Subfield: Thermodynamics

Problem T2. Kelvin water dropper

The following facts about the surface tension may turn out to be useful for this problem. For the molecules of a liquid, the positions at the liquid-air interface are less favourable as compared with the positions in the bulk of the liquid. This interface is described by the so-called surface energy, $U=\sigma S$, where $S$ is the surface area of the interface and $\sigma$ is the surface tension coefficient of the liquid. Moreover, two fragments of the liquid surface pull each other with a force $F=\sigma l$, where $l$ is the length of a straight line separating the fragments.

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A long metallic pipe with internal diameter $d$ is pointing directly downwards. Water is slowly dripping from a nozzle at its lower end, see fig. Water can be considered to be electrically conducting; its surface tension is $\sigma$ and its density is $\rho$. A droplet of radius $r$ hangs below the nozzle. The radius grows slowly in time until the droplet separates from the nozzle due to the free fall acceleration $g$. Always assume that $d \ll r$.

Part A. Single pipe
Context question:
i. Find the radius $r_{\max }$ of a drop just before it separates from the nozzle.
Context answer:
\boxed{$r_{\max }=\sqrt[3]{\frac{3 \sigma d}{4 \rho g}}$}


Context question:
ii. Relative to the far-away surroundings, the pipe's electrostatic potential is $\varphi$. Find the charge $Q$ of a drop when its radius is $r$.
Context answer:
\boxed{$Q=4 \pi \varepsilon_{0} \varphi r$}


Question: iii. Consider the situation in which $r$ is kept constant and $\varphi$ is slowly increased. The droplet becomes unstable and breaks into pieces if the hydrostatic pressure inside the droplet becomes smaller than the atmospheric pressure. Find the critical potential $\varphi_{\max }$ at which this will happen.

Solution:
Excess pressure inside the droplet is caused by the capillary pressure $2 \sigma / r$ (increases the inside pressure), and by the electrostatic pressure $\frac{1}{2} \varepsilon_{0} E^{2}=\frac{1}{2} \varepsilon_{0} \varphi^{2} / r^{2}$ (decreases the pressure). So, the sign of the excess pressure will change, if $\frac{1}{2} \varepsilon_{0} \varphi_{\max }^{2} / r^{2}=2 \sigma / r$, hence

$$
\varphi_{\max }=2 \sqrt{\sigma r / \varepsilon_{0}}
$$

The expression for the electrostatic pressure used above can be derived as follows. The electrostatic force acting on a surface charge of density $\sigma$ and surface area $S$ is given by $F=\sigma S \cdot \bar{E}$, where $\bar{E}$ is the field at the site without the field created by the surface charge element itself. Note that this force is perpendicular to the surface, so $F / S$ can be interpreted as a pressure. The surface charge gives rise to a field drop on the surface equal to $\Delta E=\sigma / \varepsilon_{0}$ (which follows from the Gauss law); inside the droplet, there is no field due to the conductivity of the droplet: $\bar{E}-\frac{1}{2} \Delta E=0 ;$ outside the droplet, there is field $E=\bar{E}+\frac{1}{2} \Delta E$, therefore $\bar{E}=\frac{1}{2} E=\frac{1}{2} \Delta E$. Bringing everything together, we obtain the expression used above.

Note that alternatively, this expression can be derived by considering a virtual displacement of a capacitor's surface and comparing the pressure work $p \Delta V$ with the change of the electrostatic field energy $\frac{1}{2} \varepsilon_{0} E^{2} \Delta V$.

Finally, the answer to the question can be also derived from the requirement that the mechanical work $d A$ done for an infinitesimal droplet inflation needs to be zero. From the energy conservation law, $d W+d W_{\mathrm{el}}=\sigma d\left(4 \pi r^{2}\right)+\frac{1}{2} \varphi_{\max }^{2} d C_{d}$, where the droplet's capacitance $C_{d}=4 \pi \varepsilon_{0} r$; the electrical work $d W_{\text {el }}=\varphi_{\max } d q=4 \pi \varepsilon_{0} \varphi_{\max }^{2} d r$. Putting $d W=0$ we obtain an equation for $\varphi_{\max }$, which recovers the earlier result.

Final answer: $\varphi_{\max }=2 \sqrt{\frac{\sigma r}{ \varepsilon_{0}}}$