From: W. Trevor King <wking@drexel.edu>
Date: Wed, 4 Apr 2012 19:24:25 +0000 (-0400)
Subject: Add solution for Serway and Jewett v8's problem 24.21.
X-Git-Url: http://git.tremily.us/?a=commitdiff_plain;h=a50ee03a8a182ee89e6f44127783139b13f160b9;p=course.git

Add solution for Serway and Jewett v8's problem 24.21.
---

diff --git a/latex/problems/Serway_and_Jewett_8/problem24.21.tex b/latex/problems/Serway_and_Jewett_8/problem24.21.tex
index b296b5f..1741cf8 100644
--- a/latex/problems/Serway_and_Jewett_8/problem24.21.tex
+++ b/latex/problems/Serway_and_Jewett_8/problem24.21.tex
@@ -34,17 +34,41 @@ a.draw();
 \end{problem*}
 
 \begin{solution}
-\Part{a}
+The electric flux through a surface $S$ is
+\begin{equation}
+  \Phi_{ES} = \int_S \vect{E}\cdot\vect{\dd A} = \int_S E \dd A\cos(\theta) \;.
+\end{equation}
+Where $\theta$ is the angle between \vect{E} and the perpendicular
+\vect{\dd A}.  For a uniform field and flat surface, $E$ and $\theta$
+are constants, so we can pull them of the integral:
+\begin{equation}
+  \Phi_{ES} = E\cos(\theta)\int_S \dd A = EA\cos(\theta) \;.
+\end{equation}
+For this problem, that means we only need to find the appropriate
+expression for $\cos(\theta)$ to solve each part.
 
-\Part{b}
-
-\Part{c}
-
-\Part{d}
-
-\Part{e}
+\begin{align}
+  \Phi_{Ea} &= \ans{EA\cos(\theta)} \\
+  \Phi_{Eb} &= \ans{-EA\sin(\theta)} \\
+  \Phi_{Ec} &= \ans{-EA\cos(\theta)} \\
+  \Phi_{Ed} &= \ans{EA\sin(\theta)} \\
+  \Phi_{E\text{top}} &= \ans{0} \\
+  \Phi_{E\text{bottom}} &= \ans{0} \;
+\end{align}
 
 \Part{f}
+Summing the flux through each face (above), we have
+\begin{equation}
+  \Phi_E = EA\cos(\theta) - EA\sin(\theta) - EA\cos(\theta) + EA\sin(\theta)
+    + 0 + 0 = \ans{0} \;.
+\end{equation}
+In other words, all the flux that comes into one part of the cube goes
+out through some other part of the cube.
 
 \Part{g}
+From Gauss's law,
+\begin{align}
+  \Phi_E &= \frac{q_\text{in}}{\varepsilon_0} \\
+  q_\text{in} &= \Phi_E \varepsilon_0 = \ans{0} \;.
+\end{align}
 \end{solution}