Electrical – ring-capacitor design

capacitancecapacitor

I created a simple design, based on the fundamentals of a parallel plate capacitor. Making a co-axial/tubular design:

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Front view:
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The "Air gap" is there to allow current flow in one-direction, from connector to end adjacent to the air gap.

I'm curious of the following:

  1. Does the connectors(to the exterior circuit) effect's the capacitance(\$C\$) of the design?

I only see the area between the plates being reduces based on the "width" of the connectors.

  1. As the capacitor charges/discharges (interacting with the exterior circuit) would the flow of current out/in via the connectors cause any effect to the plates and the electrostatic field(\$E_s\$) between them?

As for calculating the capacitance I believe this equation is still valid:

\$C= \frac{2\pi k\varepsilon_0}{ln(\frac{b}{a})} \$

Best Answer

If the "connectors" are made of a conductive material then they become just part of each plate. They will not change the capacitance of the structure unless the thickness of the dielectric is different beneath the connector (hard to tell from your drawing).

As Dave Tweed said, the air gap serves no electrical purpose.

Of course, current flowing through the connectors is going to change the electric field between the plates. That's how capacitors work. The current flowing into one plate must be balanced by an equal current flowing out of the other plate. The relative voltage of the plate with current flowing into it will increase, with respect to the other plate.

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