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My book says a capacitor is two conducts being connected by an insulator. Now let's take a parallel plate capacitor to simplify the problem I have.

Suppose I got two parallel plate capacitor in series and I hook the circuit up with a battery.

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As soon as I hook it up, electrons flow (forget conventional current for now) into on the right plate and builds up on that capacitor (and spreads on the surface of conducting plate) and remains stuck there because there is an insulator that blocks the electrons from going anywhere.

Now here is my confusion, how does the left plate of $\ C_1$ even build the positive charges and how do the current even run through the circuit if there an insulator blocking the electrons from moving? Is there even current through the circuit?

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3 Answers

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Yes current does flow until $q/V$ equals $C$.

In the case of two capacitors in series, the effective capacitance is $1/(1/C_1 + 1/C_2)$, because the voltage $V$ is effectively divided between them.

Maybe an easier way to see this is to define inverse capacitance $K = 1/C = V/q$ which is, for a particular capacitor, the amount of voltage $V$ needed to put a given charge $q$ on one plate (and $-q$ on the other).

Then it's easier to see that $K = K_1 + K_2$, because the voltages across them sum, because they are in series. The charge on the right plate of $C_1$ comes from the left plate of $C_2$.

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How does current flow in the first place when the insulator blocks it? –  sidht Apr 30 '12 at 20:21
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@jak: That's the same as asking how does current flow into a single capacitor, or even how does current flow in a conductor. It flows in a conductor because there is a potential gradient that pulls the charge carriers. In a capacitor, the potential gradient is transmitted across the gap. An open switch is just a capacitor with very small capacitance. –  Mike Dunlavey Apr 30 '12 at 20:26
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@jak: That's right. The air doesn't stop the potential from being felt across the gap, and if the plates were really really large, current could flow for quite a long while before it slowed down and stopped. –  Mike Dunlavey Apr 30 '12 at 20:47
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@jak: You can always fall back on a hydraulic analogy. Think of a capacitor inserted in a wire as a springy diaphram inserted in a water pipe. In your case, you have two of them, that's all. –  Mike Dunlavey Apr 30 '12 at 21:01
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@jak: Yes. Conductors are always full of charge, just like water in a pipe. We only say the capacitor is "charged" when enough of the carriers have been moved to one plate, and the lack of them has been moved to the other, making them want to get back, and that "wanting" is the voltage. –  Mike Dunlavey Apr 30 '12 at 21:27
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In a simple DC circuit current doesn't flow and there is no voltage difference between the left plate of C! and the right plate of C1.

Picture it as a single capcitor with an unconnected plate in the middle

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Will the system blow up if you never unhook the battery then? –  sidht Apr 30 '12 at 20:22
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The voltage across the capacitor will initially rise until it equals the battery voltage - then it just sits there –  Martin Beckett Apr 30 '12 at 20:27
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Current does flow through the circuit but only for a short time, as a transient effect, while the charge builds up on the plates, and after that current flow stops. You can think of electrons in the wire to the left as being attracted to the positive terminal on the battery, so they move towards it. However, because the circuit is open, soon enough the electrons near the plate leave a "hole" of positive charge on the plate, which soon enough becomes as attractive to the electrons in the wire as the battery, so the situation stabilizes. Needless to say, the analogous (but charge-opposite) process happens on the wire to the right, so the plate becomes negatively charged.

Simultaneously (i.e. with a delay too short to measure), the electrons on the middle wire then get subjected to the attraction of an increasingly positive charge to their left and the repulsion of an increasingly negative charge to the right, so they tend to move left and accumulate on $C_1$'s right plate (leaving a corresponding excess of positive charge on $C_2$'s left plate). After the transient, these charges exactly cancel out the charges on the other plates (the ones connected to the battery) so electrons in the wire do not feel any force and no current flows.

EDIT: Wikipedia's capacitor article has a nice section on a hydraulic analogy to a rubber membrane completely sealing a pipe, but able to stretch to a finite amount.

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+1--Transience of the current seems to go to the heart of the question asked. –  daniel Sep 18 '12 at 0:20
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