A1  ^2.50 A narrow straight channel passes through the center of a fixed cube with a side \(a\). The cube is uniformly charged, the charge density is \(\rho\). The distance from the cube center to the point of intersection of the channel and a face is \(L\). In the channel there is a particle of a mass \(m\) and a charge \(q\). Find the period of small oscillations of the particle near the center. The gravitational interaction of the particle and the cube can be neglected. The cube and the particle are oppositely charged.

1 M1 The idea of the cube decomposition is stated: either into 2 cubes ($+\rho$, $-\rho$) or into a cube and 3 plates. Идея разбиения на куб и пластины	0.20
2 M1 The thickness of the plates is correctly related to the displacement of the particle $h_x$ (either $2h_x$ on one side, or $h_x$ on both sides and with opposite charges) Одна пластина толщиной х2 или две (+ и -)	0.20
3 M1 A correct initial expression (e.g. in terms of an integral) for the electric field (or potential) induced by one of the plates at the center of the cube is given. Верные исходные уравнения для поля или потенциала пластины	0.30
4 M1 The above equation is transformed such that it becomes independent of $a$ and the orientation of the channel Получено выражение для поля пластины не зависящее от $a$ и $\alpha$ (ориентация канала)	0.30
5 M1 The resulting equation E=$\cfrac{\rho h}{6\epsilon_0}$ is obtained Итоговое выражение для поля пластины не содержащее интегралы	0.30
6 M1 The vector sum of forces is written (or shown that the electrostatic force is aligned with the channel) Сложение сил по проекциям	0.20
7 M1 The resulting equation for the force is correct (both modulus and direction) $\vec{F} = \cfrac{q\rho}{3\epsilon_0}\vec{r}$ (The direction is the key here, and must be clearly stated) Верное выражение для силы и направления. Ключевым является направление, оно должно явно следовать из выражения.	0.30
8 M2 A differential of the electrostatic field (or potential) induced by an infinitesimal volume element is given Записан дифференциал потенциала или поля от маленького элемента куба	0.20
9 M2 A correct volume integral for the field (or potential) is given Записан трехмерный интеграл для поля или потенциала	0.20
10 M2 The above integral is transformed such that it becomes independent of $a$ and the orientation of the channel Интеграл сведен к виду, не содержащему $a$ и $\alpha$ (ориентация канала)	0.60
11 M2 The resulting equation $E=\cfrac{\rho h}{6\epsilon_0}$ is obtained Верное значение напряженности или потенциала (без интеграла)	0.50
12 M2 The resulting equation for the force is correct (both modulus and direction) $\vec{F} = \cfrac{q\rho}{3\epsilon_0}\vec{r}$ (The direction is the key here, and must be clearly stated) Верное выражение для силы и направления. Ключевым является направление, оно должно явно следовать из выражения.	0.30
13 M3 The proof of centrality of the electrostatic field (e.g. $V = Ax^2 + By^2 + Cz^2$) Доказательство центральности поля	0.50
14 M3 The proof of central symmetry of the electrostatic field (e.g. $A=B=C, V=A*r^2$) Доказательство $V=Ar^2$	0.50
15 M3 The correct coefficient $A$ for the proof above is found (e.g. from Gauss law) Значение $A$	0.20
16 M3 Gauss law is written and correctly applied. Запись теоремы Гаусса	0.30
17 M3 The resulting equation for the force is correct $\vec{F} = \cfrac{q\rho}{3\epsilon_0}\vec{r}$ Выражение для силы	0.30
18 Equation of motion of an harmonic oscillator is obtained Дифференциальное уравнение колебаний	0.20
19 The resulting equation for the period is correct (up to the signs of charges): $T = 2\pi \sqrt{\cfrac{3\epsilon_0 m}{|q||\rho|}}$ (Marked only if method is correct and justified) Верное выражение для периода (с любым знаком под корнем). Оценивается только при правильном и обоснованном методе.	0.40
20 The opposite signs of the charges are considered, and, as a result, period of oscillations contains $\sqrt{-\rho q}$ Верный знак по корнем. Оценивается, даже если ответ не правильный.	0.10
21 Arithmetic error	-0.10
22 Not used	-0.30