Electronic – How capacitors connected in series are affecting the total capacitance of the network, since there is no actual flow of electricity due to dielectric

capacitorseries

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 🙂

Best Answer

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.