From 8a8c8afe8cb46811182c931340bc82182cdf02cc Mon Sep 17 00:00:00 2001 From: "W. Trevor King" Date: Wed, 4 Apr 2012 15:07:52 -0400 Subject: [PATCH] Add solution for Serway and Jewett v8's problem 24.13. --- .../Serway_and_Jewett_8/problem24.13.tex | 23 +++++++++++++++++++ 1 file changed, 23 insertions(+) diff --git a/latex/problems/Serway_and_Jewett_8/problem24.13.tex b/latex/problems/Serway_and_Jewett_8/problem24.13.tex index 2341930..5a83d2b 100644 --- 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} -- 2.26.2