Правка разделов «Statement», «Solution», «Answer»
en/14.4.8.md
+6 −31
| @@ -1,37 +1,12 @@ | |||
| ### Statement | |||
| − | $14.4.8.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | \documentclass[12pt,a4paper]{article} | ||
| − | \usepackage[english]{babel} | ||
| − | \usepackage{float} | ||
| − | \usepackage{wrapfig} | ||
| − | \usepackage{lmodern} | ||
| − | \usepackage[T1]{fontenc} | ||
| − | \usepackage[utf8]{inputenc} | ||
| − | \usepackage{microtype} | ||
| − | \usepackage{graphicx} | ||
| − | \usepackage{booktabs} | ||
| − | \usepackage{amsmath,amssymb} | ||
| − | \usepackage{hyperref} | ||
| − | \usepackage{csquotes} | ||
| − | \usepackage{geometry} | ||
| − | \usepackage{subcaption} | ||
| − | \usepackage{tikz} | ||
| − | \usepackage{array} | ||
| − | \usepackage{pgfplots} | ||
| − | \usepackage{wrapfig} | ||
| − | \usepackage{subcaption} | ||
| − | |||
| − | \begin{document} | ||
| − | |||
| $14.4.8$ How fast does an electron move around a heavy nucleus with charge $ez$ in a | |||
| circular orbit of radius $R$? | |||
| + | ### Solution | ||
| + | |||
| We can solve this problem using Newton's second law for the radial direction of the motion. | |||
| − | In this case the acceleration is: | ||
| + | In this case the radial acceleration is: | ||
| \begin{equation} | |||
| a_r = \frac{v^2}{R} | |||
| \end{equation} | |||
| The force acting on the electron is the electric force: | |||
| \begin{equation} | |||
| F_e = \frac{(ez)e}{4 \pi \varepsilon_0 R^2} | |||
| \end{equation} | |||
| This force points in the radial direction, like the acceleration of the electron. | |||
| Using Newton's second law: | |||
| \begin{equation} | |||
| F_e = m_e a_r \rightarrow \frac{m_e v^2}{R} = \frac{(ez)e}{4 \pi \varepsilon_0 R^2} | |||
| \end{equation} | |||
| Then: | |||
| \begin{equation} | |||
| @@ -57,8 +32,8 @@Solution | |||
| v = \sqrt{\frac{e^2 z}{4 \pi \varepsilon_0 m_e R}} | |||
| \end{equation} | |||
| − | \end{document} | ||
| − | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | \begin{equation} | ||
| + | v = \sqrt{\frac{e^2 z}{4 \pi \varepsilon_0 m_e R}} | ||
| + | \end{equation} | ||
| @@ -1,37 +1,12 @@ | |||
| ### Statement | ### Statement | ||
| $14.4.8.$ [Insert the problem statement] | |||
| ### Solution | |||
| \documentclass[12pt,a4paper]{article} | |||
| \usepackage[english]{babel} | |||
| \usepackage{float} | |||
| \usepackage{wrapfig} | |||
| \usepackage{lmodern} | |||
| \usepackage[T1]{fontenc} | |||
| \usepackage[utf8]{inputenc} | |||
| \usepackage{microtype} | |||
| \usepackage{graphicx} | |||
| \usepackage{booktabs} | |||
| \usepackage{amsmath,amssymb} | |||
| \usepackage{hyperref} | |||
| \usepackage{csquotes} | |||
| \usepackage{geometry} | |||
| \usepackage{subcaption} | |||
| \usepackage{tikz} | |||
| \usepackage{array} | |||
| \usepackage{pgfplots} | |||
| \usepackage{wrapfig} | |||
| \usepackage{subcaption} | |||
| \begin{document} | |||
| $14.4.8$ How fast does an electron move around a heavy nucleus with charge $ez$ in a | $14.4.8$ How fast does an electron move around a heavy nucleus with charge $ez$ in a | ||
| circular orbit of radius $R$? | circular orbit of radius $R$? | ||
| ### Solution | |||
| We can solve this problem using Newton's second law for the radial direction of the motion. | We can solve this problem using Newton's second law for the radial direction of the motion. | ||
| In this case the acceleration is: | In this case the radial acceleration is: | ||
| \begin{equation} | \begin{equation} | ||
| a_r = \frac{v^2}{R} | a_r = \frac{v^2}{R} | ||
| \end{equation} | \end{equation} | ||
| The force acting on the electron is the electric force: | The force acting on the electron is the electric force: | ||
| \begin{equation} | \begin{equation} | ||
| F_e = \frac{(ez)e}{4 \pi \varepsilon_0 R^2} | F_e = \frac{(ez)e}{4 \pi \varepsilon_0 R^2} | ||
| \end{equation} | \end{equation} | ||
| This force points in the radial direction, like the acceleration of the electron. | This force points in the radial direction, like the acceleration of the electron. | ||
| Using Newton's second law: | Using Newton's second law: | ||
| \begin{equation} | \begin{equation} | ||
| F_e = m_e a_r \rightarrow \frac{m_e v^2}{R} = \frac{(ez)e}{4 \pi \varepsilon_0 R^2} | F_e = m_e a_r \rightarrow \frac{m_e v^2}{R} = \frac{(ez)e}{4 \pi \varepsilon_0 R^2} | ||
| \end{equation} | \end{equation} | ||
| Then: | Then: | ||
| \begin{equation} | \begin{equation} | ||
| @@ -57,8 +32,8 @@Solution | |||
| v = \sqrt{\frac{e^2 z}{4 \pi \varepsilon_0 m_e R}} | v = \sqrt{\frac{e^2 z}{4 \pi \varepsilon_0 m_e R}} | ||
| \end{equation} | \end{equation} | ||
| \end{document} | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | \begin{equation} | ||
| v = \sqrt{\frac{e^2 z}{4 \pi \varepsilon_0 m_e R}} | |||
| \end{equation} | |||