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What is meant by "Electrons lose potential Energy" when going through a resistor. In a case of falling from a height you would be at a lower altitude and therefore lower potential but how is this analogous to electrons. Do they slow down? Wouldn't this mean that they have just lost kinetic energy? If it is meant that they've lost potential energy due to being closer to the positive terminal of the battery when going through a resistor wouldn't this still hold even when there isn't a resistor? If this is the case electrons should always gradually be losing potential energy when going towards the positive terminal but are said they only lose it going through resistors. How?

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  • Gravitational potential energy is stored when two masses are separated. It is a consequence of the conservative force of gravitational attraction. As they get closer, this energy is released.

  • Similarly, electric potential energy is stored when two like charges are separated. It is a consequence of the conservative electric repulsion force. As they get farther away from each other, this energy is released (vice versa for opposite charges that instead attract).

This is not about slowing down. Place a charge next to a point that has a large charge already, and it will want to move as far away as it can. It may reach high speeds while doing this if nothing else slows it down (because the released electric potential energy then is converted into kinetic energy).

When dealing with classical circuits where the free charges are negatively charged electrons, then you might have a battery with a positiv terminal and a negative terminal. Plenty of charges are acculumated at the negative terminal, and none (less) at the positive terminal. We thus call the negative terminal a point of high potential and the positive terminal a point of low potential. Any electron in the circuit is repelled from the negative terminal, i.e. the point of high potential, and attracted towards the positive terminal, i.e. the point of low potential. In short, electrons "want to" move towards points of low potential.

As it moves along with this electric "push" towards the point of low potential, the stored electric potential energy is released. It might initially be converted into kinetic energy. But the higher speeds is immediately slowed again via resistors in the circuit that absorb this kinetic energy (typically as thermal energy and release it to the air as heat).

Ideally this will not happen without resistors. Ideal wires have no resistance. The electrons would thus accelerate and accelerate and accelerate forever... until they reach the other terminal and stop. In reality, all wires do have some internal resistance. Thus, an ever increasing electron current will eventually cause so much resistance that the wire either slows the electrons down to constant speed or that the wire burns and melts.

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  • $\begingroup$ Although the potential energy of the electron is high at the negative battery terminal, the potential at the terminal is low as it is based on conventional current ( flow of positive charge). Illogical, as most current is electron flow, but we're stuck with it thanks, I believe, to Ben Franklin. $\endgroup$
    – Bob D
    Commented Dec 19, 2021 at 15:16
  • $\begingroup$ @BobD Not if I redefine it as I've done in this answer. In general, a point of attraction is associated with low potential energy (leading to the general notion that systems seek configurations of lowest possible energy). Since we have two choices when dealing with electric theory, it has become consensus to pick the choice that matches how a positive charge would behave, agreed. This is only consensus, though, and there's no problem in doing - and explaining - it differently in different contexts. Just like when explaining or working with, say, the sign of work an heat in thermodynamics. $\endgroup$
    – Steeven
    Commented Dec 19, 2021 at 17:26
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    $\begingroup$ But the positive terminal is always described as high potential and the negative terminal as low potential because of conventional current. The casual reader may not realize you are describing them in terms of electron current. Other than the reference to free charges being electrons, I'm not sure you made that clear. But that's OK. It's probably just me. BTW it wasn't I who downvoted your answer. $\endgroup$
    – Bob D
    Commented Dec 19, 2021 at 17:47
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Electrons lose potential energy when going through wires. Electrons lose potential anytime work is being done on them. They don't slow down, its the opposite much like when we fall and gain kinetic energy we lose potential energy ( they are DEFINED as the negative of another) wires have resistance (in the definition of resistance there is Length) so when electrons move through a wire they do lose potential energy

If you would connect a single wire to a battery with no load. When an electron makes a single loop, it would lose all of the voltage.

In order for electrons to move, work must be done on them. By definition making them lose potential energy

Now resistance is a bit slightly trickier because even if the electrons are moving at a constant speed( aka no NET work is being done on them) they still lose potential energy as the work done on the electrons via the E field is balanced by the negative work done by resistive forces

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Electrons lose potential energy as they move around a circuit from one terminal of a battery to another. If you have n different circuits each powered by a 4.5v battery, then the electrons in each circuit will lose 4.5v of potential overall as they move around it. The overall loss of PE is equal to the voltage difference between the battery terminals.

More generally, electrons will lose PE when they move between one point of higher potential to another point with lower potential.

If you consider the potential along the path of a circuit from one battery terminal to the other, if won't, in general, change at a constant rate. What you will find is that the rate at which the potential changes will go up or down depending on the resistivity of the material the electrons are passing through.

If you have a 4.5v circuit consisting of three equal resistors in series, then there will be a a 1.5v drop over each resistor. If there is one resistor of 8 ohms and five resistors each of 0.2 ohm, then there will be a 4v drop over the first resistor, and an 0.1v drop over each of the subsequent five resistors. Likewise if there is a 8 megohm resistor and five 0.2megohm resistors, there will again be a 4v drop over the first and 0.1v drop over each of the others.

What you can see from those examples is that the overall drop of PE is 4.5v in each case, but the 4.5v total is the sum of the contributions from each resistor along the way, and the contributions are proportional to the values of the resistances.

In practice, most circuits are made of components joined by wires. Wires are resistors too, but their resistance is very low- usually so low that it can be assumed to be zero. So by convention, people usually use the word resistor to mean a component with a resistance that is much higher than the resistance of the conducting wire, the resistance of the wire being ignored.

Given all that, when text books say that electrons lose PE when they pass through resistors, what they really mean is that the electrons also lose PE when they pass along the wires joining resistors, but the loss of PE along the wires is so small compared with the loss of PE over the resistors it can be ignored.

Of course if you don't have any resistors at all, and the battery is short circuited with thick copper wire, the electrons still lose 4.5 volts of PE over the wire.

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    $\begingroup$ The potential energy of the electron is measured in joules or electron-volts, not volts. (I'm being pedantic about this because many students get confused about the difference between potential and potential energy so we should be very careful about distinguishing them when teaching) $\endgroup$
    – The Photon
    Commented Dec 19, 2021 at 16:45
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First of all, why can we define an electric potential along the circuit? Because there is a conservative electric field along the circuit. As the electron moves along the circuit, it is accelerated by this electric field, so the electric field does work on the charge. Consequently, it's electric potential energy decreases. I think you get this part already.

So why doesn't the potential energy of an electron depend solely on its position relative to the terminals of the battery, or on the length of wire it has travelled? The reason is that the electric field along the circuit is not uniform. If the only thing connected to the battery was a wire with uniform resistance, than yes, the electric potential of the electron may decrease uniformly with distance, because the electric field along the wire is approximately constant (even this may not be true, the presence of bends and kinks along the wire may result in a non uniform electric field). But when a resistor is introduced to the circuit, the electric field along the wire will be almost zero, while there will be a large electric field in the resistor. So the electric potential will largely fall over the resistor.

But why should the electric field be 'concentrated' in resistors and not wires? To understand this, we should look at what produces the electric field in the circuit. It turns out that the presence of the battery induces a layer of charge, called surface charges, on the outer surface of the wires/resistors. The electric field strength in the wire is proportional to the gradient of the surface density of these surface charges, that is, how quickly the surface charge density changes. The resistor creates a sort of bottleneck for these surface charges, causing positive surface charge to accumulate on one side, and negative surface charge on the other. This sets up a sharp surface charge gradient over the resistor which generates a significant electric field. Along the wires, the surface charge density is nearly constant, resulting in a negligible electric field.

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What is meant by "Electrons lose potential Energy" when going through a resistor.

The electrons are given electrical potential energy at the negative terminal of the battery. At the microscopic level they move through the resistor alternatively acquiring kinetic energy from the force of the electric field and transferring kinetic energy to particles of which the resistor is made due to collisions. This raises the temperature of the resistor causing the dissipation of energy in the form of heat.

The equal gaining of kinetic energy from the field and losing of kinetic energy by collisions results in a constant average kinetic energy and constant drift velocity (constant current) through the resistor. So the electrons on average do not "slow down". They move from having high potential energy at the negative terminal to lower levels of potential energy as they move through a series circuit, while the current remains constant. So the repeated sequence of energy conversion is electrical potential energy, to kinetic energy, to heat.

The analogy with a falling body is what happens after a falling body reaches its terminal velocity due to air resistance. The velocity of the body (analogous to current) remains constant while the body loses gravitational potential energy (analogous to electrical potential energy). Kinetic energy is simultaneously transferred from the falling body to the air molecules by collisions due to air resistance (analogous to the electrical resistance).

Hope this helps.

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  • $\begingroup$ Hi Bob. I understand that falling body reaches its terminal velocity after some time. but then you say KE is transferred from the body to air molecules, but if so, KE decreases, and if so, why doesn't velocity decrease then ? by decreasing KE, velocity should decrease , hence not stay constant. If you meant KE decreases, which decreases velocity, but then KE again increases due to potential energy decrease, I might understand but then no chance you would get the speed constant as it must be decreasing/increasing $\endgroup$
    – Matt
    Commented Jun 25, 2023 at 12:24
  • $\begingroup$ @Matt The KE doesn't decrease because at the same time gravitational potential energy(GPE) continues to be converted to KE as the object continues to fall at constant velocity. $\endgroup$
    – Bob D
    Commented Jun 25, 2023 at 12:32
  • $\begingroup$ if that happens, then KE must be increasing. Since it's not increasing(if it did, velocity would even increase more), then what actually happens is KE decreases, but same amount KE decreases by, same amount of PE converts to KE and hence KE stays constant. but to repeat my point, KE is still decreasing, it's just decreasing/increasing by the same amount. Is this what you mean ? $\endgroup$
    – Matt
    Commented Jun 25, 2023 at 13:18
  • $\begingroup$ @Matt The falling object attains its terminal velocity when the downward force of gravity equals the upward force of air resistance for a net force of zero. Gravity is doing positive work on the object (since its force is in the same direction as the displacement of the object) giving it kinetic energy. $\endgroup$
    – Bob D
    Commented Jun 25, 2023 at 13:40
  • $\begingroup$ But SIMULTANEOUSLY air resistance is doing an equal amount of negative work (since its force is in the opposite direction to the displacement of the object). Its negative work takes the kinetic energy away from the object. Per the work energy theorem the net work done on an object equals its change in kinetic energy. Since the net work done on the falling object is zero at its terminal velocity, its change in kinetic energy is zero. $\endgroup$
    – Bob D
    Commented Jun 25, 2023 at 13:41
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What is meant by "Electrons lose potential Energy" when going through a resistor. In a case of falling from a height you would be at a lower altitude and therefore lower potential but how is this analogous to electrons. Do they slow down?

It would be interesting to read the exact text you are quoting, for reference.

In general, when there is a force acting on an object and the work done by this force in accelerating the object does not depend on the specific path taken, we can say that the work done is just a function of start and end points the object has moved through. We then introduce a mathematical function that gives you values for all points in space and, if you take the difference between the values at two of these points and suppose your object has moved between them, then you automatically have the work done by the force on that object.

This function is called potential energy.

Many forces of interest have an associated potential energy - they're so called conservative - Gravity and Electric are examples.

Because of the relationship between the force and the potential, which is mathematically defined, you end up having that when following the influence of the forces, objects always move towards lower potential energy.

I particularly like this answer on the topic and "why" "things try to go to lower potential energy"

In electrostatics and electric circuit theory, it is convenient to use the electric potential, which is the potential energy per unit charge. So, to really get the work done by the electric field over a potential difference of - say - 1V, you need to know how much charge has been moved between those two points. Otherwise, you just have the work done on a unit charge, i.e. one Coulomb.

Onto the resistor, what is it? A resistor is a model of a specific phenomenon happening in the electric world: the transformation of electric energy in some other form of energy. The energy supplied by a battery to a resistor, the classic example of the light bulb for instance, is exiting the electric circuit, in form of heat and light (EM radiation).

enter image description here

There is an unfortunate circumstance when dealing with electric circuits, which causes a lot of confusion when first approaching the subject. See, we like to refer all the quantities like the forces, fields, and potentials, as to what they do to a positive charge. When we say "the electric field has this direction", for instance, what we actually mean is that this is the direction of the force a positive charge would experience. A negative charge would experience just the opposite :-) Same for potential, it is referred to the force "felt" by a positive charge. Same for current, it is the flow of positive charge. It so happens, however, that conduction in metals (and resistor is a "bad" metal, in the end) happens by movement of negative charge, the electrons. So this adds a layer of confusion.

So let's imagine, for now, that our conductor has positive charges able to move inside it. We can flip everything afterwards, but it is just easier to see that, as the resistor imposes:

$$V = RI$$

That is, if you want a current $I$ to flow across the resistor ends, then there has to be a potential difference of $V/R$ imposed at its terminals. Otherwise, no current will flow.

If you accept that, then you can also accept that the charges will pass from the left hand in the drawing above, where they are at a positive potential with respect to the other end of the resistor and also of the '-' sign of the battery (we call this point 'the reference'), to the same potential as the '-' sign, which we said is our reference, when going through the bulb/resistor, exiting on the right hand side of the bulb.

Therefore, in passing through the bulb, and as required by Ohm's law V = RI, the charges go from having a potential of V with respect to the negative node of the battery, to being on the wire that connects the negative of the battery with the bulb, which is at ZERO potential with respect to itself.

Did they accelerate, because they lost potential energy? Did they slow down?

Neither of the two, and here's the catch. Remember when I said that the resistor - which I depicted as bulb for this specific reason - actually models an outward flow of electric energy? I did it to introduce the reason why the positive charges, at the end of their tour through the bulb, lose potential energy but do not gain kinetic energy. This is because, the extra kinetic energy that the charges would have gained, they gave to the outside world! That's the light and the heat you benefit from :-D

The extra energy does not go in changing the momentum of the charges, it gets out of the electric circuit, for good . The component that you use to model this phenomenon is always, invariably, his Majesty the Resistor!

If you want to do with electrons, you just have to think that the flow for the electrons is reversed, because for them the potential is also reversed. They still lose potential energy, they tend to go to lower potential energy for them, which is just the classic electric potential, but with negative numbers.

As for an analogy, I was thinking of this:

Imagine a water pump that brings the water up to a certain level, say h, a mass of water m - Now, let the water fall through a rough pipe, actually so rough that the water falls at constant speed through the pipe. In doing so, it warms the pipe up, but the water does not accelerate, despite the constant gravity force acting on it.

When it reaches the bottom, it is lifted back up by the pump, which is actually the poor boy doing the work here!

Hope this helps.

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