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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?

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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.

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  • $\begingroup$ More to the point in the NASA example, you can use a thinner, lighter weight wire and get the same voltage drop if you use silver. And weight is all important in their application. $\endgroup$
    – The Photon
    Mar 24 at 14:52
  • $\begingroup$ @ThePhoton I don't know all the considerations, but battery weight in the Apollo modules was a big consideration, so better conductivity to reduce battery reserve needs (and hence lower batter weight) seems to me a larger factor. $\endgroup$
    – Bill N
    Mar 24 at 17:02
  • $\begingroup$ but you could also accomplish that by using copper wire, just increasing the wire diameter. $\endgroup$
    – The Photon
    Mar 24 at 17:31
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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

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  • $\begingroup$ So I get that the potential difference is path independent so the length of the wire wouldn't affect the potential difference. I have another question then: Thinking about my previous scenario, if I force the electron to travel along some weird path that was NOT straight from the negative particle to the positive, the potential would stay the same correct? Okay, so how about if I extended the distance between the negative and positive particles, would that affect the potential difference? $\endgroup$ Mar 24 at 12:58
  • $\begingroup$ both would be EXACTLY the same. as potentialdifference is evaluated to the PARTICLES position relative to a charge. if that never changes then the pd across the path would be the same. moving the two terminals any distance from another and making the wire in the shape of a heart wouldnt change the pd across the path. ASLONG as the particles starting position and end position is the same relative to your charged. e.g still 1m away from charge A and still 1m away from charge B for example $\endgroup$ Mar 24 at 13:10
  • $\begingroup$ if you extend a wire in a circuit though,then more work would be done moving the charges through the wire so measuring your PD across the WIRE would read a higher pd, so the pd would read lower across any load on the circuit. as the potential atthe start of a load is lower because of the increased work done on the charges moving through the wire before reaching the load. $\endgroup$ Mar 24 at 13:18
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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.

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