Electrical – Total charge on capacitors in series

Two wires do make a capacitor. Just a very small one. For parallel plates, capacitance can be calculated as:

$$ C = \frac{\varepsilon A}{d} $$

Where:

• \$\varepsilon\$ is the permittivity of the dielectric, which is mostly air for some wires, with \$\varepsilon \approx 8.85 \cdot 10^{-12} \mathrm F/\mathrm m\$.
• \$A\$ is the area of the plates
• \$d\$ is the distance between the plates

For two ordinary wires in a circuit, \$A\$ is very small, and \$d\$ is very large, compared to the distances in your typical capacitor. Thus, the capacitance is really, really small, and we can neglect it in most cases.

As for your second question, you have to be careful about the words you use. Does charge mean electric charge or how much energy you've stored in the capacitor? I'm not the only person frustrated by the contradictory vocabulary around capacitors. I'll do my best to be clear.

The charge imbalance does spread out along the wire, in one sense. Between the battery terminals, or between any two points along the wire, or between the plates of the capacitor, you will measure the same potential difference with your voltmeter. The electric field exists not only between the plates of the capacitor, but between the two entire halves of the circuit.

Inside the capacitor, the electric field must change from the potential of one half to the potential of the other half within a very tiny distance, just the separation of the plates (\$d\$ from above: it's tiny to make a high capacitance). Thus, the field strength, measured in volts per meter, is highest inside the capacitor.

As far as where the electric charge goes, think of it this way: half of the circuit has too many electrons, and the other half of the circuit has not enough. When there are too many electrons, they want to move to someplace where there are less, because like charges repel. So for the half with too many electrons, the closest they can get to a place where there are less electrons is inside the capacitor, because it's closest to the other half of the circuit.

Not all the electrons pile up in the capacitor, mind you, because that would leave the wire with a positive charge. Rather, the electrons redistribute themselves so that the potential difference (voltage) is the same everywhere in that half of the circuit. Most of the excess electrons end up in the capacitor, precisely because this is where the electric field is strongest.

You can also think about this for the opposite half by considering the absence of an electron to be a "hole", a sort of positive charge carrier.

You can also think about how the electric charges distribute themselves this way: we already established that the wires have a very low, but non-zero capacitance. Capacitance \$C\$ is just another way of saying how much charge \$Q\$ it takes to make a voltage \$V\$ in a thing:

$$ C = \frac{Q}{V} $$

The wires, having a low capacitance, don't take much electric charge imbalance (extra or missing electrons) to make a big change in voltage. The capacitor, having a large capacitance, takes much more charge imbalance to change the voltage. Thus, to make the voltages equal across each half of the circuit, most of the imbalanced charge must end up in the capacitor, not the wires.

Your gap is too wide. Make it very, very very narrow. And rolled into a cylinder. Like a real capacitor.

Yes, +q could be less than -q, but only if the attraction/repulsion effects of electrons in the connecting wires were nearly as large as the attraction/repulsion down between the capacitor plates. (In that case the plates wouldn't be a near-perfect electrical shield for the fields produced by the wires.) But with real-world capacitors, this doesn't happen, and instead the field between the plate is totally enormous compared to the tiny fields produced by electrons in the wires. If +q only differs from -q by a millionth of a percent, we ignore it. See Engineer's capacitor vs. Physicist's capacitor, a split metal ball, versus two separate balls.

For capacitors used in circuitry, if we dump some charge on one capacitor terminal, exactly half of it will seemingly migrate to the other terminal. Weird. But "physicist-style capacitors" with small, wide-spaced plates are different, and an extra electron on the wire will make +q not equal to -q.

In detail: if the capacitance across the plates is 10,000pF, and the capacitance to Earth of each wire and plate is 0.01 pF, then the opposite plate's charges will ignore any small +q and/or -q on the connecting wires. The attraction/repulsion of electrons in the wires doesn't significantly alter the enormous +q and -q on the inner side of the capacitor plates.

Engineers use real-world components: wide capacitor plates with very narrow gaps; gaps the thickness of insulating film. But if you were a physicist, your capacitors might be metal spheres with large gaps between, or metal disks where the space between the plates was large when compared to their diameter. (Or you'd draw a capacitor symbol where the gap between plates was enormous and easy to see.) In this case the attraction/repulsion of electrons on the connecting wires would have an effect on the balance of +q -q between capacitor plates.

PS

Another weird concept: make a solid stack of thousands of disc capacitors: foil disk, dielectric disk, foil disk, etc. Use half-inch wide disks, and stack them up into a narrow foot-long rod. Now connect one end to 1,000 volts. The same kilovolt will appear on the other end! The rod is acting like a conductor. Yet its DC resistance is just about infinite. Series capacitors! Each little capacitor induces charge on the next and the next, all the way to the end.