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In my book it says that the capacitance in a capacitor with n electrodes is (n-1)*Co, Co is a capacitance of a two parallel plates capacitor. I know how to distribute charge when there are 3 parallel plates( two of them are connected to single Q,and the third which is in the middle to -Q), and I get 2*Co, but I can't figure out how to distribute charge when there are four of them.

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  • \$\begingroup\$ Assuming the plates are connected in an interleaved fashion as normal, there are n-1 gaps with n electrodes. Each one acts as a capacitor in parallel with the others. \$\endgroup\$ – Spehro Pefhany Oct 26 '14 at 14:35
  • \$\begingroup\$ Okay, I'll make it an answer and expand on it a bit. \$\endgroup\$ – Spehro Pefhany Oct 26 '14 at 14:53
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Assuming the plates are connected in an interleaved fashion as normal, there are n-1 gaps with n electrodes. Each one acts as a capacitor in parallel with the others.

Here is an image from this website.

http://www.itacanet.org/wp-content/uploads/2011/06/cap-600x294.jpg

You can see that there are 11 plates and 10 gaps resulting in effectively 10 parallel capacitors of area equal (approximately) to the area of the dielectric slabs.

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  • \$\begingroup\$ If I may ask you one more thing. What if three capacitors were long wirees, and first and third are connectd. How would that affect potential and charge distribution? \$\endgroup\$ – Desperado Oct 26 '14 at 15:28
  • \$\begingroup\$ Not sure I understand the question.. the plates are wires? \$\endgroup\$ – Spehro Pefhany Oct 26 '14 at 15:30
  • \$\begingroup\$ Yes, instead of 3 parallel plates, there are 3 parallel wires. I need capacitance of a system like that and I don't know how does this connection affects voltage. If this requires any calcutaions, ignore \$\endgroup\$ – Desperado Oct 26 '14 at 15:31
  • \$\begingroup\$ Capacitance between wires is much more complex. The middle wires won't shield the outer wires from each other the way the plates will. It's either a fairly complex calculation or a field solver simulation I think. \$\endgroup\$ – Spehro Pefhany Oct 26 '14 at 15:37
  • \$\begingroup\$ This is just entry level electronis and outer wires are connected, that means that potential difference is 0, and that implies that field between them is also 0. Right? \$\endgroup\$ – Desperado Oct 26 '14 at 15:42

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