The parabolic path with a vertical axis is given by $y=25 \frac{1}{\mathrm{~m}} \cdot x^{2}$. In the focus, there is an electric charge of $Q_{1}=10^{-7}$ coulombs. A small ball with a mass of $0.72 \mathrm{~g}$ and a charge of $Q_{2}=2 \cdot 10^{-8} \mathrm{C}$ can slide without friction along the parabolic path. Where is the equilibrium position of the ball?
The focal distance of the parabola is $f=1 /(4 a)=10^{-2} \mathrm{~m}=1 \mathrm{~cm}$. The position of the ball on the parabolic path is determined by its $x$ coordinate. Let's draw the tangent to the parabola at $Q_{2}$, then the angles $\alpha$ marked on the diagram are equal.

![](https://cdn.mathpix.com/cropped/2024_05_02_eeb1512ce4a214187679g-1.jpg?height=539&width=509&top_left_y=267&top_left_x=797)

The condition for equilibrium is that the resultant of the Coulomb force, gravitational force, and the normal force perpendicular to the path is zero, i.e., the components of the forces in the direction of the tangent satisfy:

$$
m g \cos \alpha=\frac{k Q_{1} Q_{2}}{x^{2}+\left(a x^{2}-f\right)^{2}} \cos \alpha
$$

The trivial solution to this equation is $\cos \alpha=0, \alpha=90^{\circ}$, when the charge is at the vertex of the parabola. At the vertex of the parabola, the body is in an unstable equilibrium position. For the case $\cos \alpha \neq 0$, the sought $x$ value can be obtained by solving a quadratic equation. After substituting the numerical data, $x= \pm 4 \mathrm{~cm}$ is obtained. These two equilibrium positions are stable.

Horváth László (Celldömölk, Berzsenyi D. Gimn., IV. o. t.)

II. solution. The stable equilibrium position of the ball can be determined from the minimum of the potential energy. Using the fact that the parabola is the locus of points that are equidistant from the focus and the line $y=-f$, the potential energy can be written in the following form:

$$
E=m g(r-f)+\frac{k Q_{1} Q_{2}}{r}
$$

where $r$ is the distance of the ball from the focus. The potential energy can be minimal for such an $r$ value where

$$
\frac{\mathrm{d} E}{\mathrm{~d} r}=m g-\frac{k Q_{1} Q_{2}}{r^{2}}=0
$$

Using the numerical values, $r=5 \mathrm{~cm}$ is obtained, which corresponds to the $x= \pm 4 \mathrm{~cm}$ position. It is easy to show that at this point

$$
\frac{\mathrm{d}^{2} E}{\mathrm{~d} r^{2}}>0
$$

so the energy has a local minimum here. This implies that this equilibrium position is stable.

Kertay Zoltán (Budapest, Petófi S. Gimn., IV. o. t.)