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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=dσ·A
R
Leak resistance, Ω
d
Gap, mm
A
Plate area, cm²
σ
Conductivity, µS/cm
LaTeX 
R = \frac{d}{σ \cdot A}

Work it out

Fill from a part the library

cm²

The area of one plate facing the other.

mm

The water-filled distance between the plates.

µS/cm

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

Compare with a variation

Result

R Leak resistance 8 Ω

The resistance of the water between the plates.

With your numbers
8 Ω=1 mm125 µS/cm·100 cm²
LaTeX 
8\,\mathrm{Ω} = \frac{1\,\mathrm{mm}}{125\,\mathrm{µS/cm} \cdot 100\,\mathrm{cm²}}

Worth knowing

  • A leak of 8 Ω is a short as far as step charging is concerned: across a nanofarad it discharges in 0.008 µs. Only distilled or deionised water gets the leak into the kilohms.

This result is a link — the address bar holds your numbers, so it can be pasted into a post and opened to the same answer.

What this looks like

Leak resistance against gap Gap swept from 0.5 mm to 1.5 mm with everything else held at your numbers. The dashed lines cross where you are.
Leak resistance against gapLeak resistance rises from 4 Ω to 12 Ω as gap rises from 0.5 mm to 1.5 mm. At your gap of 1 mm it is 8 Ω.46810120.50.7511.251.51 mm8 ΩGap (mm)Leak resistance (Ω)
The formula behind the curve
R=dσ·A
R
Leak resistance, Ω
d
Gap, mm
A
Plate area, cm²
σ
Conductivity, µS/cm
LaTeX 
R = \frac{d}{σ \cdot A}
What moves the answer Each input moved 10% either way, with the others held still, and the effect on leak resistance.
What moves the answerLeak resistance is most sensitive to Plate area, which moves it by about 11.1% for a 10% change. It is least sensitive to Gap, at about 10%.Change in the answer when each input moves by 10%-20%-10%10%20%Plate area±11.1Water conductivity±11.1Gap±10
The formula behind the curve
R=dσ·A
R
Leak resistance, Ω
d
Gap, mm
A
Plate area, cm²
σ
Conductivity, µS/cm
LaTeX 
R = \frac{d}{σ \cdot A}

Method

  1. Convert the gap to metres, the area to square metres and the conductivity to siemens per metre.
  2. 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.

Assumptions

  • 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.