What happens if you put an axial inductor between the plates of a cylindrical capacitor (capacitor supplied with AC signal)?
Axial inductor between the plates of a cylindrical capacitor
capacitorinductor
Related Solutions
Why you would want to have the plates different areas escapes me but here goes.
The charge on the plates will be the same.
The electric field between the plates will essentially follow the geometry as the charge on the smaller and larger plate will be evenly spread over its surface forming equipotentials parallel to the plates.
Charge = capacitance x voltage (\$Q=C\cdot V\$)
If the capacitor has a voltage across its plates and the supply is disconnected, the charge remains irrespective of the distance so, if distance increases (and capacitance falls) then voltage increases proportionally. If the plates are taken to an infinite distance, the voltage becomes infinite.
It should be noted that the energy "held" in the capacitor increases as the plates are pulled apart i.e.
Energy = \$\dfrac{CV^2}{2}\$
The increase in energy comes about because work (joules) has to be done to move the plates physically apart i.e. there is a force needed to open up the gap. This, I believe keeps all the conservation of energy and charge equations happy and smiling. Remember, that on a regular capacitor, there is an attractive force between the two oppositely charged plates and it is this force that is trying to stop the plates from being pulled-apart.
If the capacitor plates remain connected to the supply, as the distance increases the voltage must stay the same so therefore charge is reduced (because C reduces) and this pushes current back into the power source.
Best Answer
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.