Homework Template
作者:
Jennifer Byford
最近上传:
8 年前
许可:
Creative Commons CC BY 4.0
摘要:
LaTeX template I've used extensively for Engineering homeworks.
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
LaTeX template I've used extensively for Engineering homeworks.
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
\documentclass[12pt]{article}
\usepackage[margin=1in]{geometry} % For setting margins
\usepackage{amsmath} % For Math
\usepackage{fancyhdr} % For fancy header/footer
\usepackage{graphicx} % For including figure/image
\usepackage{cancel} % To use the slash to cancel out stuff in work
%%%%%%%%%%%%%%%%%%%%%%
% Set up fancy header/footer
\pagestyle{fancy}
\fancyhead[LO,L]{Jennifer Byford}
\fancyhead[CO,C]{ECE305 - Homework Example}
\fancyhead[RO,R]{\today}
\fancyfoot[LO,L]{}
\fancyfoot[CO,C]{\thepage}
\fancyfoot[RO,R]{}
\renewcommand{\headrulewidth}{0.4pt}
\renewcommand{\footrulewidth}{0.4pt}
%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\noindent \textit{A uniformly charged dielectric gel having charge density $\rho_v = \rho_o$ and dielectric constant of $\epsilon_r = 2$ is enclosed inside a dielectric shell with dielectric constant of $\epsilon_r = 5$ as shown in the figure. The dielectric shell is surrounded by free space ($\epsilon_r = 1$). Determine E, $\Phi$, D, P in all regions.}
\begin{figure}[!h]
\begin{centering}
\includegraphics[keepaspectratio = true, width = 2in]{image.PNG}
\caption{Given scenario.}
\end{centering}
\end{figure}
\noindent In this problem we can consider there to be 3 regions, (1)~$r < a$, (2)~$a \leq r < b$, and (3)~$r \geq b$. For finding the electric flux density Gauss's Law provides the most straight forward solution. Then, using the constitutive relation between electric flux density and electric field, $\vec{E}$ can be found. Thereafter $\vec{P}$ and $\Phi$ are directly related to $\vec{E}$.\\
\\
\underline{Electric Flux Density:}
\begin{align*}
\oint_S \vec{D} \cdot \hat{n} &= Q_{en} \\
\int_0^\pi \int_0^{2\pi} D_r r^2 sin\theta d\phi d\theta &= Q_{en}\\
4\pi r^2 D_r &= Q_{en}
\end{align*}
\textit{Region 1}
\begin{align*}
Q_{en} &= \int_0^\pi \int_0^{2\pi} \int_0^r \rho_o r^2 sin\theta dr d\phi d\theta \\
Q_{en} &= \frac{4\pi \rho_o r^3}{3} \\
4\pi r^2 D_r &= \frac{4\pi \rho_o r^3}{3} \\
\vec{D} &= \frac{\rho_o r^3}{3 r^2} \\
\vec{D} &= \frac{\rho_o r}{3} \hat{r}~C/m^2
\end{align*}
\textit{Region 2 and 3}
\begin{align*}
Q_{en} &= \int_0^\pi \int_0^{2\pi} \int_0^a \rho_o r^2 sin\theta dr d\phi d\theta \\
Q_{en} &= \frac{4\pi \rho_o a^3}{3} \\
4\pi r^2 D_r &= \frac{4\pi \rho_o a^3}{3} \\
\vec{D} &= \frac{\rho_o a^3}{3 r^2} \hat{r}~C/m^2
\end{align*}
\underline{Electric Field:}
\begin{align*}
\vec{D} &= \epsilon \vec{E} \\
\vec{E} &= \frac{1}{\epsilon} \vec{D} \\
\vec{E} &= \frac{1}{\epsilon_o \epsilon_r} \vec{D}
\end{align*}
\textit{Region 1}
\begin{align*}
\vec{E} &= \frac{1}{\epsilon_o \epsilon_r} \vec{D} \\
\vec{E} &= \frac{1}{\epsilon_o \epsilon_r} \frac{\rho_o r}{3} \\
\vec{E} &= \frac{1}{2 \epsilon_o} \frac{\rho_o r}{3} \hat{r}~V/m
\end{align*}
\textit{Region 2}
\begin{align*}
\vec{E} &= \frac{1}{\epsilon_o \epsilon_r} \vec{D} \\
\vec{E} &= \frac{1}{\epsilon_o \epsilon_r} \frac{\rho_o a^3}{3 r^2} \\
\vec{E} &= \frac{1}{5 \epsilon_o} \frac{\rho_o a^3}{3 r^2} \hat{r}~V/m
\end{align*}
\textit{Region 3}
\begin{align*}
\vec{E} &= \frac{1}{\epsilon_o \epsilon_r} \vec{D} \\
\vec{E} &= \frac{1}{\epsilon_o \epsilon_r} \frac{\rho_o a^3}{3 r^2} \\
\vec{E} &= \frac{1}{\epsilon_o} \frac{\rho_o a^3}{3 r^2} \hat{r}~V/m
\end{align*}
\underline{Electric Polarization:}
$$\vec{P} = (\epsilon_r - 1)\epsilon_o \vec{E}$$
\textit{Region 1}
\begin{align*}
\vec{P} &= (\epsilon_r - 1)\epsilon_o \vec{E} \\
\vec{P} &= (2 - 1) \cancel{\epsilon_o} \frac{1}{2 \cancel{\epsilon_o}} \frac{\rho_o r}{3} \\
\vec{P} &= \frac{1}{2} \frac{\rho_o r}{3} \\
\vec{P} &= \frac{\rho_o r}{6} \hat{r}~C/m^2
\end{align*}
\textit{Region 2}
\begin{align*}
\vec{P} &= (\epsilon_r - 1)\epsilon_o \vec{E} \\
\vec{P} &= (5 - 1) \cancel{\epsilon_o} \frac{1}{5 \cancel{\epsilon_o}} \frac{\rho_o a^3}{3 r^2} \\
\vec{P} &= \frac{4}{5} \frac{\rho_o a^3}{3 r^2}\\
\vec{P} &= \frac{4\rho_o a^3}{15 r^2} \hat{r}~C/m^2
\end{align*}
\textit{Region 3}
\begin{align*}
\vec{P} &= (\epsilon_r - 1)\epsilon_o \vec{E} \\
\vec{P} &= (1 - 1) \epsilon_o \vec{E} \\
\vec{P} &= 0 \hat{r}~C/m^2
\end{align*}
\underline{Electric Potential:}
$$\Phi = -\int_{\infty}^{r} \vec{E} \cdot dl$$
% e region 1 = \frac{1}{2 \epsilon_o} \frac{\rho_o r}{3} \hat{r}
% e region 2 = \frac{1}{5 \epsilon_o} \frac{\rho_o a^3}{3 r^2} \hat{r}
% e region 3 = \frac{1}{\epsilon_o} \frac{\rho_o a^3}{3 r^2} \hat{r}
\textit{Region 3}
\begin{align*}
\Phi &= -\int_{\infty}^{r} \vec{E}_3 \cdot dl \\
\Phi &= -\int_{\infty}^{r} \frac{1}{\epsilon_o} \frac{\rho_o a^3}{3 r^2} \hat{r} \cdot \hat{r} dr \\
\Phi &= -\frac{\rho_o a^3}{3 \epsilon_o} \int_{\infty}^{r} \frac{1}{r^2} dr \\
\Phi &= -\frac{\rho_o a^3}{3 \epsilon_o} \left[-\frac{1}{r}\right]_\infty^r \\
\Phi &= \frac{\rho_o a^3}{3\epsilon_o r} \\
\Phi &= \frac{\rho_o a^3}{3\epsilon_o r}~V
\end{align*}
\textit{Region 2}
\begin{align*}
\Phi &= -\int_{\infty}^{r} \vec{E} \cdot dl \\
\Phi &= -\int_{\infty}^{b} \vec{E}_3 \cdot dl - \int_b^r \vec{E}_2 \cdot dl\\
\Phi &= -\int_{\infty}^{b} \frac{1}{\epsilon_o} \frac{\rho_o a^3}{3 r^2} \hat{r} \cdot \hat{r} dr - \int_b^r \frac{1}{5 \epsilon_o} \frac{\rho_o a^3}{3 r^2} \hat{r} \cdot \hat{r} dr\\
\Phi &= -\frac{\rho_o a^3}{3 \epsilon_o} \int_{\infty}^{b} \frac{1}{r^2} dr - \frac{\rho_o a^3}{15 \epsilon_o} \int_b^r \frac{1}{r^2} dr\\
\Phi &= -\frac{\rho_o a^3}{3 \epsilon_o} \left[-\frac{1}{r}\right]_\infty^b - \frac{\rho_o a^3}{15 \epsilon_o} \left[-\frac{1}{r}\right]_b^r\\
\Phi &= \frac{\rho_o a^3}{3 \epsilon_o} \frac{1}{b} + \frac{\rho_o a^3}{15 \epsilon_o} \left[\frac{1}{r} - \frac{1}{b}\right]\\
\Phi &= \frac{\rho_o a^3}{3 \epsilon_o b} + \frac{\rho_o a^3}{15 \epsilon_o r} - \frac{\rho_o a^3}{15 \epsilon_o b}~V
\end{align*}
\textit{Region 1}
\begin{align*}
\Phi &= -\int_{\infty}^{r} \vec{E} \cdot dl \\
\Phi &= -\int_{\infty}^{b} \vec{E}_3 \cdot dl - \int_b^a \vec{E}_2 \cdot dl - \int_a^r \vec{E}_1 \cdot dl\\
\Phi &= -\int_{\infty}^{b} \frac{1}{\epsilon_o} \frac{\rho_o a^3}{3 r^2} \hat{r} \cdot \hat{r} dr - \int_b^a \frac{1}{5 \epsilon_o} \frac{\rho_o a^3}{3 r^2} \hat{r} \cdot \hat{r} dr - \int_a^r \frac{1}{2 \epsilon_o} \frac{\rho_o r}{3} \hat{r} \cdot \hat{r} dr\\
\Phi &= -\frac{\rho_o a^3}{3 \epsilon_o} \int_{\infty}^{b} \frac{1}{r^2} dr - \frac{\rho_o a^3}{15 \epsilon_o} \int_b^a \frac{1}{r^2} dr - \frac{\rho_o}{6 \epsilon_o} \int_a^r r dr\\
\Phi &= -\frac{\rho_o a^3}{3 \epsilon_o} \left[-\frac{1}{r}\right]_\infty^b - \frac{\rho_o a^3}{15 \epsilon_o} \left[-\frac{1}{r}\right]_b^a - \frac{\rho_o}{6 \epsilon_o} \left[\frac{r^2}{2}\right]_a^r\\
\Phi &= \frac{\rho_o a^3}{3 \epsilon_o} \frac{1}{b} + \frac{\rho_o a^3}{15 \epsilon_o} \left[\frac{1}{a} - \frac{1}{b}\right] - \frac{\rho_o}{12 \epsilon_o} \left[r^2 - a^2\right]\\
\Phi &= \frac{\rho_o a^3}{3 \epsilon_o b} + \frac{\rho_o a^3}{15 \epsilon_o a} - \frac{\rho_o a^3}{15 \epsilon_o b} - \frac{\rho_o r^2}{12 \epsilon_o} + \frac{\rho_o a^2}{12 \epsilon_o}\\
\Phi &= \frac{\rho_o a^3}{3 \epsilon_o b} + \frac{\rho_o a^2}{15 \epsilon_o} - \frac{\rho_o a^3}{15 \epsilon_o b} - \frac{\rho_o r^2}{12 \epsilon_o} + \frac{\rho_o a^2}{12 \epsilon_o}\\
\Phi &= \frac{\rho_o a^2}{\epsilon_o}\left[ \frac{a}{3b} + \frac{1}{15} + \frac{1}{12}\right] - \frac{\rho_o}{\epsilon_o}\left[\frac{a^3}{15b} + \frac{r^2}{12}\right]~V
\end{align*}
Final Answer Summary:
\begin{tabular}{| c | c | c | c |}
\hline
& Region 1 ($r<a$)& Region 2 ($a\leq r < b$)& Region 3 ($r\geq b$)\\ \hline
$\vec{D}~(C/m^2)$ & $\frac{\rho_o r}{3} \hat{r}$ & $\frac{\rho_o a^3}{3 r^2} \hat{r}$ & $\frac{\rho_o a^3}{3 r^2} \hat{r}$\\
$\vec{E}~(V/m)$ & $\frac{\rho_o r}{6 \epsilon_o} \hat{r}$ & $\frac{1}{5 \epsilon_o} \frac{\rho_o a^3}{3 r^2} \hat{r}$ & $\frac{1}{\epsilon_o} \frac{\rho_o a^3}{3 r^2} \hat{r}$\\
$\vec{P}~(C/m^2)$ & $\frac{\rho_o r}{6} \hat{r}$ & $\frac{4\rho_o a^3}{15 r^2} \hat{r}$ & $0 \hat{r}$\\
$\Phi~(V)$ & $\frac{\rho_o a^2}{\epsilon_o}\left[ \frac{a}{3b} + \frac{1}{15} + \frac{1}{12}\right] - \frac{\rho_o}{\epsilon_o}\left[\frac{a^3}{15b} + \frac{r^2}{12}\right]$ & $\frac{\rho_o a^3}{3 \epsilon_o b} + \frac{\rho_o a^3}{15 \epsilon_o r} - \frac{\rho_o a^3}{15 \epsilon_o b}$ & $\frac{\rho_o a^3}{3\epsilon_o r}$\\ \hline
\end{tabular}
\end{document}