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If you drop a really heavy ball the ball's gravitational potential energy will turn into kinetic energy.

If you place the same ball in the pool, the ball will still fall. A lot of kinetic energy will turn into thermal energy because of friction, but the gravitational potential energy will still be converted.

Similarly, why doesn't electricity flow without a good conductor? Why won't Electrons flow from the negative terminal to the positive terminal without a wire attaching them?

Electricity flows like a wave and metals have free electrons in the electron cloud that allows the wave to propagate, or spread. But when these free electrons aren't available to propagate the wave, why don't the electrons just "move" like the ball? Why don't the electrons just "move" through the air to the positive terminal?

A slow drift speed means that the electrons most likely will take a long time to propagate the wave of electricity, but they should still get there.

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  • $\begingroup$ Related: physics.stackexchange.com/q/38488/24774 $\endgroup$
    – fffred
    Commented Aug 12, 2013 at 18:52
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    $\begingroup$ Your analogy is good! But why do you assume that the ball will "fall" to the ground of the pool? Depending on the material, the ball can also just float on top of the water or stay somewhere in the middle and dive around. If you increase the voltage, then the air will get ionized and can also conduct electrons, therefore it only depends on the energies and materials involved, but your analogy still holds in terms of a potential field. Even in vacuum, if the energy is high enough, you can create electrons and positrons (pair production). $\endgroup$
    – Mike
    Commented Aug 12, 2013 at 20:10
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    $\begingroup$ Water isn't a very good insulator against heavy balls. If you filled the pool with sand or concrete instead, you might get different results. $\endgroup$
    – N. Virgo
    Commented Dec 30, 2019 at 11:20
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    $\begingroup$ @Nathaniel "Water isn't a very good insulator against heavy balls" saved :) $\endgroup$ Commented Dec 30, 2019 at 14:37

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To continue to use your ball analogy think of the ball as analogous to the electron. Now what if the ball were attached to a point by a spring? Would it still fall? It can oscillate about that point but it would not be able to escape the restraining effect of the spring entirely. The same is the case with bound electrons. They are more or less bound to the atom. If the gravitational field is very strong it may be able t o break the spring and rip the ball out of the spring. This happens sometimes in electricity too. In a lightning discharge, the electric field is so high that even the bound electrons are ripped out of their atoms thus ionizing the gas and creating what is known as plasma. With a pool of free electrons and positive ions available electric current can now flow freely through the plasma - you wouldn't need wires. But unless you have a high enough electric field to produce ionization(for ionization of air the field required is close to $10^6V/m$ - such high fields cannot be produced by the 100 - 250 V household voltages available in most countries) you would have to use wires made of conducting material where free electrons are readily available if you want to have electric conduction at normal voltages.

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Following discussions on dmckee's and Daniel's answers, here is my summary.

The electric potential in a solid metal is not 0 everywhere. It would be if electrons were localized exactly where the ion's nuclei are, so that the electric potential from electrons and the nuclei cancel exactly. However, quantum mechanics say that electrons have a wave-function that "spreads out" (delocalized). The electric potential that they create is somewhat wider than the nuclei's. The sum of both contributions, for one atom, might look like the following figure (that's a very crude and probably unrealistic view).

Potential from a delocalized electron

By the way, that allows the metals to bond together, the electrons providing the potential to attract the nearby nucleus.

If you sum these potentials from many of these atoms equally spaced, you may obtain the following potential.

Potential from a few atoms

Another example of such potential is depicted here. Electrons behave like waves in this potential and will bounce on each surface. That's why they do not escape easily the metal: they would have to overcome the potential wall at the metal surface. For this reason, electrons keep flowing in metallic wires.

Note that if you have electrons in excess, they will repulse each other until they are at the surface of the metal, which is why extra charge stays at the surface of conductors.

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If you define "electricity" as charge carriers in motion (which I think is reasonable), then you need free charge carriers, so you need some kind of medium from which the charge carriers can get lose.

The medium doesn't have to be metal wires, it can be gasses (as in the case of drift chambers), liquids (say a liquid argon TPC or a wet cell battery), plasma (obvious, I hope) or solids.

The atoms that make up ordinary air are not easily ionized and they recover their electrons very quickly (because of the electrostatic force). In metals man electrons are in or near the "conduction band" and can get lose quite easily and do not recombine efficiently. Electrons in the conduction band of the metal are "free" in the sense that they can move around inside the conductor easily, but it still requires energy to remove them from the metal (making them "free" in a more general sense). That energy is the "work function" you encounter in descriptions of the photoelectric effect.

The potential barrier is probably the biggest contributor to the non-flow of electricity through an open circuit.


To explain insulators in your example extend the metaphor to use cold molasses rather than water as the medium. If you are willing to wait for long enough the ball will still fall, but it will be painfully slow and you don't get appreciable useful work out of it.

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  • $\begingroup$ But aren't the electrons the charge carriers? Why can't they carry the charge through air? (even though it would be extremely slow). In other words why can't the electrons move through the air like a ball would? $\endgroup$
    – dfg
    Commented Aug 12, 2013 at 17:15
  • $\begingroup$ I guess what I'm really asking is if a ball can move through the air, why can't an electron? $\endgroup$
    – dfg
    Commented Aug 12, 2013 at 17:18
  • $\begingroup$ The electrons mostly are not lose to act as charge carriers in air. The field interacts with the whole (neutral) atom or molecule. Bulk metals have a "conduction band" which allows electrons to be free with much lower energy than in air. Drift chambers (a class of particle detectors) work by moving electrons through gas, but they require that the atoms are ionized first (and use high fields because the electrons will recombine in fairly short order). $\endgroup$ Commented Aug 12, 2013 at 17:19
  • $\begingroup$ In a galvanic cell, there are a bunch of free electrons at the negative terminal. Why can't the electrons move by themselves? Why do they need the electrons of the air particles? $\endgroup$
    – dfg
    Commented Aug 12, 2013 at 17:27
  • $\begingroup$ As soon as they leave the terminal the electrostatic force pulls them back. That is not a problem in a wire because there are free electrons all along the path and in each path segment as soon as the local electrons leave their place is taken by those from the segment before just as they take the place of those in the next segment. If there is an ionized tracking leading from the cathode to the anode then current does flow through air for a short time: that is how leaf electroscope dosimeters work. $\endgroup$ Commented Aug 12, 2013 at 17:46
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When you ask: "why don't the electrons just "move" like the ball? Why don't the electrons just "move" through the air to the positive terminal." I think you need to keep in mind that the ball is made up of neutral atoms which are themselves made up of negatively charged electrons and positively charged protons. The electrons are attracted to the positive terminal, but the protons are repelled equally. Therefore, the ball as a whole does not move. That answers one part of your question. You then might ask why the electrons don't move on their own to the positive terminal? And the answer is that they can if the voltage is high enough to overcome the attraction of the electrons to the protons that they are attached to in the ball. Lightning is a good example of this, as is static discharge.

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  • $\begingroup$ But why does the voltage need to be high for electrons to move on their own? Yeah, they need to overcome the attraction of the protons but this is countered by the repulsion of the electrons. Isn't it? $\endgroup$
    – dfg
    Commented Aug 12, 2013 at 17:25
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    $\begingroup$ @dfg the attraction to the other nuclei is largely overcome by their neighbouring electrons, yes. But attrraction to its own nucleus can still be quite strong. You're asking the equivalent of why something with tape stuck to it won't fall. There are multiple forces at play, and the attractive one wins. $\endgroup$ Commented Aug 12, 2013 at 17:32
  • $\begingroup$ The electrons are bound to the nuclei of the atoms they are a part of by their attraction to the protons in the nuclei. Remember that the protons are packed into a very dense nucleus, whereas the electrons are spread out. A bound electron in a material will feel a strong attraction to its closest nucleus and a diffuse repulsion coming from all directions due to the other electrons. The attraction to the nucleus is shielded by the other electrons and this does weaken the force of attraction and lowers the voltage required to remove it, but the attraction is still there. $\endgroup$ Commented Aug 12, 2013 at 17:33
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    $\begingroup$ To return to your original question, electricity travels more freely through wires because wires are very dense and contain a large number of highly mobile electrons. Electricity doesn't move easily through air because air is very sparse and because the electrons in air tend to be bound to the molecules that make up air. $\endgroup$ Commented Aug 12, 2013 at 17:58
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    $\begingroup$ I think I now understand Daniel's explanation. Because electrons are delocalized, they do not compensate exactly the potential from the ions. Actually they do in average, but not locally. The overall potential looks, in 2D, like a foam egg crate. Excess electrons will feel this potential, and when they cross a boundary, they have to overcome the last "hill" of potential, which is higher due to the surface effect (just as water's surface tension). This is the microscopic interpretation of @dmckee's point about the "work function". Am I correct? $\endgroup$
    – fffred
    Commented Aug 12, 2013 at 21:13
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Electrons do flow without a wire. This is exactly what is happening in a cathode ray tube. So why don't electrons flow from one conductor to another through vacuum or air if there is a potential difference? There is a minimum energy of a few eV required for an electron to exit a metal known as the work function. In a cathode ray tube this energy is minimized by selecting a material with low work function and by heating the anode.

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Try to lookup conduction and valence bands. The theory explains in terms of energies why electricity flows.

Picture

It is not like balls because electrons are bound by potential wells, which they have to climb out of for them to flow.

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  • $\begingroup$ How does the difference of conduction answer the question? Are band structures valid for air? Why wouldn't "extra" electrons flow through the insulator? $\endgroup$
    – fffred
    Commented Aug 12, 2013 at 18:16
  • $\begingroup$ Air is an insulator and as seen above, electron will not exist in it conductive band unless they climb out of a potential well. This happens at about $15 \rm{kV/in}$ of potential gradient where there is arcing. $\endgroup$ Commented Aug 12, 2013 at 18:28
  • $\begingroup$ Then, what happens to the electrons when you force them in the insulator? Are they repelled? Do they bounce off the insulator to come back in the conductor? $\endgroup$
    – fffred
    Commented Aug 12, 2013 at 18:46
  • $\begingroup$ How do you "force" them into an insulator? $\endgroup$ Commented Aug 12, 2013 at 19:22
  • $\begingroup$ First example: a capacitor that is under a given potential (the excess electrons one one side will be attracted to the other side). Second example: a spherical conductor inside an infinite insulator with excess charge on the surface of the conductor (the electrons repel each other so they tend to get push away from the sphere). $\endgroup$
    – fffred
    Commented Aug 12, 2013 at 20:27
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The answers posted by CIA and dmckee are great, and they correctly point out that textbooks usually handwave about why electrons don't easily leave the surface of conductors, but I would add that the electrons actually can move through the air, even if the electric field is not strong enough to ionize the air and form a plasma. Anyone who has tried doing quantitative electrostatics experiments knows this - put a static charge on a conductor open to the air and see how long it stays there. It depends on the material, but it's generally not that long, especially on a humid day. You usually need to keep a power supply hooked up if you want to keep a constant charge.

Also, leave a battery on the shelf long enough and you'll find that it has lost its charge. This is similar, though with ions doing the drifting rather than electrons and with the movement being through the internal insulating medium of the battery rather than through the air.

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"If you define "electricity" as charge carriers in motion (which I think is reasonable), then you need free charge carriers" This @dmckee's logic seems somewhat restrictive, as it is not applicable without caveats to alternate current, which is probably the most widely used kind of electricity. You can use bound charges for alternate current. For example, alternate current can flow through capacitors. The OP asks: "why doesn't electricity flow without a good conductor?" I would say it does, for example, in circular dielectric waveguides, which are just dielectric rods. Fiber optics is an extreme example of such phenomena, and I don't think one can dispute that there are "charge carriers in motion" in fiber optics, although it does not have to use free charge carriers.

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Succinctly: Electricity flow does happen easily in cathode rays in the famous vacuum tubes and cosmic rays flow nicely in outer space and ions flows readily in ionic solutions. So wires are not needed. The problem is to get charges into your space or air since electrons like to stay inside a material or metal as they are attracted to the positive lattice ions. Give them enough energy to get out (work function of photoelectric effect) and they will then readily accelerate and flow in whatever electric field or potential difference. There is an energy barrier to get them out. Even a charged up metal ball doesn’t throw off its excess charge in vacuum unless the charge is so high to create a large surface field to get electrons out of the metal. Heat or UV light can help jump the electrons out.

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