Kvant Physics Problem 34

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Problem

Figure 2 shows a droplet electrostatic machine (Kelvin water dropper). From a tube into a hollow insulated metallic sphere of radius $R$, water droplets fall, charged to a potential $\varphi_0$. How does the limiting potential to which the sphere can be charged depend on the height from which the droplets fall?

Setup and Assumptions

A stream of water forms identical spherical droplets of radius $a$ and density $\rho$. Each droplet falls from a tube into an insulated conducting sphere of radius $R$ through a height $h$ in a uniform gravitational field $g$. The sphere carries an electrostatic potential $\varphi$ that evolves as charge is deposited by the droplets. The droplets emerge from the tube in an initial electrostatic environment corresponding to potential $\varphi_0$ at the nozzle.

Each droplet is treated as a small isolated conductor whose capacitance is $C_d = 4\pi\varepsilon_0 a$. Air resistance is neglected, so each droplet arrives at the sphere with kinetic energy $mgh$. Mutual interactions between droplets are ignored. The limiting potential $\varphi_{\max}$ is defined as the stationary value at which further charge amplification ceases.

The droplet mass is $m = \frac{4}{3}\pi a^3 \rho$.

Physical Principles

The electrostatic energy of a charged conductor at potential $\varphi$ relative to a reference at potential $\varphi_0$ is expressed as $U_e = \frac{1}{2}C(\varphi-\varphi_0)^2$, and the incremental work associated with transferring charge $q$ through a potential difference $\Delta \varphi$ is $q,\Delta \varphi$.

For a droplet behaving as a conductor, the charge induced by a potential difference between droplet and surroundings satisfies $q = C_d(\varphi-\varphi_0)$.

Energy conservation applies to the coupled electro-mechanical process of charge separation during droplet formation and transfer, with gravitational work $mgh$ supplying the energy required for charge separation and transport.

The limiting regime corresponds to vanishing net energy gain per cycle of droplet charging and deposition, so that the gravitational energy acquired by a droplet equals the electrostatic energy required to sustain the charge transfer process.

Derivation

A droplet leaving the nozzle is influenced by the potential difference between the nozzle region at $\varphi_0$ and the sphere at potential $\varphi$. The induced charge on the droplet is taken as

$$q = C_d(\varphi - \varphi_0).$$

When this droplet falls and delivers its charge to the sphere, it moves charge through a potential difference $\varphi - \varphi_0$, so the electrical energy transferred per droplet is

$$W_e = q(\varphi - \varphi_0).$$

Substituting the expression for $q$ yields

$$W_e = C_d(\varphi - \varphi_0)^2.$$

The gravitational energy released by the droplet during the fall is

$$W_g = mgh.$$

At the limiting potential, the system reaches a steady state in which the gravitational energy supplied per droplet exactly balances the electrostatic energy required for charge separation and transport, so

$$mgh = C_d(\varphi - \varphi_0)^2.$$

Solving for the potential difference gives

$$\varphi - \varphi_0 = \sqrt{\frac{mgh}{C_d}}.$$

Substituting the droplet mass $m = \frac{4}{3}\pi a^3 \rho$ and capacitance $C_d = 4\pi\varepsilon_0 a$ produces

$$\varphi - \varphi_0 = \sqrt{\frac{\frac{4}{3}\pi a^3 \rho g h}{4\pi\varepsilon_0 a}}.$$

Simplifying the geometric factors yields

$$\varphi - \varphi_0 = \sqrt{\frac{a^2 \rho g h}{3\varepsilon_0}}.$$

Result

The limiting potential of the sphere is

$$\varphi_{\max} = \varphi_0 + \sqrt{\frac{a^2 \rho g h}{3\varepsilon_0}}.$$

The dependence on the fall height $h$ is

$$\varphi_{\max}(h) = \varphi_0 + \sqrt{\frac{a^2 \rho g}{3\varepsilon_0}},\sqrt{h}.$$

Substituting symbols without numerical values gives the final expression in SI units:

$$\boxed{\varphi_{\max} = \varphi_0 + \sqrt{\frac{a^2 \rho g h}{3\varepsilon_0}} ;\text{V}}.$$

Sanity Checks

Dimensional analysis of the radicand gives

$$\frac{a^2 \rho g h}{\varepsilon_0} \sim \frac{\text{m}^2 \cdot \text{kg m}^{-3} \cdot \text{m s}^{-2} \cdot \text{m}}{\text{F m}^{-1}} = \frac{\text{kg m}^2 \text{s}^{-2}}{\text{C}^2 \text{N}^{-1} \text{m}^{-2}} = \text{V}^2,$$

so the square root has units of volts.

The dependence $\varphi_{\max} - \varphi_0 \propto \sqrt{h}$ matches the expectation that doubling the height increases available gravitational energy per droplet by a factor of two, while the potential grows only as the square root because electrostatic energy scales quadratically with potential difference.

A sign error would most likely occur in the identification of the energy transfer term $q(\varphi-\varphi_0)$, since reversing the potential difference would incorrectly predict a decrease of the limiting potential with increasing height.