Why the charge is the same for capacitors connected in series?

Question

For example in this figure, I can understand that the charge on plates 1 and 4, and plate 2 and 3 must have the same value because of charge conservation. Is there any justification that plate 2 must have the same charge as plate 1?

Answer 1

Is there any justification that plate 2 must have the same charge as plate 1?

In a series circuit the current (charge per unit time) is the same going through all components. That means at any instant in time the positive charge supplied by the positive terminal A making plate 1 positive has to equal the positive charge exiting plate 2 making it equally negative, and so on for all the plates returning to the negative terminal B.

Hope this helps.

Answer 2

First think about why there is any charge separation on plates 2 and 3 at all. The bulk material of plates 2 and 3 is being suffused by an electric field from plates 1 and 4, which acts on charges in plates 2 and 3 (and the connecting wire). The effect is to pull negative charges to the surface of plate 2 and push positive charges to the surface of plate 3.

How much charge will accumulate on plates 2 and 3? Exactly the amount needed so that the electric field contribution from plates 1 and 4 into the bulk of plates 2 and 3 is cancelled out by the electric field from the charges on plates 2 and 3. (Remember the electric field inside the bulk needs to be 0 at electrostatic equilibrium.) Since the source of the field from plates 1 and 4 is the two charges +q and -q, the exact same amounts of charge but in reverse configuration must be pulled and pushed onto plates 2 and 3 to cancel out the field from plates 1 and 4.

This same cancelling out of field inside bulk occurs in the bulk material of a charged single capacitor at equilibrium, which may be easier to imagine. If it seems counterintuitive to think quantitatively about the two capacitor example with fields inside the bulk conductor cancelling out, instead think about a configuration of four insulating plates with charges placed on the same surfaces as the conductors. Calculate the field contributions from each plate in each direction using Gauss's law, and you'll see that the field cancels out everywhere but in the gap between the plates.

Answer 3

Refer to the posted figure, start with an uncharged capacitor and assume that the free charges in a conductor are positive. The "H"-shaped piece in the middle (from 2 to 3) has zero net charge. When the series combination is connected to the battery, it still has zero net charge because there is no path that will allow charge from the outside to flow in it.

However, the conducting piece from "A" to "1" is an equipotential at the potential of "+" terminal of the battery. The potential was raised by charges amounting to +Q that accumulated on plate "1". Since the entire circuit must have zero net charge, these charges must have come from the conducting piece "4" to "B" whose potential has decreased to match the "-" terminal of the battery. So you have charge Q removed from "4" and added to "1".

How are the free charges in the "H"-shaped piece in the middle going to respond to that? They will be attracted to "4" and pile up on "3" leaving a deficit of charges at "2". In other words, the positive charge on plate "3" is the same as the magnitude of the negative charge on plate "2". Furthermore, these magnitudes must be the same as the magnitudes of the charges on plates "1" and "4" because the entire circuit, as mentioned earlier, has zero net charge.

It's all charge conservation.