This is a very theoretical question, just for intuition.

I have the following circuit: enter image description here

I would like you to focus on the flow and movement of a single electron when the switch is closed

enter image description here

  • When the switch is closed, an electron flows from the metal wire and top plate of C1 into the DC source positive side, thereby leaving the C1 plate positively charged.
  • Since a capacitor mirrors charge, the other plate of C1 must gain an electron, thereby making this plate negatively charged.
  • Here is my question or misunderstanding - where does this electron that made C1 negative come from? If it comes from the C2 plate, then only the C2 plate would be postively charged. If it comes from C3, then only C3 would be positively charged plate. If it comes just from the metal wire, then an electron from either C2 or C3 top plate would come to fill that gap in the metal wire, but only C2 or C3 can have a positively charged plate.

My question is - if only a single electron flows into the C1 bottom plate (the one connected to C2 and C3), what will it cause for C2 and C3? Will only one of them be positively charged?

Another question. How do think about these plates? C2 and C3 and C1 negative plate are all shorted. Can I think about them like a single plate?

  • \$\begingroup\$ Batteries can source electrons. \$\endgroup\$
    – Jun Seo-He
    May 19, 2022 at 10:55
  • 1
    \$\begingroup\$ Electron motion isn't as discrete as you're thinking. If you take one electron from the left side of C1 and force it into the bottom node of the circuit, all of the other electrons in the circuit shift slightly upward, resulting in a tiny electric field in all three capacitors. The amount of the shift and the distribution of voltage depends on the relative values of the capacitors. And yes, you can think of them as a single plate. \$\endgroup\$
    – Dave Tweed
    May 19, 2022 at 11:03
  • \$\begingroup\$ @DaveTweed If I think of them as one plate, how do I justify the fact that some of the plates are positively charged and some are negatively charged? \$\endgroup\$ May 19, 2022 at 11:15
  • \$\begingroup\$ What justification do you need? That's how capacitors work! You effectively have three plates -- the top plate of C1, the bottom plate of C2+C3, and a "floating" plate in the middle. Obviously, no electrons flow into or out of that middle plate, but its voltage relative to the other two changes slightly. \$\endgroup\$
    – Dave Tweed
    May 19, 2022 at 11:17
  • 3
    \$\begingroup\$ Electrons are physically discrete, but the fields they create -- and voltage, which is just an integral of the field -- are continuous. When we talk about the "charge" of a capacitor, we're talking about the energy stored in the field between its plates. This is not directly related to the charge of a single electron. \$\endgroup\$
    – Dave Tweed
    May 19, 2022 at 11:46

2 Answers 2


Your premise that a 'single electron flows into the plate' is false. While electrons are discrete, electric fields are continuous and are made up of the influence of a very large number of electrons.

When a capacitor gets charged, a very large number of electrons move very slightly closer to one plate along its connection wire, and on the other plate, they move very slightly further away. This creates an electric field in the capacitor's dielectric layer.

If you want to put one electron's worth of charge onto the capacitor, then a gazillion electrons will move a gazillionth of some small distance towards or away from the plate.

This is of course a 'lie to children'. The electrons individually are moving about at enormous speed, and it's the average position of the gazillion that shifts slightly towards or away from the plate.

In devices where single electrons do cross a junction, like a diode for instance, you get 'shot noise'. Crudely this is when an electron leaves the group on one electrode whose average position controls the electric field in the junction, and crosses a junction to join the averaging group on the other side. Some components suffer from shot noise, some do not. Semiconductor diodes do, and obviously from the above discussion, capacitors do not. Vacuum tube diodes sometimes do, sometimes don't, depending on whether the flow is space charge limited, which field modifies the time electrons leave the cathode and arrive at the anode, reducing the discrete effect of the individual electrons.

  • 1
    \$\begingroup\$ Love it: "a gazillionth of some small distance" -- thank you for quantifying my "slightly" comment! :-) \$\endgroup\$
    – Dave Tweed
    May 19, 2022 at 15:22

Electrons chug along fairly slowly and so to get, for example, a bulb to light almost instantly when a switch is closed requires that many, many electrons are being delivered to the bulb at any one instant in order to achieve the required current flow. This is not difficult to visualize when we consider the thickness of the wires compared to the size of individual electrons.

In your capacitor circuit, the minimum theoretical limit on the number of electrons flowing into the positive battery terminal and out of the negative terminal at any one instant is two, that is to say in pairs. This then causes one electron to flow off of each of the top plates of C2 & C3 and onto the bottom plate of C1.

As I think you already realise, when a capacitor is charging or discharging, the currents in its two leads must always be identical to each other at any instant in magnitude and in the same direction, that is to say one current flows in and the other identical current flows out. This gives the illusion of the current actually flowing through the capacitor but in reality no conduction current passes directly between a capacitor's plates except for a tiny leakage current.

So in your circuit, when the switch closes, as with any series circuit the current at all points must be identical in magnitude and in the same direction. This results in a pulse of current flowing off of the top of C2 & C3 and on to the bottom of C1. The flow of charge results in voltage increases across all 3 capacitors.

So even though the tops of C2 & C3 and the bottom of C1 are at identical voltages, a flow of charge does flow between them when the switch is closed. This can be understood as being no different to a flow of electrons on to the bottom of C2 & C3 even though the voltage at that point doesn't change. This flow of charge has the effect of increasing the voltages across C2 & C3 as an identical amount of charge flows off of the top of C2 & C3 and onto the bottom of C1 increasing C1's voltage as electrons are removed back to the batteries positive terminal from the top of C1.

And so, the minimalist concept is that two electrons must flow in and out of the battery (in pairs) to cater for one electron being removed from the top of each of C2 & C3.

To consider a single electron flowing from the top of C1 back to the battery resulting in a single electron being removed from the top of C2 or C3 would result in C2 and C3 having different voltages across them which is not theoretically possible as they are shorted together although in reality there could be voltage drops across the finite resistances of the leads etc.

If you were to consider the concept of a single electron flowing from the top of C1 back to the batteries positive terminal then you would probably be forced into considering the concept of fractions of electrons flowing off of the top of C2 & C3 onto the bottom of C1 and I wouldn't like to make any further comment about the validity of the concept of fractional electron current flow!


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