Leak resistance across a plate cell

Through what resistance does the charge on these plates leak away, given the water's conductivity?

The formula

R = \frac{d}{σ \cdot A}

LaTeX

R = \frac{d}{σ \cdot A}

Work it out

Fill from a part the library

A Plate area

cm²

The area of one plate facing the other.

d Gap

The water-filled distance between the plates.

σ Water conductivity

µS/cm

What a conductivity meter reads. Distilled is 1–5 µS/cm; tap water 200–800.

or see a worked example with typical numbers

Convert the gap to metres, the area to square metres and the conductivity to siemens per metre.
The resistance of a slab of conductor is its length over its conductivity times its cross-section: R = d ÷ (σA). The water between the plates is that slab.

The field and the current are uniform between the plates and there is no fringing at the edges. A real pair of plates leaks a little more around its rim.
The water's conductivity is the meter reading and is uniform — see the tubular leak calculation for what a high field does to that.
DC ionic conduction only; dielectric loss at the pulse frequency is the Cole-Cole calculation's.