Homework Template
Author
Jennifer Byford
Last Updated
7年前
License
Creative Commons CC BY 4.0
Abstract
LaTeX template I've used extensively for Engineering homeworks.
LaTeX template I've used extensively for Engineering homeworks.
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\fancyhead[CO,C]{ECE305 - Homework Example}
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\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}