We blow a soap bubble with a surface tension of $\alpha$ from a soap solution, and then, leaving the bubble on a tube made of insulating material, we charge it to a potential $U$ (e.g., with a comb). What will be the radius of the bubble if we allow the air to flow freely through the tube? (Data: $\alpha=4 \cdot 10^{-2} \mathrm{~N} / \mathrm{m}, r_{0}=2 \cdot 10^{-2} \mathrm{~m}, U=6,7 \cdot 10^{3} \mathrm{~V}$.)
The task will be solved using the principle of virtual work. The total surface energy stored on the outer and inner surfaces of the sphere is \( W_{1}=2 \cdot 4 r^{2} \pi \alpha \), where the radius of the sphere is \( r \). The electrical energy can be calculated by determining the work done during the charging of the sphere with radius \( r \).

Let the final charge on the sphere be \( Q_{0} \), and let us add a charge \( \Delta Q=\frac{Q_{0}}{n} \) to its surface in one step. As we know, the potential on the surface of the sphere (relative to \( R=\infty \)) is:

\[ U=k \frac{Q}{r} \]

Thus, when adding the \( i \)-th \( \Delta Q \) charge, the work done is \( \Delta W_{i}=k \frac{Q}{r} \Delta Q \), and \( Q_{i}=\frac{i}{n} Q_{0} \). Substituting, we get \( \Delta W_{i}=k \frac{Q_{0}^{2}}{r} \frac{i}{n^{2}} \).

The total work done, which is the electrical energy of a sphere with radius \( r \) and charge \( Q_{0} \):

\[ W_{\mathrm{e}}=\sum_{i=1}^{n} \Delta W_{i}=k \frac{Q_{0}^{2}}{r} \sum_{i=1}^{n} \frac{i}{n^{2}} \text { in the limit as } n \rightarrow \infty \]

Since

\[ \begin{gathered}
\sum_{i=1}^{n} \frac{i}{n^{2}}=\frac{1}{n^{2}} \sum_{i=1}^{n} i=\frac{1}{n^{2}} \cdot \frac{n(n+1)}{2}=\frac{1}{2}\left(1+\frac{1}{n}\right) \rightarrow \frac{1}{2}, \text { as } n \rightarrow \infty, \text { therefore } \\
W_{e}=\frac{1}{2} k \frac{Q_{0}^{2}}{r}
\end{gathered} \]

For \( Q_{0} \) in the problem, it is constant and its magnitude is given by the initial conditions:

\[ Q_{0}=\frac{U r_{0}}{k} \]

The total energy of the bubble is:

\[ W=W_{f}+W_{e}=8 \pi \alpha r^{2}+\frac{1}{2} k \frac{Q_{0}^{2}}{r}=8 \pi \alpha r^{2}+\frac{1}{2} \frac{U^{2} r_{0}^{2}}{k \cdot r} \]

According to the principle of virtual work, equilibrium occurs at the value of \( r \) such that when it is changed by \( \Delta r \), the work done is zero, or more precisely, the work done divided by \( \Delta r \) approaches zero as \( \Delta r \rightarrow 0 \):

\[ \lim _{\Delta r \rightarrow 0} \frac{\Delta W}{\Delta r}=0 \]

This precisely means that the derivative of \( W \) with respect to \( r \) at the given value of \( r \) is 0. Differentiating \( W \) with respect to \( r \), we get the following condition:

\[ \begin{aligned}
& 16 \pi r \alpha-\frac{U^{2} r_{0}^{2}}{2 k} \frac{1}{r^{2}}=0 . \quad \text { Solving for } r: \\
& r=\sqrt[3]{\frac{U^{2} r_{0}^{2}}{32 k \pi \alpha}}, \quad \text { with our data } r=0.71 \text{ cm}
\end{aligned} \]

Terlaky Edit (Kaposvár, Táncsics M. Gymnasium, 4th grade)

Remarks. 1. We can observe that when equilibrium is established, the total energy of the bubble will be less than initially (with the principle of virtual work, we are essentially finding the minimum of the potential energy). The question is, where did the lost energy go? Part of it is used to move (blow out or suck in) the air, and part of it is converted to heat due to internal friction. If these losses are small, the bubble will oscillate around the equilibrium position for a long time. The behavior of such a bubble can model the behavior of larger atomic nuclei in certain aspects.

Those who, ignoring kinetic energies, wrote the law of conservation of energy, derived a quadratic equation from which they essentially calculated the two extreme positions of the simplest oscillatory state and thus gave an incorrect answer to the question.

2. Many tried to solve the problem directly by balancing the forces that pull the surface together and the electrostatic repulsive forces. This often failed because, using the observation that a charged sphere acts on an external charge as if all its charge were at its center, they assumed that the force on the surface charge \( \Delta Q \) is \( F=k \frac{Q \Delta Q}{r^{2}} \). The flaw in this reasoning can be easily seen if we realize that we could equally well state that the force is 0. (Indeed, the electric field inside the sphere is 0.)

Using virtual displacements or integrating the effects of the other charges, it can be shown that the force on the surface charge \( \Delta Q \) is \( F=\frac{1}{2} k \frac{Q \Delta Q}{r^{2}} \), so the distance dependence of the electric field of a charged conducting sphere is best illustrated by the diagram.

![](https://cdn.mathpix.com/cropped/2024_05_02_44295c9f5430cdecf400g-2.jpg?height=258&width=523&top_left_y=145&top_left_x=796)

The peculiar structure of this function can be explained by the fact that in the investigation of a charged conducting sphere, we assumed that the entire charge is located in a layer of thickness \( 0 \text{ cm} \) on the surface of the sphere.