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When a capacitor is charged, electrons flow from the battery terminal to the plate of the capacitor, as the electrons flow through the wire, they must be having some kinetic energy?

So, when the electrons finally stop at the plates of the capacitor, what happens to their kinetic energy? We don't normally assume the kinetic energy to get stored in the capacitor, we only consider the energy stored in the capacitor to be due to the decrease in potential energy of the charge as it flows from one terminal of battery to the other, then where does the kinetic energy go?

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  • $\begingroup$ It gets converted to heat and EM radiations $\endgroup$ Commented Oct 25, 2023 at 16:05
  • $\begingroup$ The energy stored in the capacitor comes from the chemical energy stored in the battery by way of the kinetic energy of the electrons. This is all a very simplified picture, but the idea is that the battery uses chemical energy to create a potential difference, which then accelerates the electrons to move them through the wire (turning potential energy into kinetic energy), and the electrons carry this KE with them to the capacitor, at which point they stop due to other electrons already in the capacitor slowing them down, converting KE back to PE stored in the capacitor. $\endgroup$
    – march
    Commented Oct 25, 2023 at 16:07
  • $\begingroup$ Heat is released when the current flows across a resistor, but we consider this heat only due to the change in potential across the resistor, and not kinetic energy, isn't it? $\endgroup$ Commented Oct 25, 2023 at 16:08
  • $\begingroup$ @march , whan the electron return from the other plate of the capacitor, to the +ve terminal of battery, suppose there is no resistance attached to this path, it then stops at the -ve terminal of the battery, then where does its kinetic energy go at that time? $\endgroup$ Commented Oct 25, 2023 at 16:15
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    $\begingroup$ @march The KE of the electrons due to the small field in the wire is small and dissipates in a few micrometers. It does not participate in the delivery of energy to the capacitor. It's an energy loss mechanism: energy carried by the electrons heats the wire. The electromagnetic field outside the wire is what delivers the energy. The wire guides that flow, but does not contain it. $\endgroup$
    – John Doty
    Commented Oct 25, 2023 at 17:41

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To expand slightly on the answers by @The Photon and @Dheeraj Gujrathi, the dominant contribution to the kinetic energy of the electrons in the circuit is the thermal motion of the electrons. That energy typically increases in resistances in the circuit (resistors, long wires, etc.) and we track it as thermal energy, not as kinetic energy.

The additional motion of the electrons due to what is happening in the battery is what we call the "drift velocity" of the electrons. In any typical circuit (copper wires, resistances of a few Ohms or more) the drift speeds are tiny. Values of mm/s or even $\mu$m/s are pretty typical. As a result, the additional kinetic energy in the circuit due to the drift motion of the electrons is negligible. So, it is rare to even talk about any increase of kinetic energy in a circuit. We talk about chemical energy in a battery being converted to thermal energy in resistances, and being stored as electric potential energy in capacitors (and magnetic potential energy in inductors). The only time we would have a circuit with non-negligible kinetic energy would be if some circuit element is basically a particle accelerator, such as a cathode ray tube (old-style TV screen), a fluorescent light, an X-ray emitter, or something like that.

You replied below with

suppose a capacitor is being charged with a resistor between the +ve terminal of battery and one of the capacitor plates, as some charge flows through the resistor, heat is produced, some of it is lost to the surroundings, while some of it increases the kinetic energy of the electrons(vibrational energy), so as these electrons then go and stop at the capacitor plates, where does there thermal energy go? Do the electrons still retain their vibrational(thermal) energy at the capacitor plates?

I tried to reply with an additional comment but found that I couldn't make the comment short enough to fit. So here's my answer to your comment below.

That increase in the kinetic energy of the electrons which you are calling "vibrational energy" is thermal energy. It didn't go anywhere, other than perhaps into the environment as the wires and resistor equilibrate to the temperature of the air around them. Typically once energy has been converted to thermal energy it is difficult to convert it to any other form. Difficult, but not impossible. That's what a lot of a thermodynamics course deals with. But in this situation all the energy that becomes thermal energy will remain as thermal energy.

So, let's track (simplified...) the energy conversions which occur in the circuit.

In the battery chemical energy is being converted to electric potential energy.

As the electrons move from one plate of the capacitor to the other they pass through the resistor. Some of the electric potential energy they have gained will be converted to thermal energy in the resistor. Once it is thermal energy it stays as thermal energy. It will now be transferred gradually to the surroundings, but it won't convert to any other form of energy.

Some of the electric potential energy that the battery produced will be stored "in" the capacitor. It is probably more correct to think of it as being stored "in" the electric field between the capacitor plates. Alternatively, you could think of it as being stored "in" the position of the electrons in the same sense that a mass raised to a larger height has more gravitational potential energy due to its position.

That's it. That's all of the energy transformations going on here.

However, all of the picture I've painted above is, in some sense, "lies told to children". The picture of the energy being "carried" from the battery to the resistor and the capacitor by the electrons has some major flaws. A much more correct picture is that electric and magnetic fields are produced in the vicinity of the circuit, and the energy transfers from the battery to the resistor and the capacitor via these fields. This does result in small motions of the electrons which collide with atoms in the resistor causing the conversion to thermal energy there, and small rearrangements of the electrons which produce the E-field in the capacitor, resulting in storage of electric potential energy there. To get a nicer idea of this more correct picture, have a look at the YouTube video that Veritasium did a while back. That video is a bit controversial, but I would say not controversial among people who really understand electrodynamics.

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    $\begingroup$ suppose a capacitor is being charged with a resistor between the +ve terminal of battery and one of the capacitor plates, as some charge flows through the resistor, heat is produced, some of it is lost to the surroundings, while some of it increases the kinetic energy of the electrons(vibrational energy), so as these electrons then go and stop at the capacitor plates, where does there thermal energy go? Do the electrons still retain their vibrational(thermal) energy at the capacitor plates? $\endgroup$ Commented Oct 25, 2023 at 18:00
  • $\begingroup$ also...I know this is unrelated to the question, but can you please explain what are the assumptions that we make, that should be kept in mind when a current is flowing through an ideal wire of 0 resistance? And what is the speed of the electrons in an ideal wire, if I have a resistor connected to a ideal battery via a ideal wire, is the speed of electrons in the ideal wire the same as the speed of the electrons as they pass through the resistor? $\endgroup$ Commented Oct 25, 2023 at 20:05
  • $\begingroup$ @AdityaMukherjee that really should go into a separate question. Comments are not for new questions $\endgroup$
    – Dale
    Commented Oct 25, 2023 at 21:19
  • $\begingroup$ @Dale , ok I am asking it seperately. $\endgroup$ Commented Oct 26, 2023 at 3:07
  • $\begingroup$ @gleedadswell ,thanks, my doubt is clear now $\endgroup$ Commented Oct 26, 2023 at 5:59
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as the electrons flow through the wire, they must be having some kinetic energy?

They have, but because the electrons are frequently interacting with the fixed charges (protons, but shielded by the bound electrons) in the wire molecules, their motion is highly randomized. They are moving in all directions, with only a small "bias" or tendency to move more in the direction opposite to the current.

This means that their kinetic energy is mostly considered as simply thermal energy. That is, kinetic energy associated with random thermal motion of the electrons. Only a small portion of the kinetic energy is associated with the motion that actually produces current.

So, when the electrons finally stop at the plates of the capacitor, what happens to their kinetic energy?

The electrons continue to move about randomly due to their thermal energy. The small bias in their direction of motion is eliminated, but they continue to move randomly.

This happens throughout the wires in the system, not just at the capacitor plates.

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  • $\begingroup$ so can I say that the kinetic energy is negligible, so we ignore it? $\endgroup$ Commented Oct 25, 2023 at 16:31
  • $\begingroup$ @AdityaMukherjee, depending what question you are answering, probably. It's normal in modeling conduction through materials (as opposed to, say, in a vacuum tube) to assume the motion of the electrons is randomized so that the currents are controlled solely by electrical potential (and its gradient) rather than by the kinetic energy or momentum of the electrons. $\endgroup$
    – The Photon
    Commented Oct 25, 2023 at 16:34
  • $\begingroup$ but in an ideal wire, there is no resistance, so what happens in that case? We neglect the kinetic energy of the electrons even in case of ideal wires. I saw some other posts on this site that say that an electron travels at light speed in an ideal wire, will its kinetic energy not be significant in that case? $\endgroup$ Commented Oct 25, 2023 at 16:43
  • $\begingroup$ @AdityaMukherjee, please link to the Q&A that claimed electrons travel at lightspeed. The "ideal wire" is an abstraction used in circuit theory. It's not meant to be a physical description of how real (copper or gold or ...) wires behave. It doesn't even really apply to superconductors. $\endgroup$
    – The Photon
    Commented Oct 25, 2023 at 16:49
  • $\begingroup$ physics.stackexchange.com/a/491932/377374 $\endgroup$ Commented Oct 25, 2023 at 16:58
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As a reply to your earlier comment and a Answer to the question, here is the argument :

What do you think, the average kinetic energy due to drift of Electrons comes from? When heat is produced, It is produced as follows, As soon as potential difference creates electric field, Electrons which were on average immobile, start to drift, causing current, When it happens, Electron bump to other fixed charges like positive nucleus, In the time gap where electron starts moving after first bump and second bump, it has moved forward, where its potential energy has decreased(because electric field did a small work on electron) and kinetic energy has increased, but as soon as it bump onto another nucleus, it looses the some of gained kinetic energy, this happens repeatedly and randomly

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It is but a very small effect as the drift speed of the electrons is of the order $10\,\mu\rm m/s$ and the speed due to the thermal motion of the electrons is of the order of $1\,\rm km/s$ a factor of $10^8$.
Noting that kinetic energy if proportional to the square of the speed the difference in kinetic energy is of the order of $10^{16}$ and so can be neglected.

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