Atomic description of what happens in a capacitor.

Main Question or Discussion Point

I've examined in lab how a capacitor acts within a circuit. In a case where we charge a capacitor in series with a bulb and then another capacitor, I get the circuit in charging.png. So lets say electrons in the circuit flow from the positive end to the negative end and the circuit flows like it is labeled in Charging electron flow.png, the circuits become charged and the two plates of a capacitor are polarized, thus storing charge. Once the circuit is activated, the bulb lights up briefly and then it dims. Why does this happen at the atomic level? Next, we removed the battery and connected the circuit as it is with the charged capacitors as shown in discharging.png. Again, the bulb light up briefly and then dimmed shortly after. What happened to the charge? Is any charge or electricity lost during this connection?

My knowledge up to this point is basic interaction of charges, Coulomb's law, Gauss's law, and potential (voltage) calculations. I assume that whenever a circuit is completely charged, it will repel incoming electrons from the battery and therefore current will not flow. Am I right or wrong?

My thoughts of the discharging circuit are that some of the charge on the negative plates will travel to the positive plates and spread equally throughout the wire. Since they will be in equilibrium, there will be no flow and therefore the capacitors will be neutrally charged. Am I also right or wrong in this?

To summarize my questions:

In the charging circuit: If flow really does stop when the capacitor is charged, then why, atomically, does it stop?

In the discharging circuit: Is any charge lost during the discharging of the capacitors? If so, how?

Are the directions of current flow opposite in the charging and discharging circuit?

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Related Classical Physics News on Phys.org
* Yes the current stops when the capacitor becomes charged because the charged capacitor produces a voltage that counters that of the battery

* What do you mean by charges being lost? The question is not clear (to me)
* Yes the directions of current flow in opposite directions in the charging and discharging circuit

*How do the voltages cancel out? Is there a way to explain this in terms of interaction of particles?

*I mean that since the capacitor discharges and only lights up the light for a brief moment, does the charge go somewhere outside of the circuit? I don't think so, but I could be wrong. If not, why does the charge not flow? Was my reasoning correct?
*Makes sense! thanks!

Also, did I post a question like this in the right forum? This concerns Electromagnetism, which to my knowledge is part of classical physics... However, if questions like this belong in another section, please let me know! Thanks!

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Let's ignore the circuit for a minute and just talk about capacitors.

When a capacitor is charged by a battery, there are extra electrons sitting on one of the plates, and electrons are missing from the other plate, so one plate is negatively charged and one is positively charged. Overall, the capacitor is neutral -- it isn't positive or negative -- but one plate is positive and the other is negative.

In order to charge the capacitor up more, you will have to either push more electrons onto the negative plate or pull more electrons off the positive plate. This requires a battery or other power source. The potential difference required to push electrons onto the negative plate gets bigger the more negative charge is already on that plate, so for a given battery, the plate will "fill up" -- which just means the battery isn't strong enough to push any more electrons onto the plate. If you replaced the battery with a battery with a higher voltage, you could charge the capacitor up a bit more.

Now let's look at the circuit. The battery pushes electrons onto the negative (left-hand-side) plate of the bottom capacitor. (Your second and third diagrams are not correct -- electrons flow from the negative terminal of the battery, not the positive terminal, so you have the charges backwards.) These electrons repel the electrons that are on the right-hand plate of that capacitor. They are pushed down the wire, through the bulb (from bottom to top) and into the negative (right-hand, this time) plate of the top capacitor. The missing electrons mean that the right-hand plate of the bottom capacitor is positively charged.

Meanwhile, the extra electrons in the right-hand plate of the top capacitor repel the electrons in the left-hand plate of that capacitor, pushing them down the wire and into the positive terminal of the battery.

At first this whole process is very easy, and the capacitor starts to charge up, but as the capacitors get charged up, the electrons gathered on the negative plates start to push back against the battery. Eventually, there are so many electrons on the capacitors that they push back as hard as the battery is pushing forward. This makes the circuit "fully charged".

When you disconnect the battery and replace it with a wire, the battery is no longer pushing back against the electrons. So they are free to do whatever they want. Because your wire is a conductor, the electrons at the left side of the bottom capacitor are able to flow through the left wire and fill up the holes on the left side of the top capacitor. Similarly, the electrons at the right side of the top capacitor are able to flow through the wire, through the bulb (which lights it), and fill up the holes at the right side of the bottom capacitor.

gegen
Great response! Thank you for clearing that up!

Thanks eigenperson! I literally just signed up only to thank yourself, Your the ONLY person who has given a decent explanation so far (throughout the whole of the internet).

Seriously, you should become a teacher or something. You actually explained capacitors clearer than who actually have their own electronics/physics webpage. Cheers bud!

Warm wishes,
-James