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1. How is the electric field set up in a wire by a battery so that we can assume that potential at a node is constant? 2. What happens to the electric field across a resistance which makes voltage 'drop'?


Assume the following circuit:

s

The battery sets up some electric field throughout the wire, with a component along (parallel to) the wire (otherwise, there is no current). Zooming into the wire, enter image description here

Here I display only the component of the electric field and electron velocity that is along the wire. As the electron moves a distance d, the work done on it by the electric field is Eqd. This work is non-zero. In other words, electric potential energy of the electron varies with the distance it travels along the wire - and thus so does potential. How, then, is potential on a node the same?


Now, add a resistive load:

[1]: https://i.stack.imgur.com/dEb

The potential at all points on the wire above the battery is 10V, below it - 0V. Considering the above - how is this so?


NOTE: I am aware that a non-ideal wire has resistance. My question lies with what resistance has to do with a difference in potentials in the first place - and why voltage drop is not CAUSED BY a charge's distance from the terminals of the battery.


Exact duplicate it is.

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closed as unclear what you're asking by sammy gerbil, John Rennie, user259412, Yashas, David Hammen Jun 22 '17 at 15:37

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ If there's no satisfactory answer, can you please explain what about our answers didn't cover what you asked? $\endgroup$ – probably_someone Jun 20 '17 at 0:32
  • $\begingroup$ Please don't add meta-information to the title of the post, editing will bump it to the main page for more viewability. As probably_someone said, you need to edit your post with the details you are looking for. $\endgroup$ – Kyle Kanos Jun 20 '17 at 0:47
  • $\begingroup$ It's clear that no one who has viewed this thread is able and/or willing to provide what I'm looking for. Now this question's visibility will be limited - making it less likely that such a person(s) would respond. Consider this thread closed (but not resolved). $\endgroup$ – OverLordGoldDragon Jun 20 '17 at 1:38
  • $\begingroup$ @OverLordGoldDragon see my answer below. I feel your pain ;) because this whole topic was once a bit "taboo" in undergrad physics, ignored and misunderstood, until a pair of physicists became frustrated and created an entire undergrad textbook, one where basic EM and circuitry could be analyzed from a heretical viewpoint: using e-fields and "static electricity." Even a simple flashlight is an electrostatic device, and the full explanation of a simple piece of wire involves surface charge and e-fields. $\endgroup$ – wbeaty Jun 22 '17 at 2:16
  • $\begingroup$ @OverLordGoldDragon The site is full with physicists. You can't ask anything what they couldn't have answered 20 years ago. You can't imagine such a question. The problem is dual: 1) your question looks crap. After understanding it, I think it is not crap, particularly after your edit, but it still looks so on the first spot. And the site gets a lot of really crap questions every day. 2) It was really unclear (before your question). If somebody is unable to ask a meaningful question, why is it wanted from anybody to answer it? | I give you a reopen vote and try to fix it. $\endgroup$ – user259412 Jun 22 '17 at 19:01
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First, to avoid confusion and off-topic threads, replace your "wire" with a resistive conductor such as a heating element, a carbon rod, etc. (In other words, don't let your wire be melted or vaporized before a discussion even begins!)

Q: Inside the above non-ideal conductor, why is the e-field the same everywhere, regardless of electron position?

A: it's because electric circuits are based on electrostatics, where all the patterns of currents are determined by surface charge. A wire is very different than the insulating gap between the plates of a capacitor.

With currents in any closed circuit, a conductor will automatically develop a dynamically-adjusting pattern of surface-charge, where this charge distributes itself in such a way to guarantee that the conductor's internal e-field (inside each straight section of the conductor) is constant, and is directed axially along the conductor. The uniform current in a wire is caused by the pattern of e-fields produced by the gradient of surface-charge on that wire. The particular pattern of surface-charge arises from a complicated give-and-take feedback process, where the charge is determined by patterns of resistance and current-density in the conductor (and, the current-density in turn is determined by the patterns of surface-charge!) Simple calculations don't exist for such things, so in order to view/understand the charge-distribution and circuit-fields, we'd normally use FEMM Finite-Element modelling software designed for E&M. (That, or run away screaming! Go hide yourself under Ohm's Law and simple waveguide-mechanics, where the full complexity of the actual real-world physics is traditionally ignored.)

Note: the e-field in any curved sections of your conductors won't be constant. Instead, the surface charge will distribute itself so that any charges drifting inside the conductor are forced to take a curved path which follows the conductor.


An excellent and detailed treatment of this topic is available on the undergrad textbook site for MATTER AND INTERACTIONS:

.pdf: A Unified Treatment of Electrostatics and Circuits, 1999 Chabay/Sherwood

In introductory electricity courses, electrostatic and circuit phenomena are usually treated as separate and unrelated. By emphasizing the crucial role played by charges on the surfaces of circuit elements, it is possible to describe circuit behavior directly in terms of charge and electric field. This more fundamental description of circuits makes it possible to unify the treatment of electrostatics and circuits.


Quite a bit of circuit-physics is revealed in this paper about E&M misconceptions and textbook failure, Sefton 2002, Understanding Electricity and Circuits: What the Text Books Don't Tell You


Another source is R. Morrison's textbook "The Fields of Electronics: understanding electronics using basic physics." Find a cheap used copy, or get free pdf downloads from library or university locations using this Wiley online link.

Morrison complains about the near-complete lack of physics resources for learning far more circuit-theory than found in basic intro classes, see http://web.archive.org/web/20120315092836/http://www.ralphmorrison.com/Ralph_Morrison/An_Open_Letter.html

Also see numerous essays at http://web.archive.org/web/20120519181208/http://www.ralphmorrison.com/Ralph_Morrison/Short_papers_on_interference_topics.html

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  • $\begingroup$ Thank you for the reading suggestion - I'll look into it. Hope to make more sense of your response with the full document. $\endgroup$ – OverLordGoldDragon Jun 22 '17 at 2:38
  • $\begingroup$ @OverLordGoldDragon yep, that paper reveals a lot. It explains simple circuits by surface-charge analysis, Kirchoff's physics technique from the 1800s. But his trick became lost when node/loop and Ohm's law analysis came to dominate. If we want explain the patterns of volts/amps throughout a circuit, we must look at the ringlike patterns of surface charges on the wires. Ohm's law doesn't answer our questions or explain circuit-physics, it just gives numbers while concealing all the details. $\endgroup$ – wbeaty Jun 22 '17 at 2:57
  • $\begingroup$ Liking the reading so far. A question: puu.sh/wrwma/c6d3d35c61.png How does the electric field drive electrons into 'both ends of the left bend'? Is the left wall's (-) due to electrons ramming into it from their travels along the wire (right to left, against the field)? Then what about electrons on the right wall? Also, how is E_capacitor next to E_surface charge pointing to the right? The capacitor plates set up the field in only one direction. $\endgroup$ – OverLordGoldDragon Jun 22 '17 at 17:39
  • $\begingroup$ @OverLordGoldDragon Possible confusing diagram. They mean: imagine the capacitor's E(cap) fields with no wire present. Then suddenly the wire appears. The E(cap) field causes the wire's mobile charges to briefly shift. This shift creates the rightbend/leftbend net charge imbalances, which produce the reversed field at E(surface charge.) All this is a first-stage event, and happens in the picoseconds before the pattern of surface-charge has shifted to its final state, happening long before the normal "closed-circuit currents" arise throughout the wire. $\endgroup$ – wbeaty Jun 23 '17 at 1:52
  • $\begingroup$ @OverLordGoldDragon can you access this free PDF book? onlinelibrary.wiley.com/book/10.1002/0471433934 The link might only work if you're at a computer inside a library. It's a physics text about basic electrical engineering, with all the stuff that never appears in physics class or in engineering school, since the material sits between the two. $\endgroup$ – wbeaty Jun 23 '17 at 2:24
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When we say "there is no voltage drop in an ideal wire", what we really mean is "In many circuits, a wire may be approximated as having no voltage drop because the drop across the wire is insignificant compared to the drops across other elements in the circuit."

In your circuit, there is no other element aside from the source, so this approximation is invalid.

Therefore you must treat your wire as a low-valued resistor rather than an ideal wire with no voltage drop.

Then, for any one electron, isn't potential - with whatever reference - dependent on its POSITION in the wire?

Yes, the potential in a resistor depends on the position within the resistor.

To be as clear as I know how: In your circuit the potential does vary with the position along the wire just as you say it should. In this circuit you cannot say the wire defines a node with equal potential at all points. The wire acts as a low-valued resistor and the potential is indeed different at different points in the wire.

And - what does it mean for voltage to 'drop' across a resistor - and WHY does it drop - in terms of ELECTRIC FIELDS?

Electrostatic potential difference between points $a$ and $b$ is defined as

$$V_{ab} = -\int_b^a \vec{E}\cdot{\rm d}\vec{\ell},$$

where $\ell$ is some path between the points.

As you correctly stated, when a source is applied to a resistor, a field is developed in the resistor, which exerts a force on charged particles, causing some of them to move, producing a current.

Since there is a field in the resistor, a potential difference is developed between its two terminals.

Edit

In comments you asked,

I'm asking why do we need a resistor at all for there to be a voltage drop?

You don't. If you just have a battery sitting on a table top with nothing connected to it, there will be a potential difference (aka "voltage drop") between its two terminals.

Why, then, with an 'ideal wire' and a resistor, does potential at a point in the wire does not with distance from the terminals of the battery?

In this situation, you no longer have uniform field strength along the path from one battery terminal to the other.

The E-field in the wire will be much weaker than it is in the resistor. If it were higher, a large current would flow in the wire (because it has very low resistivity). But this current would also have to flow through the resistor (because of conservation of charge), which would require a large potential across the resistor. Things naturally balance out so that most of the potential drop is across the resistor. And very often that means the drop across the wire is negligible and we can model it as an ideal wire.

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  • $\begingroup$ I'm asking why do we need a resistor at all for there to be a voltage drop? With parallel plates, the electron is in a vacuum, and its potential is strictly dependent upon its position between the plates. As long as there's an electric field, whether there are plates or not is irrelevant - in a circuit, the electric field is set up by a battery. Why, then, with an 'ideal wire' and a resistor, does potential at a point in the wire does not with distance from the terminals of the battery? The expression you wrote clearly shows that this should be the case - as E is dotted with a length increment $\endgroup$ – OverLordGoldDragon Jun 19 '17 at 16:52
  • $\begingroup$ Your question shows a schematic with a wire connecting two terminals of a source. Then you say something about a system with a parallel plate capacitor. I'm not getting what you think the parallel plate capacitor has to do with a circuit containing only a source and resistors. $\endgroup$ – The Photon Jun 19 '17 at 17:07
  • $\begingroup$ @OverLordGoldDragon your questions are not about "real world" components. They are about an ideal voltage source and an ideal connecting wire, which are purely theoretical concepts (though they are very useful approximations to real-world components, in many situations). But you are asking about a circuit constructed from ideal components, which can't physically exist. Your questions don't have any "real world" answer, any more than you can answer the question "solve the equation $x = x + 1$ for $x$." (Note, an answer like "$x$ = infinity" doesn't even make any mathematical sense.) $\endgroup$ – alephzero Jun 19 '17 at 17:16
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Your issues with the answer here largely revolve around the concept of an "ideal wire." So let's explore it here.

An ideal wire is supposed to be a circuit element that connects other elements with zero resistance. So the simplest way to picture it is as a wire of length zero - that is, the two elements are actually directly connected together, but it's simpler on a diagram to separate them.

If you don't like this view, then let's look at why the voltage drop across a resistor is higher than the voltage drop along a (real) wire. When the wire is connected to the source, at $t=0$, there will be an (assumed constant) electric field across it that is approximately equal to $V/L$, where $V$ is the potential difference across the battery terminals and $L$ is the length of the wire. But since this is a conductor, the electric field immediately moves charges around, pushing positive charges toward the positive end of the battery and negative charges toward the positive end. So if we examine the wire in the steady state, we see a buildup of negative charge toward the positive end of the wire and positive charge toward the negative end, which lowers the electric field "seen" by the inside of the wire and therefore lowers the potential difference across it.

In contrast, resistors are usually materials that are bad conductors, where charges are not as free to move around. Therefore, they can't reduce the electric field as much as a conductor does, meaning the voltage drop across them is higher.

So an "ideal wire" is a wire that is a perfect conductor, in which any electric field is immediately canceled by effectively instantaneous movement of charges, so there can't be a voltage drop across an ideal wire. Likewise, you can picture an ideal capacitor as an infinite-ohm resistor - charge is not permitted to move at all, so the electric field provided by the battery is completely unaltered.

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    $\begingroup$ It's worth pointing out that the schematic drawing showing the ideal source with its two terminals connected by an ideal wire is just as much nonsense as writing an equation like "10 = 0". $\endgroup$ – The Photon Jun 19 '17 at 17:31
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    $\begingroup$ @The-Photon Using the "length-zero" explanation, it would be like building a battery with the two terminals touching, which is obviously unusable. That's why I like that explanation: it makes thinking about these things intuitively obvious. $\endgroup$ – probably_someone Jun 19 '17 at 17:35

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