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Can someone explain how is it possible to calculate the value of caps connected in series?

I am not asking about the maths, I just try to understand how can such a network be created since there is the gap(open) of the circuit, because of the dielectric(insulator).

I can understand how the capacitance works in one capacitor because of the AC, in case of DC I know that the cap will charge and then if we short-circuit the legs of the capacitor, it will discharge.

But what about when we have 2 or more capacitors in series? In my understanding it will be used only the one plate of the first cap(the plate closer to -) and the other plate of the other cap(closer to the +) of the power supply. How the in-between caps will affect the total capacitance since they are not even "connected" to anything due to the dielectrics of each one? I mean they are insulated if there is no current between them.

Any insights much appreciated! Please make it more visual, use some kind of analogy if possible :)

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    \$\begingroup\$ I think that you're misunderstanding capacitor charging process. Capaciyor is already „full" when you get it. When you „charge" it, electrical charge is redistributed in it and it stores energy. You have current flow out of one terminal and into another. In case of series connecrion, capacitors in the middle aren't disconnected, they're connected to other capacitors. \$\endgroup\$
    – AndrejaKo
    Nov 20, 2014 at 12:47
  • \$\begingroup\$ Hi Andrejako, thanks for your input. How the current flows from one terminal into the other? On what I have read so far there is no no actual flow of electrons between the two terminals, there is flow of current between the source terminals and respectively each of the capacitor plates. \$\endgroup\$
    – Electro Jo
    Nov 20, 2014 at 16:51

3 Answers 3

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If you have a bunch of caps in series then the act of charging them is to apply a varying electric field across the ends of the bank of capacitors. Whether this varying electric field is created near-instantaneously (a direct connection to a battery with very short leads) or via a resistor, there will be a rate of change of electric field applied to the bank. This is not just a DC phenomenon but an AC phenomenon as well.

You say: -

I can understand how the capacitance works in one capacitor because of the AC

If you do understand what you say you understand then it's not a big step to realize that current isn't just the transfer of charge (conduction currents) but it can also be a "displacement current" and this is due to the change of electric field. Displacement currents are also responsible for EM wave propagation (no conductors in space to carry ordinary conduction current of course).

In a single capacitor, it is displacement current that passes thru the insulation and becomes conduction current on leaving the capacitor plate.

With several equal value caps in series, the final static electric field across the lot of them is applied voltage divided by total accumulated width of the dielectrics i.e. volts per metre. With equal value caps this field is shared equally. It is also shared equally when charging and hence the total capacitance of n equal-value series capacitors is C/n where C is the capacitance of an individual capacitor.

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  • \$\begingroup\$ Thank for your reply Andy. So in reality there is current flowing through the dielectric in the form of displacement current? I have to make clear -as if it is not well obvious :) - that I am trying to figure out how electronics work by reading books and online. What I have read so far says that there is no movement of electrons through the insulation and the only movement happens between the source terminals and each plate on it's own. I might loose something, please correct me where is needed. \$\endgroup\$
    – Electro Jo
    Nov 20, 2014 at 16:46
  • \$\begingroup\$ It's tricky for sure. There are no flow of electrons thru the dielectric but there is "this" current flow nonetheless just as in rf propagation thru a vacuum. You can measure the effect of "this" current because it produces a magnetic field inside the cap just as an EM wave propagates. Maybe a little more reading by you and me is in order. Sometimes valid questions from folk like you expose weaknesses in my knowledge so I'll brush up on this later to see if I can give a better explanation. \$\endgroup\$
    – Andy aka
    Nov 20, 2014 at 18:53
  • \$\begingroup\$ really appreciate your effort to explain \$\endgroup\$
    – Electro Jo
    Nov 20, 2014 at 19:47
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It's quite simple really:

One plate of the first capacitor is negatively charged - that is it has excess electrons. Those electrons repel the electrons in it's opposite plate.

Those electrons have to go somewhere, so they go towards the first plate of the second capacitor - so that plate then has excess electrons, which then repels the electrons in the second plate of the second capacitor.

And of course those repelled electrons have to go somewhere ... ad infinitum.

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  • \$\begingroup\$ (part A) Hi Majenko, in the simple example of one capacitor and a power source, to make it easier a DC source(battery), the difference in electrons of the two plates of a capacitor is generated due to the difference of electrons on the power source. That means the excess electrons of the battery terminal are pushed on one of the plates of the capacitor and they "stay" there because they cannot go over the dielectric, at the same time the positive terminal of the battery attracts the negative charges(electrons) from the other plate of the capacitor. \$\endgroup\$
    – Electro Jo
    Nov 20, 2014 at 16:37
  • \$\begingroup\$ (part B) So, is it really repulsion? (and if yes through which route) or is it just movement of electrons from and towards the cap plates because of the potential difference of the power source and that happens through the circuit? Electrons do not move through the dielectric, right? \$\endgroup\$
    – Electro Jo
    Nov 20, 2014 at 16:37
  • \$\begingroup\$ It is both repulsion and "sucking" by the power supply. The PSU pushes the electrons to the first capacitor, and sucks them out of the second, but in between the two it's then repulsion (and attraction) that moves the electrons around. Some electrons can pass through the dielectric - that's the leakage current - but very few. \$\endgroup\$
    – Majenko
    Nov 20, 2014 at 22:09
  • \$\begingroup\$ (Part A) All fine about the repulsion and "sucking" of PSU on the outermost plates of the first and last capacitor. What happens on the "inner" plates is what makes me wonder, the inner plates are almost completely insulated from the PSU repulsion and sucking forces -due to the dielectric, except the tiny current that leaks as you said-. \$\endgroup\$
    – Electro Jo
    Nov 21, 2014 at 7:09
  • \$\begingroup\$ (Part B) In my view the repulsion (and attraction) between them could only happen if there is some kind of imbalance on their each "plate-load" so if we use completely discharged capacitors to connect in series it should not be any kind of move on electrons between the inner plates of capacitors connected in series. Many thanks again about your replies! \$\endgroup\$
    – Electro Jo
    Nov 21, 2014 at 7:10
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I did some more realllllly basic reading and is all about electrostatic force that makes the existence of charge in between the plates of the capacitor.

So when electrons are building on one plate, they repel the electrons of the other plate, they do not "physically" push them but due to the elementary principal of the "repulsion of same charges and attraction of the opposite" this is what happens, just like the charges between the huge capacitor of our natural space. Sky and earth where the dielectric is the air in between.

This is a video that explains it better https://www.youtube.com/watch?v=3TmuYAz2_B8&list=PLJVQe4CzhfPaPLOZri-DVziUF6pmtxKKe

Really appreciate everyone who answered my question!

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