Fall 2007 20 Misc

In the absence of other forces, surface tension causes a liquid droplet to assume a spherical shape. Lord Rayleigh has shown that this is no longer true for an electrified droplet of a sufficiently large charge Q. (This instability has found a practical application in ink-jet printers.) Compute the corresponding critical charge Qc for a droplet of radius R and surface tension $\sigma$.
Hint: The capacitance C of a nearly spherical object is related to its surface area S by the Aichi-Russel formula

(1)
\begin{align} C = \sqrt{S/4\pi} \mbox{ (Gaussian units) } \end{align}

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The total energy of the droplet is

(2)
\begin{align} E(S) = \frac{Q^2}{2 C} + \sigma S = \sqrt{\frac{\pi}{S}} Q^2 + \sigma S \end{align}

Function E(S) has the minimum at S = Sc,

(3)
\begin{align} S_c = \left ( \frac{\sqrt{\pi}}{2} \frac{Q^2}{\sigma}\right ) ^ {2/3} \end{align}

However, since the sphere has the mimimal surface area for a given fixed volume, S cannot be smaller than $4 \pi R^2$. As a result, for small Q the sphere remains the optimal shape. The critical charge is determined by the condition $S_c = 4 \pi R^2$, which gives:

(4)
\begin{align} Q_c = 4 \sqrt{\pi \sigma R^3} \approx 7.1 \sqrt{\sigma R^3} \end{align}

in agreement with Lord Rayleigh (1882). At Q somewhat larger than Qc the droplet deforms into a prolate ellipsoid. This is the answer the student is expected to give for this problem.

Actually, an astute reader may realize that this answer may be incomplete. In principle, the droplet can also change its shape discontinuosly, e.g., by splitting into two smaller droplets. Let us examine this "first-order transition" scenario assuming the new droplets are also spherical and equal in size.

For the droplet of charge q and radius r the energy is

(5)
\begin{align} E = \frac{q^2}{2 r} + 4 \pi \sigma r^2 \end{align}

Comparing the energies of one droplet with q = Q and r = R with that of two droplets with q = Q/2 and r = R/21/3, we conclude that the first-order instability occurs at

(6)
\begin{align} Q > Q_m = \left (8 \pi \frac{2^{1/3}-1}{1 - 2^{-5/3}} \right )^{1/2} \sqrt{\sigma R^3} \approx 3.1 \sqrt{\sigma R^3} \end{align}

We see that Qm < Qc, and so the first order transiton wins. More precisely, the spherical droplet with charge Qm < Q < Qc is metastable, and so in practice it still may have a long lifetime.