What happens if you put an axial inductor between the plates of a cylindrical capacitor (capacitor supplied with AC signal)?
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\$\begingroup\$ It depends what the inductor is connected to and how that relates to the voltage on the plates. Capacitors are all about electric fields, charge is stored on the plates because of the field between them. A conductor between the plates will affect the field if it is connected to something (otherwise it will just float). \$\endgroup\$– WillApr 6, 2014 at 20:48
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If the inductor is of negligible size and is connected to the plates (one side of the inductor to one plate, the other to the other plate), it forms a "tank" circuit, resonant at frequency \$f_0\$= \$1\over 2\pi\sqrt{LC}\$.
It doesn't really matter topologically whether the connection to the plates is at any particular place, so for ideal components, there's no difference.
In reality, the inductor would have some effect on the capacitance and a real inductor will have some distributed capacitance internally (between turns).
The physical arrangement imagined here is a metal tube with a coaxial rod inside (cylindrical capacitor) with the inductor connected between the inside of the tube and the external surface of the rod.
If it was not connected, it would depend on the way it was constructed, but mostly likely for most frequencies it would just increase the capacitance a bit because it is reducing the effective separation of the plates by virtue of inserting a conductor.
answered Apr 6, 2014 at 21:50
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\$\begingroup\$ Actually, I think OP envisions the inductor "between the plates" as axial, as though wound around (though insulated from) the center conductor. And not necessarily connected (by wire) to the plates. \$\endgroup\$– gwidemanApr 7, 2014 at 2:14
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\$\begingroup\$ @gwideman Okay, best not ask why. Anyway, I think that would fit with my second scenario. Little effect except for an increase in capacitance. \$\endgroup\$ Apr 7, 2014 at 2:20
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\$\begingroup\$ I'm thinking that if the inductor coil has windings that leave no airgap, then it's essentially interposing an additional "capacitor plate" between the inner and outer plate. If unconnected it just floats like Will suggested. If it's connected in some way then it has a complicated distributed inductor-capacitor relationship to each of the outer/inner plates. And in that scenario the inner and outer plates no longer have a capacitive relationship with each other. But yeah, not sure how this is useful :-). \$\endgroup\$– gwidemanApr 7, 2014 at 2:39