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I need some explanations on the green underlined sentences.

1) "Must continue to have zero net charge..." What if the $\ C_2$ had +2Q on the left plate and -Q its right plate? The net charge between $C_1$ and $C_2$ obviously wouldn't be zero. So how does one go to deal with that situation? Also what is preventing the negative charges from staying on the plate of $C_2$? Sure you might say "well that's because like-charges repel!", why can't it stay and the capacitor have a negative net charge?

2) Yes we see the picture, but it doesn't explain WHY the potential difference is different. If we use hydraulic analogy, how could the membrane be at different pressure if they are in the same pipe? The pipe should have the same pressure throughout.

  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/24653/2451 $\endgroup$
    – Qmechanic
    May 2, 2012 at 23:29
  • $\begingroup$ That one is about current, this one is about potential difference and the net charge existing between the two capacitors. $\endgroup$
    – Lemon
    May 2, 2012 at 23:38

2 Answers 2


1) You probably know how to solve simple electrostatics situations where you have a bunch of parallel plates (plates, not capacitors), with different charges. What happens here is that the excess charge ($+Q$) distributes on the outer surface of both plates. You then get $\pm\frac{3Q}{2}$ on the inner plates. The charge on the outer surface cannot be changed unless you supply it externally or ground the outer surface. It will not change in a circuit, since it is easier for the charges to accumulate on the inner surfaces. So the textbooks is slightly more valid if it says "must continue to have the same net charge".

Note that the charges on the inner surfaces must be equal and opposite, otherwise the net electric field inside the plate itself will be zero. This point solves your "why can't the charges stay" confusion as well, if I understood correctly.

2) A single wire has the same current, not potential. You probably know that we can analogify resistors with rocks and capacitors with diaphragms in hydrodynamics. Well, rocks can reduce the pressure, but not the speed (otherwise we'd have a buildup of water in the rocks). So in this case itself we have different pressures in one branch. With a capacitor, this is more evident since a diaphragm obviously exerts an extra force, giving rise to extra pressure.

As time goes on, you have to wean yourself off the hydrostatics analogy--it doesn't explain everything (in fact, the analogy oftentimes is used the other way around)

Actually, you seem to have a poor understanding of pressure(don't worry, many people do), so I would shy away from the analogy till you do.

  • $\begingroup$ 1) So if the charges do stay, then both plates will have the same charge in sign and will have 0 E-field in between the plates and this gives a 0 potential. What is wrong with that? $\endgroup$
    – Lemon
    May 3, 2012 at 1:20
  • $\begingroup$ @jak the E field isn't zero.. If you have opposing charges, then they reinforce not cancel each other--they are on opposite sides and E is a vector. Put a negative charge in between and note that both plates pull/push it in the same direction. $\endgroup$ May 3, 2012 at 1:27
  • $\begingroup$ *same magnitude. $\endgroup$
    – Lemon
    May 3, 2012 at 2:35

Ok, I am so happy to read your question number 1. You asked it PERFECTLY and I really like how you included the page of the text book... It makes me realize I'm not alone in thinking about this stuff and that it is not OBVIOUS! In fact, I've been thinking A LOT lately about why a capacitor in series with other capacitors can't have, say, +2Q on one plate and -Q on another... For example, what if you have an asymmetric capacitor where one of the plates is much bigger than the other?... It seems reasonable to assume that such a capacitor could have uneven charges on its plates when connected to a circuit in series with other capacitors and still totally satisfy the conservation of energy, etc.

I think I have it (almost) figured it out!!! This question has led me to thinking very deeply about circuits... I think the best reference I've found so far that you NEED to read is: B. A. Sherwood and R. W. Chabay, ‘‘A unified treatment of electrostatics and circuits,’’ http://www4.ncsu.edu:8030/􏴯rwchabay/mi/ circuit.pdf. To try to answer your specific question, here is the deal:

  1. While it is rarely if ever explicitly stated in circuit books, there are surface charges (regardless of the presence of the capacitor) on the connecting wires and bends which sets up "secondary" electric fields in the circuit that in turn guide the electrons along the wires. As soon as you flip a switch or connect the capacitor, these charges are set up very fast (at the speed of light). These surface charges get configured in this perfect manner due to feedback (read the Sherwood and Chabay reference above). It's amazing that it works out this way, but I see it has to do with the electric field produced by the source battery, charge conservation, perfect electric conductivity of conductors, perfect insulation of vacuum, and that regions with excess surface charges create their own electric fields so all the charges in the circuit are interacting (pushing and pulling on each other...)... It is assumed that the current you solve for in a circuit (even when solving for transients) is due to charges moving down the axes of the wires AFTER the initial feedback process is complete.

  2. Now for the kicker... Check out this scan from the book Introductory Electromagnetics... You will notice the authors answer your question... well, sort of... If you read it, they say that "normally capacitors are made so that there is no field outside them if they are charged with equal but opposite charges"... and that THESE are types of lumped capacitors you assume in circuit theory that will have equal and opposite charges on their plates. However, I don't buy this argument... True, the infinite parallel plate capacitor does have zero field outside the capacitor because it is infinite... but any FINITE capacitor (real capacitor) does have field outside the capacitor... THEY MUST since that path integral of the electric field along ANY path (say, from one plate to the other) is equal to a constant number (the voltage difference)... since you can take any path you want, there must be a field outside the capacitor... I think the real answer is that there must be equal and opposite charges on the two plates of the capacitor (total charge equals zero) because in such a case the ELECTRIC FLUX will equal zero (same number of electric field lines leaving the capacitor as entering it)... If the flux was not zero EVEN FOR AN ASSYMETRIC CAPACITOR then there would be, say, more electric field lines pointing towards the capacitor than away from it... and charges from the surrounding wires would respond to this misbalanced field and flow into the capacitor until the charges on each do exactly balance... I'm not 100% sure of this argument... but it is the best I could come up with so far! Oh I should also say that this argument assumes the capacitor is not not electrically "coupled" to other parts of the circuit except through its leads....

UPDATE (DAY LATER): I think I got it!!! My problem was that I thinking about it only in terms of quasi-static forces and charges, when I should have been thinking dynamics and fields. The charges on both plates of capacitors (even an asymmetric capacitors) connected in series are the equal and opposite due to 1. charge conservation and 2. continuity of displacement current with conduction current. The charges had to get to the capacitor somehow in the first place... which is a electrodynamics problem involving displacement current.

It's not, as the book Introductory Electromagnetics states, because the E-field is zero outside the capacitor. The thing that makes it all work out is the assumption that the E-field lines from the positive plate must all terminate at the negative plate. Boundary conditions say that where those E-fields terminate at the surface of the conductor you have charge. So, in the end, this assumption is equivalent to saying the electric flux across the boundary of a capacitor is zero. If the flux was not zero then you would model the points in the circuit between areas where electric lines begin and terminate with another lumped capacitor! capacitor!

  • 1
    $\begingroup$ Hello Aaron, Please be specific on your answer. I'd say that, "It's a way too CHATTY". Also, please reduce the usage of symbols. (It's too much for your answer). At least, try to make your answer a bit clear :-) $\endgroup$ Oct 25, 2012 at 18:58

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