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Add solution for Serway and Jewett v8's problem 24.11.
author
W. Trevor King
<wking@drexel.edu>
Wed, 4 Apr 2012 18:51:45 +0000
(14:51 -0400)
committer
W. Trevor King
<wking@drexel.edu>
Wed, 4 Apr 2012 18:54:19 +0000
(14:54 -0400)
latex/problems/Serway_and_Jewett_8/problem24.11.tex
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diff --git
a/latex/problems/Serway_and_Jewett_8/problem24.11.tex
b/latex/problems/Serway_and_Jewett_8/problem24.11.tex
index e2b43901b2103aee93aadfe3a908958b9cc187c4..c9109cf5b6feeb39845b8c541c7baf5836094082 100644
(file)
--- a/
latex/problems/Serway_and_Jewett_8/problem24.11.tex
+++ b/
latex/problems/Serway_and_Jewett_8/problem24.11.tex
@@
-16,4
+16,17
@@
electric flux through each surface.
\end{problem*}
\begin{solution}
+The flux through any surface is given by Gauss's law:
+\begin{equation}
+ \Phi_E = \frac{q_\text{in}}{\varepsilon_0} \;,
+\end{equation}
+so
+\begin{align}
+ \Phi_{E1} &= \frac{-2Q + Q}{\varepsilon_0}
+ = \ans{\frac{-Q}{\varepsilon_0}} \\
+ \Phi_{E2} &= \frac{Q - Q}{\varepsilon_0} = \ans{0} \\
+ \Phi_{E1} &= \frac{-2Q + Q - Q}{\varepsilon_0}
+ = \ans{\frac{-2Q}{\varepsilon_0}} \\
+ \Phi_{E1} &= \frac{0}{\varepsilon_0} = \ans{0} \;.
+\end{align}
\end{solution}