasing wire length does NOT decrease potential as the relative start position and end position for a charge moving through the wire is invariant

edit: ofcourse as commenter above stated... if the material has a polarisation to it as in a capacitor, then the resulting E field will decrease causing a lower potential difference, for a different reason to your first point stated

edit 2: commenter above is actually talking about losses in a wire due to heat. this is a pd difference across the wires. if your talking about the circuit then the pd across the Battery TERMINALS is invariant to wire length (ignoring polarisation

Answer 3

If the distance is directly correlated to the potential difference for the equation V = E*dr, how come this does not apply to a wire in a circuit?

It does apply, although it should be $dV$ rather than $V$. Inside the wire $dV \approx 0$ and since $dr \ne 0$ that implies $E \approx 0$ inside the wire.

Now let's say I replace this path between the particles with a wire, shouldn't the potential difference be the same and depend upon the length of the wire?

No, the wire alters the potential dramatically. The wire has surface charges which will gather on points near the external charge. These surface charges will cancel the external field inside the wire so that $E \approx 0$ inside the wire. The surface charges will also alter the external field as needed to achieve that.