Kvant Physics Problem 34
A hollow insulated conducting sphere of radius $R$ is charged by water drops falling from a vertical tube.
Verified: yes
Verdicts: PASS + PASS
Solve time: 22m45s
Source on kvant.digital
Problem
Figure 2 shows a drop electrostatic machine (Kelvin generator). Water drops charged to a potential of $\varphi_0$ fall from a tube into a hollow insulated metal sphere of radius $R$. How does the limiting potential to which the sphere can be charged depend on the height through which the drops fall?
Setup and Assumptions
A hollow insulated conducting sphere of radius $R$ is charged by water drops falling from a vertical tube. Each drop carries a fixed electric charge $q$ that was established during its preparation. The drops fall through a vertical height $h$ before reaching the sphere. The drop’s charge is assumed constant during the fall, independent of the sphere’s potential, consistent with the problem statement specifying pre-charged drops.
The goal is to determine the limiting potential $\varphi_{\max}$ of the sphere, defined as the potential at which the electrostatic repulsion of the already accumulated charge prevents further drops from reaching the sphere. The sphere is treated as an ideal conductor, and the drops are small enough that their self-capacitance and mutual interactions are negligible. Air resistance is ignored, and the potential at infinity is taken as zero. Leakage currents and electrical breakdown of air are neglected.
The symbols used are
R \quad \text{sphere radius (m)}, \quad h \quad \text{fall height of a drop (m)}, \quad m \quad \text{mass of a drop (kg)}, \quad q \quad \text{charge carried by a drop (C)}, \quad \varphi \quad \text{potential of the sphere (V)}, \quad g \quad \text{acceleration due to gravity (m/s^2)}.
The drop potential $\varphi_0$ is the preparation potential of the drop, which fixes $q$, but it does not contribute directly to the drop’s motion in the external field.
Physical Principles
The drop moves under the influence of gravity and the electrostatic field of the charged sphere. Its mechanical energy consists of gravitational potential energy and electrostatic potential energy due to interaction with the sphere. For a drop with mass $m$ and charge $q$ at a position where the sphere potential is $\varphi$, the electrostatic potential energy is
$$U_e = q \varphi.$$
The gravitational potential energy relative to the sphere’s center at the final position is
$$U_g = m g h.$$
The total energy of the drop is conserved in the absence of dissipative forces. The drop can reach the sphere as long as the total mechanical energy is sufficient to overcome the electrostatic repulsion. The limiting potential $\varphi_{\max}$ is attained when the drop arrives with zero speed, so the gravitational energy gained during the fall is exactly equal to the increase in electrostatic potential energy due to the sphere’s charge.
Derivation
Let the drop begin at height $h$ above the sphere. The drop’s total mechanical energy relative to the sphere is purely gravitational initially, since the drop experiences negligible electrostatic potential from the sphere at the starting point, which is far from the charged sphere. Thus, the initial energy is
$$E_i = m g h.$$
When the drop reaches the sphere’s surface, its gravitational potential energy is zero, and its electrostatic potential energy in the field of the sphere is
$$E_f = q \varphi.$$
Energy conservation gives
$$m g h = q \varphi + \frac{1}{2} m v^2.$$
The limiting potential occurs when the drop arrives with zero speed, $v=0$, yielding
$$m g h = q \varphi_{\max}.$$
Solving for the limiting potential gives
$$\varphi_{\max} = \frac{m g h}{q}.$$
This formulation correctly accounts only for the interaction of the drop with the external potential of the sphere and the work done by gravity. The initial drop potential $\varphi_0$ enters only indirectly, through the drop charge $q$, which is fixed by the preparation of the drop.
Result
The limiting potential of the receiving sphere is
$$\boxed{\varphi_{\max} = \frac{m g h}{q}}.$$
The increase of potential above the external reference is
$$\boxed{\Delta \varphi = \varphi_{\max}}.$$
The limiting potential is linear in the fall height $h$ and inversely proportional to the drop charge $q$. The dependence on the drop’s preparation potential $\varphi_0$ appears only insofar as $\varphi_0$ determines the value of $q$; it does not directly add to the limiting sphere potential.
Sanity Checks
The ratio $m g h / q$ has dimensions
$$\frac{\mathrm{kg , m^2 / s^2}}{\mathrm{C}} = \frac{\mathrm{J}}{\mathrm{C}} = \mathrm{V},$$
confirming dimensional consistency. If $h = 0$, then $\varphi_{\max} = 0$, consistent with the physical expectation that a sphere cannot be charged by drops unless gravitational energy assists their approach. Increasing the fall height increases the gravitational work per unit charge, thereby raising the attainable sphere potential. This formulation relies only on the energy balance between gravity and electrostatic repulsion and does not assume any specific drop capacitance, making it consistent with the physical situation of pre-charged drops.