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Leak resistance across a plate cell, two ways

The same calculation run twice, side by side. Change anything on either side and the difference is shown output by output. This page is a link: both sets of numbers are in the address bar.

Side A open alone
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.

Side B open alone
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.

Swap sides

What changed

The two sides are the same. Change something on either.

Output A B B against A
R Leak resistance 600 Ω 600 Ω same

Side A, with its numbers

600 Ω=3 mm5 µS/cm·100 cm²
LaTeX 
600\,\mathrm{Ω} = \frac{3\,\mathrm{mm}}{5\,\mathrm{µS/cm} \cdot 100\,\mathrm{cm²}}
  • A leak of 600 Ω is a short as far as step charging is concerned: across a nanofarad it discharges in 0.6 µs. Only distilled or deionised water gets the leak into the kilohms.

Side B, with its numbers

600 Ω=3 mm5 µS/cm·100 cm²
LaTeX 
600\,\mathrm{Ω} = \frac{3\,\mathrm{mm}}{5\,\mathrm{µS/cm} \cdot 100\,\mathrm{cm²}}
  • A leak of 600 Ω is a short as far as step charging is concerned: across a nanofarad it discharges in 0.6 µs. Only distilled or deionised water gets the leak into the kilohms.