Physicists, please help a humble electronics engineer understand his electrons better!

What I was taught in my recent electronics degree-> A voltage drop is indicated over a component (e.g resistor) the higher potential end is labelled (+) and the lower (-) The potential energy difference causes charged particles to 'fall' through the electric field created by the potential difference, conventional current falls from + to - and electrons from - to +

This is where I am confused; one end of a resistor doesn't have more charge that the other i.e one end isn't more charge-positive than the other like a battery or capacitor, the same charge carriers are at different energetic states after experiencing the resistor and electrical energy is different to charge (gravitational energy and mass analogy), so is this +/- labeling used differently to mean both charge and potential difference just to confuse me?

And if so, does this mean that electrons are falling uphill to the higher energy state? or is this where the gravitational energy analogy breaks down, i.e is high and low energy purely relative in electrical domain like +/- charge are relative to eachother and it;s just a matter of defining 'ground', so high potential for a positive charge is low potential for a negative charge?

  • $\begingroup$ Remember that the charge of the electrons determines which direction it moves then but in an electrostatic potential. $\endgroup$ Jan 12, 2017 at 16:44
  • $\begingroup$ Thanks for your comment Mikael! However, I don't mean to be rude but I don't understand your sentence... I know that electrons move to + charge. '..it moves then but in an electrostatic potential.' <- what does this mean? Could you relate you answer to specific parts of my question and perhaps i'll understand what you are trying to say? $\endgroup$ Jan 12, 2017 at 16:56
  • $\begingroup$ Potential is due to an accumulation of charge. A + potential is a lack of electrons (or abundance of "holes"). A - potential is an abundance of electrons (or lack of "holes"). Each electron carries a small individual charge, it get's "pushed away" by nearby like charges, i.e. from other electrons collaborating to create a potential. Also attracted by opposite charges or potential. The gravity analogy goes a long way, I like to consider water flow for that $\endgroup$
    – JMLCarter
    Jan 12, 2017 at 16:57
  • $\begingroup$ Thank you for your comment JMLCarter! However, this is only partially true, an electromagnetic field in free space has no charges yet has a potential difference which defines it, so how do you account for this? If you were correct we couldn't have radio. And charge carries electrical energy and difference between charges in this energy creates potential difference, as I said in my question using the resistor example, if you were correct the resistor would have to be capacitive to have a voltage across it, do you not agree? $\endgroup$ Jan 12, 2017 at 17:00
  • $\begingroup$ I thought your question was an electric circuit question. Electromagnetic waves are oscillations of potential that travel without the charge, and they are created by oscillating charges (in electronics using antenna). The effect of one charge on another that I described earlier is induced by exchange of an EM wave, which is the force carrier. $\endgroup$
    – JMLCarter
    Jan 12, 2017 at 17:06

2 Answers 2


The labeling has perfect physical meaning.

Consider we have fixed positive charge which creates electric field $\vec E$. Electron in this field will be attracted to positive charge with force $\vec F = q\vec E$, where $q$ is electric charge (with sign!). From the latter we can see that force and field have opposite directions for electron. In particular, it follows that $\vec E$ points from positive to negative charge.

From the point of view of energy gain, the electron has lowest energy close to positive charge. The energy of the former is related to the electric field potential $\phi$ (integral of force) as $\epsilon = q\phi$ and is the lower the higher $\phi$ is. Coulomb force drags electron towards positive charge, i.e. in direction opposite to electric field $\vec E$. This is the physical illustration of the relation $\vec E = -\mathrm{grad}\phi$ (gradient of some scalar field is the vector pointing towards the direction of increasing field).

The density of current passing through a conductor cross-section can be written as $\vec j = qn\vec v$, where $q$ is charge, $n$ - density of charge carriers, $v$ - speed. The latter is linked to applied electric field as $\vec v = \mu \vec E$. Note that electrons move in the direction opposite to electric field, as was shown above. Here $\mu$ is carrier mobility and is defined as $\mu = q\tau/m^*$, where $\tau$ is an average scattering time and $m^*$ is "effective" mass of the carrier. Using the above one can finally arrive to the differential form of the Ohm's law: $\vec j = (q^2n\tau/m^*)\vec E$, where the quantity inside the brackets is electric conductivity or inverse resistivity.

This allows to answer the current "sign" question.

  1. Current and electric field directions always coincide, i.e. independent of carriers sign.
  2. For electrons the direction of current is inverse with respect to electrons velocity. For holes (positively charges carriers) the directions of current and velocity coincide.

Voltage drop between points 1 and 2 is the difference of electric field potential in this points, $V_{12} = \phi_1 - \phi_2$. It is positive for $\phi_1>\phi_2$, hence it is opposite in sign to electric field. So here is the final conclusion: current flows from highest potential to lowest.

As you can see, you do not need any conventions, and all can be figured out from mechanics. Hope this clarifies the confusion about signs.

Concerning charges, non-zero electric field means charge imbalance. It also means, there should be current as a reaction to remove this charge imbalance. When you are building energy diagram in this case, keep in mind that for positive charges the higher the field potential, the higher the energy, so that their energy diagram is inverted compared to that of electrons.

UPDATE: The energy diagram difference for positive and negative charges follow from the energy definition $\epsilon = q\phi$. While an electron has lower energy near positive charge $Q$, a hole (positive charge) will have larger energy near positive charge $Q$, whereas potential $\phi$ of the field created by $Q$ stays the same.

This is explicitly illustrated in p-n diode (see picture). p-doped region has negatively charged ions, and n-doped region has positively charged ions. Here the energy well for electrons looks as usual: electrons are accumulated in the energy well, since they indeed have lower energy near positively charged ions in the n-region and larger energy in p-doped region near negatively charged ions.

On the other hand, holes are accumulated near the the negatively charged ions in the p-region, where they have lower energy, and not in the n-region near positively charged ions, where the have higher energy. If you look at their energy diagram from electronic point of view, it looks counterintuitive: holes sit at higher energy. However, if you reverse energy axis, it will make perfect sense: holes sit in their energy well.

Vertical axis - energy, horizontal axis - distance. White dots - negative ions, black dots - positive ions. Shaded areas show population with carriers.

Image legend: Vertical axis - energy, horizontal axis - distance. White dots - negative ions, black dots - positive ions. Shaded areas show population with carriers. Image is derived from here.

  • $\begingroup$ Thank you for you answer drYG! I am still confused about potential energy drop; going back to my question, if we look at a resistor in isolation(hypothetically-I know that this breaks circuit laws) with a voltage drop across it, the voltage drop across it isn't caused by charge imbalance..and in general there doesn't need to be a charge imbalance for there to be a potential difference, in fact batteries and capacitors are the only scenarios I can think of where this is really the case.. $\endgroup$ Jan 13, 2017 at 10:41
  • $\begingroup$ ..using the steam/fluid analogy which I think your answer depends on - in the electrical world there doesn't need to be 'water particles' to have 'pressure' i.e em waves have no requirement for charge to carry them..So in the resistor with the same current before an after (which is in line with KCL) it isn't a charge balence that creates the field it is a potential energy difference between charge carriers either side (using the person waling down steps analog as the electron going through a resistor) $\endgroup$ Jan 13, 2017 at 10:41
  • $\begingroup$ "When you are building energy diagram in this case, keep in mind that for positive charges the higher the field potential, the higher the energy, so that their energy diagram is inverted compared to that of electrons." I think THIS is the answer I am looking for! Could you clarify what you mean by this please? :-) $\endgroup$ Jan 13, 2017 at 10:42
  • $\begingroup$ I have updated the answer to illustrate the inversion of energy diagram. $\endgroup$
    – drYG
    Jan 13, 2017 at 11:37
  • $\begingroup$ Ohm's law has indeed fluid flow analogy, but it can also be derived from Boltzmann equation for carrier distribution function, so it's safe. $\endgroup$
    – drYG
    Jan 13, 2017 at 11:51

I think this has got to do with the arbitrary convention of assigning the direction of "current" to be the opposite to the direction in which electrons flow (most probably due to historical reasons, though I'm not 100% sure). Current flows from a higher (+) potential to a lower (-) potential. But remember the crucial difference between potential and potential energy. Electrons are falling down from higher potential energy to a lower potential energy state. It seems counter-intuitive because negative mass doesn't exist in a gravitational analogy.

  • $\begingroup$ Thank you Arkya! I think you may have answered my question. So you are saying that if we had a negative mass particle equivalent to the electron, it would fly of the earth at 9.8m/s^2 rather than towards it and would in doing so reduce its potential energy converting into kinetic energy? $\endgroup$ Jan 12, 2017 at 17:27
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    $\begingroup$ @AndrewDavis Don't fail to accept this answer if it satisfies you. $\endgroup$
    – garyp
    Jan 12, 2017 at 18:03
  • $\begingroup$ @garyp I am simply seeking clarification that I have comprehended this answer $\endgroup$ Jan 13, 2017 at 10:26
  • $\begingroup$ @garyp exactly so.. $\endgroup$
    – Arkya
    Jan 13, 2017 at 20:13

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