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Leak resistance across a spherical cell

Through what resistance does the charge on a sphere-in-sphere cell leak away?

The formula
R=12π·σ(1a−1b)
R
Leak resistance, Ω
a
Inner diameter, in
b
Outer diameter, in
σ
Conductivity, µS/cm
LaTeX 
R = \frac{1}{2\pi \cdot σ} \left( \frac{1}{a} - \frac{1}{b} \right)

Work it out

Fill from a part the library

in

The inner electrode's outside diameter.

in

The outer electrode's inside diameter.

µS/cm

What a conductivity meter reads.

Compare with a variation

Result

R Leak resistance 4.177 kΩ

The resistance of the water between the spheres.

With your numbers
4.177 kΩ=12π·5 µS/cm(11.5 in−13 in)
LaTeX 
4.177\,\mathrm{kΩ} = \frac{1}{2\pi \cdot 5\,\mathrm{µS/cm}} \left( \frac{1}{1.5\,\mathrm{in}} - \frac{1}{3\,\mathrm{in}} \right)

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 inner sphere diameter Inner sphere diameter swept from 0.75 in to 2.25 in with everything else held at your numbers. The dashed lines cross where you are.
Leak resistance against inner sphere diameterLeak resistance falls from 12.5 kΩ to 1.39 kΩ as inner sphere diameter rises from 0.75 in to 2.25 in. At your inner sphere diameter of 1.5 in it is 4.18 kΩ.0510150.511.522.51.5 in4.18 kΩInner sphere diameter (in)Leak resistance (kΩ)
The formula behind the curve
R=12π·σ(1a−1b)
R
Leak resistance, Ω
a
Inner diameter, in
b
Outer diameter, in
σ
Conductivity, µS/cm
LaTeX 
R = \frac{1}{2\pi \cdot σ} \left( \frac{1}{a} - \frac{1}{b} \right)
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 Inner sphere diameter, which moves it by about 22.2% for a 10% change. It is least sensitive to Water conductivity, at about 11.1%.Change in the answer when each input moves by 10%-40%-20%20%40%Inner sphere diameter±22.2Outer sphere diameter±11.1Water conductivity±11.1
The formula behind the curve
R=12π·σ(1a−1b)
R
Leak resistance, Ω
a
Inner diameter, in
b
Outer diameter, in
σ
Conductivity, µS/cm
LaTeX 
R = \frac{1}{2\pi \cdot σ} \left( \frac{1}{a} - \frac{1}{b} \right)

Method

  1. Convert the diameters to metres and the conductivity to siemens per metre.
  2. The resistance between concentric spheres of radii r_a and r_b through a conductor is (1/(4πσ))(1/r_a − 1/r_b). Written in diameters the 4π becomes 2π, which is the form above.

Assumptions

  • Two complete concentric spheres of uniform water; the stem and filler port are ignored.
  • The meter's conductivity, uniform and at DC — see the tubular leak calculation.