I recently heard the expression that "Voltage is the amount of energy per electrical particle". I am confused about what it means. Isn't every particle have a certain amount of electrical energy?

How can an electron have a different amount of energy.

What is the difference of charge and particle energy ? Is there an analogy between this and a water circuit or other physical system ?

  • \$\begingroup\$ Potential and potential energy share the same relationship as field and force: electric/gravitational potential is electric/gravitational energy per unit charge/mass, while electric/gravitational field is electric/gravitational force per unit charge/mass. The concepts of potential and field are introduced to make them independent from the probing mass or charge you use to 'gedanken-measure' them. \$\endgroup\$ – Sredni Vashtar Nov 8 '16 at 8:43

Isn't every particle have a certain amount of electrical energy?

No. Electrically charged particles have more or less energy depending on where they are located relative to other charged particles.

To push two electrons close together requires energy. To pull an electron away from a proton (or positively charged atomic nucleus) requires energy.

How can an electron have a different amount of energy.

As an analogy, think about gravitational potential energy.

A bowling ball at the bottom of a hill has less g.p.e. than one at the top of a hill. To push the ball from the bottom to the top of the hill, you will have to do work on it, that is, apply a force as it moves from one place to another. The work you do on the ball adds to its g.p.e.

If you wanted to push a tennis ball from the bottom to the top of the same hill, it would take less energy, because the tennis ball has less mass.

With this in mind, even if there's no ball around, you can talk about the gravitational potential function of the space where the hill is. The potential is higher at the top of the hill and lower at the bottom of the hill. This lets you predict how much energy it will take to move a mass from one place to another, even though we don't know yet whether the mass we will want to move is going to be a bowling ball or a tennis ball.

Same thing in a circuit. If the potential difference between two places is 1 V, we know that if we want to move 1 C of charge from the lower potential location to the higher potential location it will require 1 J of energy. If we want to move 10 C between those two places it will take 10 J (assuming that moving all this charge around doesn't change the potential, like when we're charging a capacitor).

  • \$\begingroup\$ so the electrical charge is like the mass and the distance from the other particles (with opposite charge) is like the distance from the ground (i.e potential gravitational energy) ? Did I understand corectly? \$\endgroup\$ – yoyo_fun Nov 7 '16 at 23:52
  • \$\begingroup\$ You shouldn't worry about the distance from the ground. Just that there's a function of position, the gravitational potential function, that lets you predict how much energy it will take to move a mass from one place to the other. The fact that this function depends on the distance from the center of mass (not the surface) of some object (the earth) is secondary. \$\endgroup\$ – The Photon Nov 8 '16 at 0:27
  • \$\begingroup\$ The big difference between electrical potential and gravitational potential is that there's only one kind of mass, but two kinds of charge, so don't try to go too far with this analogy. \$\endgroup\$ – The Photon Nov 8 '16 at 0:28

If its a free roaming electron it can have kinetic energy \$ \frac{1}{2}m_ev^2 \$ this is expressed in electron-volts

In physics, the electronvolt1 (symbol eV, also written electron volt) is a unit of energy equal to approximately 160 zeptojoules (10−21 joules, symbol zJ) or 1.6×10−19 joules (symbol J). By definition, it is the amount of energy gained (or lost) by the charge of a single electron moving across an electric potential difference of one volt.

Electron Volts is just another way to express energy. In the world of particle physics its easier to run calcuations than to express things in Joules. You could find out how many electron volts you would have moving down the hallway but it wouldn't have much meaning.

Inside of an atom depending on where the electron is it has more or less energy.


"Voltage is the amount of energy per electrical particle"

Yes this Voltage \$U\$ applies to static charges in particles, Q & q separated by distance r. \$U=kQq/r\$ . The energy stored depends on the dielectric of the charges and thus the Capacitance C such that \$Q=CV\$ and \$E=\frac{1}{2}CV^2\$.

Then current is the flow rate of charged particles \$I=dQ/dt\$

Particle Energy is the sum of the Einstein mass energy and kinetic speed energy equations. This applies to chemical reactions and detonation of particles and many other non-typical electronic things.

Normally Voltage is stored in a battery dielectric with large C and a chemical reaction voltage or converted from some other power source. The electrical energy is E = VIt [joules=watt-seconds]

You may be familiar with battery capacity in amp-hours

  • (Ah or mAh) * Vbat = Wh
  • where units of measurement can be watt-hours or converted (X3600) to watt-seconds [joules]

"Energy is the ultimate convertible currency." Brian Greene


I think the very simplest way to understand what a volt is/does, and how it relates to energy and charge, is to just do the simple thought experiment where you imagine two plates in outer space (vacuum) with an imposed, fixed potential difference between them. (One is positive, relative to the other.)

Now imagine that you release a passel (technical term) of electrons right next to the plate that is negatively charged relative to the other plate. Suppose one Coulomb of them. Now, these electrons will be attracted by the positive plate and will accelerate towards it. (The magnitude of that acceleration will depend upon the distance between the plates.) Eventually, they will hit the positive plate and when they do there will be a certain total kinetic energy at the moment of that impact.

It turns out that no matter how far you separate the plates, which changes the magnitude of the acceleration but also changes the distance over which it operates, the final velocity at the moment of impact will always be the same. And the kinetic energy computed by \$\tfrac{1}{2}m v^2\$ will always be the same, no matter the separation distance, given the same potential difference between the plates and the same number of charges.

Now, if you use positively charged ions (which will be much more massive) and placed those very close to the relatively positive charged plate instead, then the acceleration will be still less than it would be for electrons, which are fairly light. But at the moment of impact, while the velocity \$v\$ will be lower, the mass \$m\$ will be enough higher to cause the same resulting kinetic energy.

In short, a specific potential difference between two plates implies a certain number of Joules of kinetic energy given a certain number of Coulombs of charge. It sets up (or is a consequence of) a proportionality relationship between Energy and Charge: \$E=V\cdot Q\$. As mentioned already, an electron-volt is just that -- one volt times one electron's charge. It's dimension is energy.

  • \$\begingroup\$ for 1/2 pt. explain why \$ \frac{1}{2}\$ is used rather than \$E=mv^2+mc^2\$ in particle energy \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 8 '16 at 0:26
  • \$\begingroup\$ Your explanation is good, but I think you're explaining what is electrical potential energy, not electrical potential. I think beginners have a hard time separating these two ideas, so we should be careful when answering this question to be clear which one we're talking about. \$\endgroup\$ – The Photon Nov 8 '16 at 0:30
  • \$\begingroup\$ @TonyStewart.EEsince'75 You're suggesting a direction where mass isn't invariant, I think. I am discussing the case where it can be treated as an invariant and it's enough to take the integral of momentum with respect to velocity. \$\endgroup\$ – jonk Nov 8 '16 at 0:31
  • \$\begingroup\$ @ThePhoton Just trying to give a gut feeling for the idea and why the the dimensions are entirely based only on energy and charge, rather than bringing in other dimensions (which aren't needed.) It was what worked best for me, many years ago when I first wondered about it. \$\endgroup\$ – jonk Nov 8 '16 at 0:33
  • 1
    \$\begingroup\$ @TonyStewart.EEsince'75 Or were you wanting to suggest the principle of least action, momentum, conjugate momentum, and the Euler-Lagrange equation? Oh. And how you are you going to give me half a point?? Round-down, I bet! \$\endgroup\$ – jonk Nov 8 '16 at 0:39

Think like this: if we have a charged capacitor, and we physically pull the plates farther and farther apart, then the charges on the plates end up with more and more energy. (That's actually how Wimshurst and VDG machines work.) And, if we let the plates fall all the way together and touch, then the opposite charges cancel out, and the energy becomes zero. So, the energy in an electron (and in a proton) doesn't stick with the particles themselves. Instead, the energy depends on how far we've pulled the opposite charges away from each other.

Here's another example: if we have two uncharged capacitor plates, and we pull an electron from one plate and drop it onto the other one, then both plates now have opposite charges. And something weird: the entire surface of both plates becomes energized! Even with just one electron involved, the entire metal surfaces now have small excess charge. Also, the added energy was shared between every particle on the plate surface. (It's like pouring a bucket of water into a pond. The entire water level rises everywhere. The hydro-energy didn't stick with the added bucket-water.)

Or, drop one electron onto a metal ball, and the electron stays in place but the "excess charge" as well as the stored energy spreads almost instantly across the entire ball surface. It happens because the electrons that were already there in the metal, they must jostle around and slightly spread out to maintain a uniform distribution. The "electricity water-level" rose a bit over the entire metal surface, even when adding just one electron to the ball.

So, electrical energy doesn't stick to individual electrons and protons, any more than hydro energy sticks to a cup of water. When a dam pestock drains a lake to run a turbine, the entire lake's water level falls simultaneously, and energy is "sucked" nearly instantly from the entire lake, even though the water itself moves sideways quite slowly. And, if we progressively discharge a capacitor, the charges move across the metal plates quite slowly, yet the energy in the entire capacitor is "sucked" almost instantly across the plates and to the terminals.

"Voltage is the amount of energy per electrical particle"

This isn't right. It also appears to be an extremely widespread misconception.

Where's the error? Well, the statement is like insisting that "Gravity is the amount of energy per lifted kilogram." No, wrong, because gravity is still there, even when there's no boulder being lifted up to store some energy.

The same is true of voltage: the voltage is a way of measuring an e-field, and isn't a measure of particle energy. E-fields and b-fields can exist in empty space, with no need to be lifting any charges or magnet-poles while calculating energy. Look at the top of your 9V battery. There is voltage-flux in the space between the terminals. Half way between, there's 4.5V, hanging in empty space. Voltage is like magnetism and gravity: it's a field.

Here's a much better version of that statement:


Simple? It reverses the wrong statement by stating that energy is determined by charge and voltage. (Then we just have to explain "charge" and "voltage." Heh.)

In similar fashion, it makes sense to say that gravitational energy is stored when work is done to transport a kilogram upwards against gravity. We don't turn this inside out, and insist that gravity itself is nothing but the energy per kilogram being moved!

I think this misconception about voltage has a definite origin.

Michael Faraday proposed the existence of "fields," of magnetic and electrostatic fields. Unfortunately the physics community of the time hated this, and completely rejected it. They stuck to their older concept called "instantaneous action at a distance," where forces and energy exist, but fields (and voltage) do not. Faraday died, still being ridiculed for his fields concept. Then JC Maxwell published his famous work, totally vindicated Faraday's "fields," and made them the basis of all EM science, at the same time discovering EM waves and the electrical nature behind the speed of light.

Yet some textbooks today still are sticking with the anti-Faraday viewpoint, and teaching the "distant-action" philosophy which insists that only forces and charges exist, but "fields" do not. Supposedly "fields" are nothing but an abstract mathematical gimmick: the energy per coulomb. (Heh, JC Maxwell says that magnetism, voltage, and radio waves flying through space aren't mathematical gimmicks, they're the stuff of pure EM fields!)

Yet we're left with deeper questions: We know what electrical energy is, given charge and voltage-difference it's just 0.5*Q*V^2. But what is the Q, the Electric Charge? What is the V, the Voltage? These are extremely important questions, but we don't even ask them if we've been taught that voltage doesn't really exist, and is nothing but the energy per unit charge.

What is voltage itself? Hmm. I can tell you what voltage looks like. Imagine that a charged particle is surrounded by a radial "puff" of flux lines of the e-field, like a dandilion-puff or a sea-urchin. In that case the "electrical potential" is a bunch of concentric spheres centered on the same particle, and are always perpendicular to the flux-lines of the electric field. If the lines of e-flux are like a bunch of hair, then the voltage is like a stack of paper, where every sheet of paper has flux-hairs poking through it at 90deg.

If e-field flux is an invisible star-shape with a charged particle in the center, then voltage is an "invisible onion" with a charged particle at the center. Both flux-lines and voltage (the "equipotential planes" or "shells") are descriptions of the e-field found in the space around electric.

So whenever two wires have a potential difference, this "voltage-stuff" actually extends away from the metal surfaces, and fills the space between wires, as if the wires were a pair of capacitor plates.


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