1

Here we have a wire. At both ends there is an equal and opposite field caused by a chemical reaction. So, if we decrease or increase the distance between the two points, the strength of the field increases/decreases with the square of distance. Why then is resistance linearly proportional to distance then?

Andres Salas
  • 658
  • 2
    Because the inverse-square law only applies in free space. The resistor is not looking at space, or even the field strength. It's looking at the potential difference. – Carl Witthoft Jul 14 '14 at 17:05
  • No offense, yet electric fields penetrate all space, and potential energy difference is caused by an electric field, so it certainly matters what the field strength is. – Andres Salas Jul 14 '14 at 17:09
  • No, it doesn't. The resistor sees only the potential difference, not the "volts/meter" of the field. – Carl Witthoft Jul 14 '14 at 17:10
  • So you believe that the field strength and the potential difference, while similar issues, do not equal each other? I agree with that. After a while the electric field will cause induction which will cause charge buildups which will naturalize the field to a constant level throughout. The specific relationship between the field and the net potential difference confuses me. Obviously we go from an inverse square field to a constant potential, how weird. Appreciate the help – Andres Salas Jul 14 '14 at 17:21
  • A battery isn't a pair of point charges. The basic nature of what a battery is is that it maintains a constant potential difference between the terminals. – Zo the Relativist Jul 14 '14 at 17:45
  • While it is true that the net effect of an imposed electric field by a battery results in a constant electric field and potential difference, perhaps you'll understand what it is that I'm asking by checking this question http://physics.stackexchange.com/questions/102930/why-doesnt-the-electric-field-inside-a-wire-in-a-circuit-fall-off-with-distance?rq=1 (which admittedly does a good job of answering this question; perhaps I should close this eh). What I'm asking for is a mathematical treatment of the initial supplied field and the final, constant field. Thanks though – Andres Salas Jul 14 '14 at 17:56
  • I was about to post an answer, but the answer of that other question you linked to does a better job of explaining it. Something you may want to look up is the idea of capacitance. If the charge on the battery gives an electric field that's inconsistent with the voltage the battery tends to operate at, this can be modeled as a capacitance. – David Jul 14 '14 at 18:59
  • Bruce Sherwood has made an excellent simulation of field in a circuit. It's here. Does not run in all browsers. This simulation is an actual dynamic simulation of the motion of charges under the influence of an EMF, not a calculation. The demo shows a lot of stuff, and takes quite some time to fully understand. Bottom line: select "snaky circuit", and select only the button for E_net. You will see the electric field both inside and outside the circuit. Note that the field within the wire is constant! – garyp Jul 14 '14 at 20:14
  • Ah. Thank you, for some reason the program won't compile, but I think I've seen something like this in my intro to physics textbook – Andres Salas Jul 14 '14 at 21:31

2 Answers2

1

“[A]n equal and opposite field” is an oxymoron and gibberish. Henceforth I’ll assume that we have a chemical voltage source there.

The Coulomb’s law is applicable in the case of a charged body, a body that has a constant electric charge. A chemical voltage source is a completely different situation: there is no pre-defined distribution of charges (as they move easily through conductors, electrolytes, and so), but there is a pre-defined difference of electrostatic potential on the ends of the source. How is it compatible with the Coulomb’s law? I already answered: charge carriers can move and adjust their position to create and maintain the potential difference (for the question why do they do it address the Battery and current confusion? topic).

If a wire with uniform cross-section and resistivity is attached to the source, then a dynamic equilibrium is formed: charge carriers drift along the wire with a constant average speed, but there is no change in the charge density in it. Okay, why the electrostatic potential distribution in the wire is such that the field strength is constant, that the voltage drop is constant for a unit of length? Because it is an equilibrium! If somewhere in the wire the field will become weaker, then electrons will drift slower in this section, so density of electrons will change (I’ll not explain it in details, but it is easy to imagine) in the way that the difference in the field strength will defuse quickly. Really such fluctuations take place, but variations in electrons’ density are insignificant because even a small such variation would produce a strong field.

Community
  • 1
Incnis Mrsi
  • 823
1

The answer-as requested by HDE. So my foolish misconstruct of the electric field is that since it accelerates charged particles more or less depending on the distance from the source, it should create less motion in electrons a further distance from the source.

I'm not sure why I thought resistance had much to do with it-on further analysis resistance makes linear sense, since it takes twice as much energy in twice as much wire given that a particle gives off heat energy dependent on how much wire it travels through.

The problem with my initial idea-which at least was a valid question- is that electrons self-balance. If a pocket of electrons spreads out more near the source and less further from the source, the clumping further from the source will try to spread out, and the spread out part will come together to achieve coulombic balance.

Overarching idea-balance of charge.

My problem was the act of clustering of charge that initially occurs, to which I was like: why does this not become an issue? A valid question at least.

Andres Salas
  • 658