Coaxial cell capacitance

What is the capacitance of a tube-in-tube cell with water between the walls?

C = \frac{2\pi\varepsilon_0 \cdot ε \cdot l}{\ln\!\left(b / a\right)}

Method

1. Convert the diameters and the length from inches to metres. The formula is in SI and one inch is exactly 0.0254 m.
2. Find the water’s relative permittivity at 20 °C from its type, then adjust it for temperature — about 0.4 % lower per degree above 20 °C.
3. Take the natural logarithm of the ratio of the diameters. Because it is a ratio, using radii instead of diameters gives the same answer; using the gap instead does not.
4. Multiply 2π by the permittivity of free space (8.854 × 10⁻¹² F/m), by the relative permittivity, and by the overlapping length.
5. Divide by the logarithm. The result is in farads; for a cell of these dimensions expect hundreds of picofarads to a few nanofarads.

Assumptions

• The electrodes are perfectly concentric and the whole overlapping length is submerged. A tube sitting off-centre has a higher capacitance than this, because the close side contributes more than the far side loses.
• End effects are ignored. The field bulges outward past the ends of the electrodes, which adds a little capacitance — negligible for a long cell, not for a short one.
• The water is a pure dielectric. It is not: it conducts, and above a few kilohertz the loss is significant. This gives the capacitance, not the impedance — see the Cole-Cole calculation.
• No gas is present. Bubbles on the plates displace water with something whose permittivity is 1 rather than 80, so a cell in full production has measurably less capacitance than a still one.
• The permittivity is the static value. Water holds ~80 all the way up through the megahertz range, so this is safe for any VIC frequency.