Новое решение

JAMF правка от 13 июн. 2026 г., 17:18

правка #18720 предыдущая #18719 ← раньше позже →

en/14.3.10.md +75 −0

		@@ -0,0 +1,75 @@
		+	### Statement
		+
		+	$14.3.10.$ [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}
		+
		+	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}
		+
		+	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
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,75 @@
			### Statement

			$14.3.10.$ [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}

			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}

			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

			[Insert a concise answer or boxed result]