Add solution for Serway and Jewett v8's problem 24.13.

diff --git a/latex/problems/Serway_and_Jewett_8/problem24.13.tex b/latex/problems/Serway_and_Jewett_8/problem24.13.tex
index 2341930470767be9f44386bc54087d236f3429d7..5a83d2ba76752392bfcd0784cb92c0c7a8339b72 100644 (file)
--- a/latex/problems/Serway_and_Jewett_8/problem24.13.tex
+++ b/latex/problems/Serway_and_Jewett_8/problem24.13.tex
@@ -7,4 +7,27 @@ air between these two elevations?  Is it positive or negative?
 \end{problem*}
 
 \begin{solution}
+Lets take a vertical cylinder of this air as our gaussian surface.
+Because the electric field is directed downward, there is no flux
+through the walls of the cylinder.  All the flux crosses the cylinder
+at the end caps.  If the area of the end cap is $A$, that flux is
+\begin{equation}
+  \Phi_E \equiv \oint_S \vect{E}\cdot\vect{dA} = E_{500}A - E_{600}A \;,
+\end{equation}
+where $E_{500}=120\U{N/C}$ and $E_{600}=100\U{N/C}$.  From Gauss's
+law,
+\begin{align}
+  \Phi_E &= \frac{q_\text{in}}{\varepsilon_0} = (E_{500}-E{600})A \\
+  q_\text{in} &= (E_{500}-E{600})A\varepsilon_0 \;.
+\end{align}
+This gives an average volume charge density of
+\begin{equation}
+  \rho \equiv \frac{q_\text{in}}{V}
+    = \frac{(E_{500}-E{600})A\varepsilon_0}{A(h_{600}-h_{500})}
+    = \frac{E_{500}-E{600}}{h_{600}-h_{500}}\varepsilon_0
+    = \frac{20\U{N/C}}{100\U{m}}\cdot8.85\E{-12}\U{C$^2$/N$\cdot$m$^2$}
+    = \ans{1.77\E{-12}\U{C/m$^3$}} \;.
+\end{equation}
+Because there is a net flux out of the gaussian cylinder, the net
+charge contained by the cylinder is positive.
 \end{solution}