-1

I learn by visualizing ideas and have a problem visualizing ac or dc negative voltage- what happens? for ac, do the electrons reverse direction? for dc, what happens? why dont we have simple textbooks answering these things in plain English? with diagrams to booth? i have read a couple of answers about this but it doesn't help when trying to visualize these terms. please forgive my insistence, but i do not want an analogy, i want a straight , dumb answer. again my apology for insisting, but i do not want analogies, i want an answer in electron and Hole terms.

someone edited the question while i was away and removed the key part mistakenly- the most important part for me is the fact about what is physically going on, not an analogy, i dont want to imagine it like something else, i want to imagine it as it is in real life.

kind regards

Z.Some
  • 25
  • 3
  • Not everything in physics can be visualized, so your learning method is pretty limited. Voltage is a potential difference. Difference can be either negative or positive depending on what you are taking as a reference. Potential present when there is a potential field around. The field can be generated by charged particles or something else. Not necessarily by electrons. – Eugene Sh. Jul 12 '17 at 15:54
  • 3
    Ignore the electrons. Thinking about electrons will mostly confuse you. For "what is voltage", start with: https://electronics.stackexchange.com/questions/245610/is-voltage-the-speed-of-electrons/245621#245621 – pjc50 Jul 12 '17 at 15:56
  • 3
    Voltage only matters when comparing two different points, if you have more electrons at one location than the other, they'll want to flow to the other location. – Voltage Spike Jul 12 '17 at 15:56
  • 2
    To understand negative voltage, first you have to eliminate any prior misconceptions. For example, can one point in a circuit have a voltage? No, that's wrong. A *single* point can't have a voltage, EVER. That's not how voltage works. – wbeaty Jul 12 '17 at 17:27
  • 1
    Sigh, I don't like down votes on questions like this. It may be a bit of a vaguely written question, but the question itself is a good one. – Trevor_G Jul 13 '17 at 14:01
  • Please remove the stuff after "why don't we have simple textbooks..." until "i do not want analogies,..." it does not help describing the problem _any further_ and lowers the value of your question in my opinion. The part before is clear enough so that it can be answered. Pointing out you don't want analogies is OK through. I second Trevor's opinion. – try-catch-finally Jul 16 '17 at 09:50

2 Answers2

2

In order to understand this you have to think about how atoms work. Atoms consist of a "positively" charged core surrounded "layers" of negatively charged electrons. Normally, this is balanced and the amount of electron charge equals the nucleus charge.

In a conductor, the outer layer of electrons is only loosely attached since the layer itself is not "full". When you line up a bunch of these atams and apply an electric field along the chain, electrons will tend to be driven by that electric field and will leap-frog to the next atom. This then causes the next atom to have an extra electron so it quickly looses that electron to the next atom... and on and on.

enter image description here

The electrons then move in a kind of chain through the entire circuit path.

Notice I said an electron is driven to the next atom. This effectively leaves a "hole" in the outer layer of that atom and the atom will become positively charged.

Now you can think of it as cars lines up in a traffic jam on the highway. As one car moves forward it leaves a gap behind it. The car behind fills that gap and the gap seems to move in the opposite direction to the traffic back down the traffic jam. This is the same as what we call hole migration in a conductor. Note nothing is actually moving, it's the absence of something that moves.

If you apply an AC electric field to the conductor, the electrons do indeed move forwards and backwards as you imagine.

Trevor_G
  • 46,364
  • 8
  • 68
  • 151
  • thanks, your last paragraph explains gets to the heart of the matter- do you have any references i could read up about this please? – Z.Some Jul 13 '17 at 07:12
  • @Z.Some, there are a myriad of documents on line that try to explain this at the levels you are thinking, some of them you need to be a theoretical physicist, or mathematical don to understand. In the end it does not really matter unless you plan on developing something like a phantom computer. Most of us only deal with the macroscopic level of electricity where we work with currents and try not to think about electrons themselves. It's a theoretical model that works well for almost everything we do. – Trevor_G Jul 13 '17 at 13:57
2

I think the simplest way to visualize voltage is to imagine two metal plates (like a capacitor's) that are in outer space and separated by some distance, \$d=10\:\textrm{m}\$. Label the plates \$A\$ and \$B\$. You place a voltage source across the two plates. Let's say this is \$V=10\:\textrm{V}\$.

Let's make \$A\$ the positive plate and \$B\$ the negative plate.

Before I go on, let's now add a third plate, \$C\$. It is suspended in outer space as well and is \$10\:\textrm{m}\$ away from \$B\$ such that plate \$B\$ is directly in between \$A\$ and \$C\$.

I need to point out that when I say that \$A\$ is the positive plate, I only mean that it is positive with respect to \$B\$. Not plate \$C\$. In fact, with this above arrangement using a voltage source across plates \$A\$ and \$B\$, there is no particular way to say what the voltage of \$C\$ is. It's not connected to anything and is just floating out there in space and could have any arbitrary voltage with respect to either of the other two plates. It's "unknown." But because we've applied a voltage source to \$A\$ and \$B\$, we do know the voltage between those two plates.

Now. Let's take one Coulomb of electrons and magically place them right next to (but not precisely touching) the surface of plate \$B\$ that faces plate \$A\$ (in short, in between the two plates but at plate \$B\$) with an initial velocity (relate to both of these plates) of zero. These electrons will be repelled by \$B\$ and attracted to plate \$A\$ and will start to move towards plate \$A\$. This force accelerates the Coulomb's worth of electrons (a specific number you can look up) and their velocity increases until they finally impact plate \$A\$. (Not unlike an asteroid impacting the moon.)

A certain amount of energy will be related to this impact. In this case, the energy will be exactly \$10\:\textrm{J}\$ (Joules.) This is because a Volt just happens to be one Joule per Coulomb. Since we released one Coulomb and subjected it to the acceleration of ten Volts, there will be ten Joules of energy due to the impact.

Now, I said the distance between the plates was \$d=10\:\textrm{m}\$. Was that important in figuring out the energy? No. Had the distance been ten times as much, the resulting energy at the moment of the impact would be the exact same. The acceleration would have been slower, but there would have been now more distance for that acceleration to be continuously applied, and the resulting velocity of the Coulomb of electrons would have been exactly the same. So distance has no bearing on it. (Volts has to do with the energy per Coulomb, only.)

(What about plate \$C\$? I don't know. No one else knows, either. It's just floating out there.)

Voltage is always measured between two points. In between those points, there is an electric field strength. Sometimes, this field strength is uniform (as it is in the above example.) Sometimes, it varies (in which case the acceleration of the electrons would have varied as well during their travel time.)

Another way to look at the above example, turning it "inside out" in a sense, is to ignore the voltage source that I applied for a moment and to instead just think about the Coulomb of electrons themselves as they traveled from \$B\$ to \$A\$. In the set up I gave, there was a uniform (constant) acceleration they experienced throughout their travel time. Each moment, the electrons moved from one point in space to another and still felt the same acceleration there, as well. So you can imagine that there are an infinite number of points along their travel and that there was an "accelerating" field at each point that continued to push them along, the same as before. Think of each of these points in space as possessing an "electric field strength" (or arrow) that points in some direction (from one of the plates to the other, in this case.) And that the electrons "follow" that arrow, moving to a new point, where a new "arrow" points them again. etc.

Looked at this way, the voltage between two points is really just the sum of all these arrows that exist in between the two plates. It has nothing really to do with the plates themselves, but about the field strength in the space that exists between the plates.

Because we've been stuck with "conventional" currents that flow from positive to negative (relatively speaking), an electron accelerating from a negative plate to a positive plate (relatively speaking, again) is instead considered to be positive charge moving in the other direction.

Oh, well. We're stuck with it.

Physicists consider the voltage (if positive) to be the work required to move a positive charge against the electric field. In other words, rather than just letting the charge move with the field to "impact" on the other side, they want to know how much work it takes to go the other way -- the work needed to move a charge against each and every electric field strength "arrow" as they move a charge from one place to another. If it takes work to do that, they consider it "positive." If it gives work to do that, they consider it negative. (This is just book-keeping to let you know if energy is released or required.)

So, the potential of \$B\$ with respect to \$A\$ is the work done against that field in between the two plates, when taking a positive charge from \$A\$ to \$B\$:

$$V=-\int_{x_A}^{x_B} \mathscr{E}_x\:\textrm{d}x$$

This can also be written as:

$$V=\int_{x_A}^\infty \mathscr{E}_x\:\textrm{d}x-\int_{x_B}^\infty \mathscr{E}_x\:\textrm{d}x$$

where \$\mathscr{E}_x\$ is the electric field strength at each point in space between the two plates.

That's a mathy way of saying everything else I said before. Also, the above is all "linear." But it has the same meaning in 3D space, except that now you have more coordinates and the little pointing electric field arrow can point in any direction along the way. So the path of the electron(s) (or positively charged ions) can follow a crazy-looking path at times. (A corkscrew path isn't uncommon!)

As far as knowing what is the ultimate reality of what is physically happening in the universe? None of us knows. We have models. Some models work well at one level of consideration and work terribly at another. (Newtonion mechanics works good here on Earth, but terribly when velocities are near the speed of light.) We have atomic models, "electrons as a gas" models, etc. But the absolute physical reality is still a work in progress. Frankly, I don't believe we will ever really know what it is. But we will have better models and they will continue to improve their reach into new situations.

jonk
  • 77,059
  • 6
  • 73
  • 185
  • +1 for the last paragraph. You are right it's all just the best theory that meets the observations and the math appears to approximate it in the majority of practical cases. But tomorrow is another day... – Trevor_G Jul 13 '17 at 13:52
  • Also. the capacitor analogy is a sound one. In fact, when you boil it down that's exactly what a conductor is, at least according to the valence model, a bunch of atomic capacitors with electrons jumping between them. – Trevor_G Jul 13 '17 at 14:03