# Is this Voltage Variable Capacitor available commercially?

#### Leo Freeman

Each parallel mesh lies on an equipotential plane in the dielectric. All you are doing is building a stack of higher value capacitors in series that must be switched.
You might consider a variable capacitance multiplier, or maybe using something like a gyrator to simulate a capacitor.
Thanks for the tip about the gyrator! This is turning into a learning experience for a non EE like me.
I'm not looking so much for a variable capacitor, but to simulate a "super-dielectric", using applied potentials to force a capacitor soak up a huge amount of charge, by canceling any field within the plates.

Why do I really need a variable capacitor?
I have an intuitive hunch about a possible circuit, that is based on Kelvin's Thunderstorm apparatus.
It might be able to soak up the positive charges in the atmosphere. I have to sit down and try to sketch out this vague idea.

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#### Leo Freeman

A capacitor with a central conductor (a mesh like you describe) is equivalent to two capacitors in series. Add a wire to the connection point and you have a three terminal device. It's not very interesting device, however, because it's linear. What makes the patent interesting (for some value of interest) is the non-linearity of the mythical dielectric material.
Thank you for this info. I can see how a single mesh would make two capacitors in series, but what about two meshes?
Two meshes forming an inner capacitor, with a polarity that's the reverse of that of the outer plates?

Discrete capacitors in circuits have their own self-contained electric fields that are not visible to the outer circuit. But I'm still uneasy about the scenario where the field of one capacitor directly influences the field of another.
If I can settle this, you would have taught me more than all my Uni electronics courses combined.

#### Baluncore

When you insert a mesh into a capacitor you do not increase the energy storage because you must also reduce the voltage. The total energy that can be stored is proportional to the volume of the dielectric.

Capacitance is proportional to the area of the dielectric and inversely proportional to the thickness. Unfortunately, the thickness of the dielectric also decides the voltage rating. There lies the problem.
The energy stored in a capacitor is E = ½ C V2.

Split a capacitor along the V/2 equipotential and you get two capacitors.
They have twice the capacitance each, but only half the voltage rating.
The total possible energy stored after the split is then; E = 2 * ½ * 2C * (V/2)2
Which is unfortunately the same as the original E = ½ C V2.

Likewise, if you cut the capacitor in half, parallel with the field lines you make no change to the voltage, the total capacitance or the energy stored.

I can see how a single mesh would make two capacitors in series, but what about two meshes?

Two meshes forming an inner capacitor, with a polarity that's the reverse of that of the outer plates?
The voltage rating becomes V/3, the capacitance is 3C, and there are then three capacitors so;
E = 3 * ½ * 3C * (V/3)2 = ½ C V2 again.

#### Leo Freeman

When you insert a mesh into a capacitor you do not increase the energy storage because you must also reduce the voltage. The total energy that can be stored is proportional to the volume of the dielectric.
...
The voltage rating becomes V/3, the capacitance is 3C, and there are then three capacitors so;
E = 3 * ½ * 3C * (V/3)2 = ½ C V2 again.
I get it now. Thank you for explaining this so patiently!
So the fact that the inner mesh plates are actively charged by an external voltage, it still doesn't have any effect on the capacitance of the outer plates? How about using an electret as a dielectric?

Well, if Nature is not going to give me something for nothing, then I'll just have to take it by force! 