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 'speff' PefhanyCommented Oct 26, 2014 at 14:35
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\$\begingroup\$ Okay, I'll make it an answer and expand on it a bit. \$\endgroup\$– Spehro 'speff' PefhanyCommented Oct 26, 2014 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.
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.
answered Oct 26, 2014 at 14:59
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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\$ Commented Oct 26, 2014 at 15:28
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\$\begingroup\$ Not sure I understand the question.. the plates are wires? \$\endgroup\$ Commented Oct 26, 2014 at 15:30
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\$\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\$ Commented Oct 26, 2014 at 15:31
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\$\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\$ Commented Oct 26, 2014 at 15:37
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\$\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\$ Commented Oct 26, 2014 at 15:42