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IdPhO 2020, Теория — Задача 1A. Part A. Cubic oscillations

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Всички материали от IdPhO 2020

Theory — IdPhO 2020 — Q1 — Problem mix (10 points) · 2,5 т.

Условие

Part A. Cubic oscillations (2.5 points)

A.1 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. [2,5 т.]

A wire-frame cube with the edge labelled a; a straight channel passes through the cube centre, where a particle (black dot) sits, and the segment from the centre to the face is labelled L.

Отговори

Решение

We will use a coordinate system with axes parallel to the cube’s edges, with the origin set at the cube’s center. Assuming the particle is at coordinates $(x, y, z)$ $x \ll a$ , $y \ll a$ , $z \ll a$ , we will find the force $\vec{F}$ acting on the particle, by splitting the cube $a \times a \times a$ into a rectangular cuboid $(a - 2x) \times (a - 2y) \times (a - 2z)$ and three square plates of thickness $2x$ , $2y$ and $2z$ .

Hand-drawn cube with coordinate axes through the centre O; the particle at (x; y; z) with the force vector F; dashed lines cut off three plates of thicknesses 2x, 2y, 2z from the faces; half-edges marked a/2.

The particle is in the center of the cuboid, so there is no force from the cuboid.
Let us find the force between a particle with a charge $q$ and a uniformly charged square plate of small thickness $h$ and edge length $a$ . The plate’s charge density is $\rho$ , the particle is placed above the center of the plate at distance $a/2$ .

Hand-drawn square plate of thickness h, edge a and charge density ρ, with a charge q located at distance a/2 above the centre of the plate.

Due to symmetry and Gauss’s law, the flux of the particle’s electric field through the plate is

\Phi = \frac{q}{6\varepsilon_0}.

Hence, the force

F = \sigma\Phi = \frac{q\rho h}{6\varepsilon_0},

where $\sigma = \rho h$ is plate’s surface charge density.
Three square plates act on the particle with forces $\vec{F}_1 = \frac{q\rho x}{3\varepsilon_0}\hat{x}$ , $\vec{F}_2 = \frac{q\rho y}{3\varepsilon_0}\hat{y}$ , and $\vec{F}_3 = \frac{q\rho z}{3\varepsilon_0}\hat{z}$ . Net force $\vec{F} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \frac{q\rho}{3\varepsilon_0}\vec{r}$ , where $\vec{r}$ is the position vector of the particle.

The particle’s equation of motion

m\ddot{\vec{r}} = \frac{q\rho}{3\varepsilon_0}\vec{r}

is an equation of simple harmonic motion with period

T = 2\pi\sqrt{\frac{3m\varepsilon_0}{q(-\rho)}}.

Оригинал в Архива: IPhO_2020_Q1.pdf · официални решения: IPhO_2020_S1.pdf