I Computing the capacitance of a strange material

[fluidistic]

I have a 2D material which has the shape of a quarter annulus. That is, in polar coordinates $(r, \theta)$, $\theta$ ranges from $0$ to $\pi/2$ and $r$ ranges from $r_i$ to $r_o$. When I apply a thermal perturbation to the material, i.e. when I keep the inner and outer curved boundaries at different temperatures (corresponding to the curves at $r=r_i$ and $r=r_o$), a rather complicated electrostatics potential field setups. None of the boundaries of the material are at a same potential. This electrostatics field creates a charge separation. If we consider that we have electrodes at the inner and outer boundaries (where the temperature is fixed), then each electrode would have 0 net charge, although along them there is a charge separation. From $\theta \in [0,\pi/4]$, a net charge $-q$ builds up, while for the rest of the curve, that is, $\theta \in [\pi/4,\pi/2]$ a charge $+q$ builds up.

If I were to use this material in a circuit, I guess it would have a capacitance $C$. How would I compute it? Would it be 0 because the net charge along those curve vanishes? Then choosing the electrodes to be at the $\theta=0$ and $\theta=\pi/2$ curves would fix this, in that the charge wouldn't be 0. But there is an additional feature, is that if this material is placed in an electric circuit, then the electrostatics potential field might greatly differ from the one I obtained in this open circuit. Would $C$ change? Or is this computed only in open circuit conditions?

 

[Baluncore]

I have a 2D material which has the shape of a quarter annulus.

You want to know the capacitance between two filamentary arcs. The voltage gradient close to the filaments will be infinite because they have a radius of zero.

Since in 2D, you allow no third dimension, there will be either no area, or no separation, so you cannot calculate a capacitance.

You will need to assign non-zero radii to the two filaments, making them cylinders.

 

[fluidistic]

I will post later a picture. I can define a distance between 2 boundaries where opposite charges are present. The problem being the electroststics potential being non constsnt there. However I might define.a sort of unusual capacitance, where instead of a ΔV, a ΔT creates the charge separation. There is a direct link beteen the 2 and no ambiguity as to what each vsriable mean. The question is whetger this C behaves as the usual C in a circuit or not.

More details to come.

 

[fluidistic]

Here are V and $\vec E$, followed by a picture of the charge distribution.

While the material is in 2D, its $\vec E$ field extends in 3D. It contains energy, so there's a direct relationship between the energy stored in the $\vec E$ field and $\Delta T$ which generates this charge separation. So I can define a sort of capacitance $C'$ that possesses different units than the usual one, and links $\Delta T$ to $U_E$ rather than $\Delta V$ to $U_E$.