 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-344/3/f/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

For a fixed center height

h

, the radius decreases according to the preceding current enhancement. Thus

τ_{h} (R_{0}) = \frac{2 ϕ _{B}^{2}}{M γ} \int_{0}^{R_{0}} \frac{R ^{2}}{c ( R / h )} d R .

Within the stipulated logarithmic approximation,

c (y) = 1 + y /2 + O (y^{2})

, so

1/ c (y) = 1 - y /2 + O (y^{2})

. Expanding uniformly over

0 \leq R \leq R_{0} ≪ h

and integrating gives

τ_{h} = \frac{2 ϕ _{B}^{2}}{M γ} [\frac{R _{0}^{3}}{3} - \frac{R _{0}^{4}}{8 h} + O (\frac{R _{0}^{5}}{h ^{2}})] .

Therefore the leading correction to the isolated-droplet lifetime is

τ_{h} (R_{0}) = τ_{\infty} (R_{0}) [1 - \frac{3 R _{0}}{8 h} + O ((R_{0} / h)^{2})], Δ τ = - \frac{ϕ _{B}^{2} R _{0}^{4}}{4 M γh} .

The negative correction reflects faster evaporation into the nearby reservoir. This leading coefficient also agrees with the small-

R / h

expansion of the exact sphere-plane capacitance; the stipulated logarithmic approximation need not match higher orders.

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