Solution

Statement

$2.1.54.$ Electrons moving in a circle of any radius around a charged thread have the same speed $v$. The mass of an electron is $m_e$. How does the force acting on the electron from the thread depend on the distance between the electron and the thread? Describe qualitatively the initial segment of the trajectory along which the electron will move if its speed during circular motion suddenly becomes slightly less than $v$? slightly greater than $v$?

Solution

1. A stationary circular orbit of an electron (a negatively charged particle with rest mass $m_e$) occurs when the magnitude of the Coulomb attractive force equals the required centripetal force (or inertial centrifugal force in a rotating frame):

$$F_k = F_i$$

2. This force is directly proportional to the square of the linear velocity of the particle and inversely proportional to the distance to the axis of rotation:

$$F_i = \frac{m_e v^2}{r}$$

Thus, since the speed $v$ is constant for any radius by the problem's condition, the Coulomb force is inversely proportional to the distance between the electron and the charged thread.

3. If the speed of the electron suddenly decreases ($v' < v$), the required centripetal force becomes less than the actual attractive Coulomb force:

$$F_k > \frac{m_e v'^2}{r}$$

Due to this excess attractive force, the electron will be pulled inward, decreasing its orbital radius. Conversely, if the speed suddenly increases, the required force will exceed the Coulomb force, and the electron will move outward. Since the force field is locally uniform over a very short initial distance, the initial segments of these trajectories will resemble parabolas tangent to the original circle from the inside (for decreased speed) and from the outside (for increased speed).