Edits to “Statement”, “Solution”, “Answer”
en/14.3.10.md
+7 −36
| @@ -1,44 +1,11 @@ | |||
| ### Statement | |||
| − | $14.3.10.$ | ||
| + | $14.3.10.$ How many times will the potential difference and capacitance of a long cylindrical capacitor change when it moves at a speed $\beta c$ along the axis? | ||
| ### 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} | ||
| − | |||
| − | How many times will the potential difference and capacitance of a long cylindrical capacitor change when it moves at a speed $\beta c$ along the axis? | ||
| − | |||
| − | \begin{center} | ||
| − | Solution | ||
| − | \end{center} | ||
| − | |||
| There are two ways to solve this problem; we will see both. | |||
| − | \begin{center} | ||
| − | \Large \textbf{First Way} | ||
| − | \end{center} | ||
| − | |||
| Because charge is a conserved quantity, we can say that $Q = C V \rightarrow C V = \text{constant}$ for all frames. | |||
| Suppose that in the frame of the capacitor, the capacitor has a height $h$ and a surface area $S$; then the capacitance in the moving frame is: | |||
| \begin{equation} | |||
| C = \frac{\varepsilon_0 S}{h} | |||
| \end{equation} | |||
| In the Earth frame, the capacitor is moving with velocity $\beta c$ along its axis; then in the Earth frame the surface area of the capacitor will be the same, | |||
| but the height will be: | |||
| \begin{equation} | |||
| h' = h \sqrt{1-\beta^2} | |||
| \end{equation} | |||
| Thus, the capacitance will be: | |||
| \begin{equation} | |||
| C' = \frac{\varepsilon_0 S}{h'} = \frac{\varepsilon_0 S}{h \sqrt{1-\beta^2}} = \frac{C}{\sqrt{1-\beta^2}} | |||
| \end{equation} | |||
| Using the conserved quantity $Q$, we can obtain the voltage $V$: | |||
| \begin{equation} | |||
| V'C' = V C \rightarrow V' = V \sqrt{1-\beta^2} | |||
| \end{equation} | |||
| @@ -68,8 +35,12 @@Solution | |||
| On the other hand, we can use the equation for the voltage of the capacitor $V = E h$. Using it, we see that because the distance between plates in the Earth frame will be $\sqrt{1-\beta^2}$ times smaller, and the field does not change between frames, | |||
| the voltage in the Earth frame will be $\sqrt{1-\beta^2}$ times smaller. | |||
| − | \end{document} | ||
| #### Answer | |||
| + | \begin{equation} | ||
| + | C' = \frac{C}{\sqrt{1-\beta^2}} | ||
| + | \end{equation} | ||
| − | [Insert a concise answer or boxed result] | ||
| + | \begin{equation} | ||
| + | V' = V \sqrt{1-\beta^2} | ||
| + | \end{equation} | ||
| @@ -1,44 +1,11 @@ | |||
| ### Statement | ### Statement | ||
| $14.3.10.$ |
$14.3.10.$ How many times will the potential difference and capacitance of a long cylindrical capacitor change when it moves at a speed $\beta c$ along the axis? | ||
| ### Solution | ### 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} | |||
| How many times will the potential difference and capacitance of a long cylindrical capacitor change when it moves at a speed $\beta c$ along the axis? | |||
| \begin{center} | |||
| Solution | |||
| \end{center} | |||
| There are two ways to solve this problem; we will see both. | There are two ways to solve this problem; we will see both. | ||
| \begin{center} | |||
| \Large \textbf{First Way} | |||
| \end{center} | |||
| Because charge is a conserved quantity, we can say that $Q = C V \rightarrow C V = \text{constant}$ for all frames. | Because charge is a conserved quantity, we can say that $Q = C V \rightarrow C V = \text{constant}$ for all frames. | ||
| Suppose that in the frame of the capacitor, the capacitor has a height $h$ and a surface area $S$; then the capacitance in the moving frame is: | Suppose that in the frame of the capacitor, the capacitor has a height $h$ and a surface area $S$; then the capacitance in the moving frame is: | ||
| \begin{equation} | \begin{equation} | ||
| C = \frac{\varepsilon_0 S}{h} | C = \frac{\varepsilon_0 S}{h} | ||
| \end{equation} | \end{equation} | ||
| In the Earth frame, the capacitor is moving with velocity $\beta c$ along its axis; then in the Earth frame the surface area of the capacitor will be the same, | In the Earth frame, the capacitor is moving with velocity $\beta c$ along its axis; then in the Earth frame the surface area of the capacitor will be the same, | ||
| but the height will be: | but the height will be: | ||
| \begin{equation} | \begin{equation} | ||
| h' = h \sqrt{1-\beta^2} | h' = h \sqrt{1-\beta^2} | ||
| \end{equation} | \end{equation} | ||
| Thus, the capacitance will be: | Thus, the capacitance will be: | ||
| \begin{equation} | \begin{equation} | ||
| C' = \frac{\varepsilon_0 S}{h'} = \frac{\varepsilon_0 S}{h \sqrt{1-\beta^2}} = \frac{C}{\sqrt{1-\beta^2}} | C' = \frac{\varepsilon_0 S}{h'} = \frac{\varepsilon_0 S}{h \sqrt{1-\beta^2}} = \frac{C}{\sqrt{1-\beta^2}} | ||
| \end{equation} | \end{equation} | ||
| Using the conserved quantity $Q$, we can obtain the voltage $V$: | Using the conserved quantity $Q$, we can obtain the voltage $V$: | ||
| \begin{equation} | \begin{equation} | ||
| V'C' = V C \rightarrow V' = V \sqrt{1-\beta^2} | V'C' = V C \rightarrow V' = V \sqrt{1-\beta^2} | ||
| \end{equation} | \end{equation} | ||
| @@ -68,8 +35,12 @@Solution | |||
| On the other hand, we can use the equation for the voltage of the capacitor $V = E h$. Using it, we see that because the distance between plates in the Earth frame will be $\sqrt{1-\beta^2}$ times smaller, and the field does not change between frames, | On the other hand, we can use the equation for the voltage of the capacitor $V = E h$. Using it, we see that because the distance between plates in the Earth frame will be $\sqrt{1-\beta^2}$ times smaller, and the field does not change between frames, | ||
| the voltage in the Earth frame will be $\sqrt{1-\beta^2}$ times smaller. | the voltage in the Earth frame will be $\sqrt{1-\beta^2}$ times smaller. | ||
| \end{document} | |||
| #### Answer | #### Answer | ||
| \begin{equation} | |||
| C' = \frac{C}{\sqrt{1-\beta^2}} | |||
| \end{equation} | |||
| [Insert a concise answer or boxed result] | \begin{equation} | ||
| V' = V \sqrt{1-\beta^2} | |||
| \end{equation} | |||