From a50ee03a8a182ee89e6f44127783139b13f160b9 Mon Sep 17 00:00:00 2001 From: "W. Trevor King" Date: Wed, 4 Apr 2012 15:24:25 -0400 Subject: [PATCH] Add solution for Serway and Jewett v8's problem 24.21. --- .../Serway_and_Jewett_8/problem24.21.tex | 40 +++++++++++++++---- 1 file changed, 32 insertions(+), 8 deletions(-) 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} -- 2.26.2