Flat-plate cell capacitance and field

What is the capacitance of a flat-plate cell, and how hard is the field across the water?

C = \frac{\varepsilon_0\varepsilon_r \cdot A \cdot p}{d} \qquad E = \frac{V}{d}

Method

1. Convert the plate area from square centimetres to square metres (divide by 10,000) and the gap from millimetres to metres (divide by 1,000). Both conversions are places where a factor of ten commonly goes missing.
2. Count the active gaps: a stack of n interleaved plates has n − 1 gaps, and they sit electrically in parallel, so they add.
3. Multiply the permittivity of free space (8.854 × 10⁻¹² F/m) by the water’s relative permittivity, by the facing area, and by the number of gaps.
4. Divide by the gap. Capacitance goes as one over the gap here — unlike the coaxial cell, where it goes as one over a logarithm — so halving the gap really does double the capacitance.
5. For the field, divide the applied voltage by the gap. Compare it against the breakdown strength of what is in the gap, which for distilled water is around 65–70 kV/mm and for anything with ions in it is much less.

Assumptions

• The plates are parallel and evenly spaced. An adjustable cell that has been set by hand rarely is, and the closest pair carries a disproportionate share of the field.
• Fringing at the plate edges is ignored. For plates whose width is many times the gap this is a small underestimate; for a wide gap it is not.
• Every gap in the stack is identical. If they are not, the stack is a chain of unequal capacitors and the smallest gap dominates both the capacitance and the breakdown risk.
• The field figure is the average across the gap. Real field concentrates at edges, burrs and corners, so breakdown starts well below the average figure this reports.
• No gas in the gap. Bubbles are both a lower permittivity and a much lower breakdown strength, which is why a cell in production arcs at voltages a still cell tolerates.