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enter image description here

I'm a bit confused on the sign of charges in circuits like these.

The bottom (negative) plate of C1 has -Q charge and the top plate (positive) of C2 has +Q charge. However, these two plates are at the same node, so they are the same plate essentially.

How can they have a different charge each???

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The +/- represents the voltage relative only to the capacitor's other plate only.

If I am standing on your head, our feet are always below our heads, but that doesn't mean my feet and your head can't be at the same height.

The voltage potential at the bottom plate of C1 is not different than the potential in the top plate of C2. This is different than the charge on each plate.

If I have a large and small air tank and connect them together, their pressure (voltage) will be the same but their air volume (charge) won't be. In the same way, the larger plate of a larger cap needs more charge to produce the same voltage than a smaller plate.

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  • \$\begingroup\$ I see, so you're saying that even though the two capacitors share a plate, they can have a different charge? as long as the potential is the same? \$\endgroup\$ – AlfroJang80 Feb 2 at 3:28
  • \$\begingroup\$ @AlfroJang80 Yes, except they don't share a plate. The plates are different but share the same potential. It is no different than the situation when two capacitors in parallel with top (or bottom) plates connected together. \$\endgroup\$ – DKNguyen Feb 2 at 3:30
  • \$\begingroup\$ Ah. So like in the parallel case where in the fully charged state of e.g 5V, capacitor C1 could have a larger charge on it's plates than C2 (depending on size), but the voltage is the same. \$\endgroup\$ – AlfroJang80 Feb 2 at 3:34
  • \$\begingroup\$ Yeah. You can't really have a truly shared plate the way you might have been picturing it because of the influence of the opposite plate. It takes two plates to hold a charge and they hold it equally but opposite.If you had three plates in a series with the middle being a shared plate like a siamese twin capacitor, the two outer plates end up holding the same but opposite charge while the middle plate doesn't do much (I think). \$\endgroup\$ – DKNguyen Feb 2 at 3:41
  • \$\begingroup\$ Awesome. Thank you very much. \$\endgroup\$ – AlfroJang80 Feb 2 at 17:59
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It is an electrostatic behavior as shown below,

enter image description here

Does this make sense? Comparing to

enter image description here

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  • \$\begingroup\$ I have no idea what this is. \$\endgroup\$ – AlfroJang80 Feb 2 at 3:18
  • \$\begingroup\$ The lower plate of C1, the wire, and the upper plate of C2 together form the conductor in the first picture. Does this make more sense to you? \$\endgroup\$ – X J Feb 2 at 3:22
  • \$\begingroup\$ @AlfroJang80, just because a metallic object is an equipotential doesn't mean the charge is evenly distributed over its volume (or surface). If there's an external field, then the charge must redistribute (some positve charge over here, some negative charge over there) in order for the object to remain an equipotential. \$\endgroup\$ – The Photon Feb 2 at 17:05
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The C1 and C2 definitely share a common node. The charge on lower plate of C1 will be negative with respect to the top plate of the C1. The same plate can be treated as positive when the lower plate of C2 is taken as reference.

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Just remember that in electronics, when you see this type of notation, it means "with respect to the other side of the device". So the pluses on C1 are in comparison to the minuses on C1, not necessarily anything else. By looking at the design of the circuit, however, it is implied that the plus side of C1 is positive in comparison to the plus or the minus side of C2, for example.

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The bottom (negative) plate of C1 has -Q charge and the top plate (positive) of C2 has +Q charge. However, these two plates are at the same node, so they are the same plate essentially.

How can they have a different charge each???

In electrostatics, a conductive material will be an equipotential (the potential will be the same throughout its volume).

That doesn't mean that the charge in the material will be uniformly distributed.

If (like in this case) there's an external electric field, the charge will distribute unevenly. In fact, the charge has to re-distribute for the object to remain an equipotential.

In this case you have a metallic object (two plates joined by a wire), so the potential is the same throughout the object. But some negative charge has moved to the lower plate part of the object, leaving a net positive charge on the upper plate part of the object. This is necessary or else the whole object wouldn't be at the same potential.

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