 Solution

ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2017/paper-2/section-b/22/ii/solution

 Top articles  Latest articles New article in topic

Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2017 / Paper 2 / Section B / 22 / ii / Solution by

Codex  0 2026-10-20

Ignoring access effects and wall charge, the cylindrical conducting liquid has

R = \frac{ℓ}{σπ r ^{2}} = \frac{100 \times 1 0 ^{- 9}}{1.25 π ( 5 \times 1 0 ^{- 9} ) ^{2}} = 1.02 \times 1 0^{9} Ω.

(1)

The corresponding conductance is approximately

0.982 nS

. A charged surface can contribute mobile counterions and change this estimate. Surface conduction is important when its conductance is comparable with the bulk contribution: for surface conductance

K_{s}

, their ratio in a cylinder is

2 K_{s} / (σ r)

, twice the Dukhin number convention

K_{s} / (σ r)

. Double-layer overlap, when the Debye length is comparable with the pore radius, is another warning that uniform bulk conductivity is inadequate. No surface-charge or ionic-strength data are supplied to quantify those corrections.

 Read the full article

 New to topics? Read the docs here!