From f7c69e195aad989757ef2d987f54d330fd8ef460 Mon Sep 17 00:00:00 2001
From: "W. Trevor King" <wking@drexel.edu>
Date: Wed, 1 Jul 2009 17:25:50 -0400
Subject: [PATCH] Added week two recitation solutions.

---
 .../Young_and_Freedman_12/problem22.16.tex    | 32 ++++++++
 .../Young_and_Freedman_12/problem22.18.tex    | 32 ++++++++
 .../Young_and_Freedman_12/problem22.23.tex    | 28 +++++++
 .../Young_and_Freedman_12/problem22.25.tex    | 28 +++++++
 .../Young_and_Freedman_12/problem22.37.tex    | 77 ++++++++++++++++++-
 5 files changed, 193 insertions(+), 4 deletions(-)

diff --git a/latex/problems/Young_and_Freedman_12/problem22.16.tex b/latex/problems/Young_and_Freedman_12/problem22.16.tex
index d0a64b9..694a6be 100644
--- a/latex/problems/Young_and_Freedman_12/problem22.16.tex
+++ b/latex/problems/Young_and_Freedman_12/problem22.16.tex
@@ -6,4 +6,36 @@ point inside the sphere, $0.100\U{m}$ below the surface.
 \end{problem*}
 
 \begin{solution}
+\Part{a}
+From Gauss' law
+\begin{equation}
+  \Phi_E=\oint_S\vect{E}\cdot\vect{dA}=\frac{q_e}{\varepsilon_0}
+\end{equation}
+we have (in cases of spherical symmetry)
+\begin{align}
+  \Phi_E &= EA = 4\pi r^2 E = \frac{q_e}{\varepsilon_0} \\
+  E &= \frac{q_e}{4\pi\varepsilon_0 r^2} = \frac{k q_e}{r^2}
+\end{align}
+which looks just like the electric field from a point charge, except
+the $q_e$ is only the charge \emph{enclosed} by the gaussian sphere of
+radius $r$.
+
+In the case of $r_a = R+0.1\U{m}$ outside the sphere, all of the
+charge is enclosed ($q_e=q$), so
+\begin{equation}
+  E_a = \frac{k q}{r_a^2}
+    = \frac{8.99e9\U{Nm$^2$/C$^2$} \cdot 0.250\E{-9}\U{C}}{(0.550\U{m})^2}
+    = \ans{7.43\U{N/C}}
+\end{equation}
+
+\Part{b}
+Inside a steady-state conductor, $\ans{E=0}$, otherwise charges would
+be moving around, and the system would not actually be in a steady
+state.
+
+Another way to think about it is that all the excess positive charges
+in the sphere want to get as far away from each other as possible, so
+they end up evenly distributed in a thin shell around the outside of
+the sphere.  This leaves no net charge inside any gaussian surface of
+$r<R$, so $q_e=0$, and $E=0$.
 \end{solution}
diff --git a/latex/problems/Young_and_Freedman_12/problem22.18.tex b/latex/problems/Young_and_Freedman_12/problem22.18.tex
index 88df491..3e0c435 100644
--- a/latex/problems/Young_and_Freedman_12/problem22.18.tex
+++ b/latex/problems/Young_and_Freedman_12/problem22.18.tex
@@ -10,4 +10,36 @@ the planet's surface.
 \end{problem*}
 
 \begin{solution}
+\Part{a}
+Normally Gauss' law problems involve some sort of symmetry argument to
+make the integral in
+\begin{equation}
+  \Phi_E=\oint_S\vect{E}\cdot\vect{dA}=\frac{q_e}{\varepsilon_0}
+\end{equation}
+easy to deal with.  In \Part{a} though, we don't need any of that,
+since they give you the result of the integral $\Phi_E$ directly.
+Finding $q$ is simple algebra
+\begin{equation}
+  q = \varepsilon_0\Phi_E
+    = 8.885\U{C$^2$/Nm$^2$} \cdot 3.63\E{16}\U{Nm$^2$/C}
+    = \ans{3.21\E{5}\U{C}} \;.
+\end{equation}
+
+\Part{b}
+Now the spherical symmetry comes in, and we use the formula for
+electric field outside a spherically symmetric body (with $q_e=q$).
+\begin{equation}
+  E = \frac{k q}{R^2} = \frac{\Phi_E}{A} = \frac{\Phi_E}{4\pi R^2}
+    = \ans{250\U{N/C}} \;,
+\end{equation}
+where we looked up the radius of Mars $R=3.40\E{6}\U{m}$.
+
+\Part{c}
+The surface charge density $\sigma$ is the charge per unit area.  This
+is the same as the total charge on Mars divided by its total surface
+area.
+\begin{equation}
+  \sigma = \frac{q}{A} = \frac{q}{4\pi R^2}
+    = \ans{2.21\E{-9}\U{C/m$^2$}} \;.
+\end{equation}
 \end{solution}
diff --git a/latex/problems/Young_and_Freedman_12/problem22.23.tex b/latex/problems/Young_and_Freedman_12/problem22.23.tex
index e42e132..fba4ce9 100644
--- a/latex/problems/Young_and_Freedman_12/problem22.23.tex
+++ b/latex/problems/Young_and_Freedman_12/problem22.23.tex
@@ -10,4 +10,32 @@ the inner surface of the sphere?
 \end{problem*}
 
 \begin{solution}
+\Part{a}
+The total charge on the shell is $Q=4\pi R_o^2 \sigma=5.00\U{$\mu$C}$.
+
+Once the $q=-0.500\U{$\mu$C}$ charge is placed inside the shell,
+$q_i=-q$ accumulates on the shell's inner wall.  This ensures that the
+enclosed charge $q_e=0$ for any gaussian surface lying inside the body
+of the conducting shell, which must be the case, because $E=0$ inside
+steady-state conductors.  The left over outer-wall charge
+\begin{equation}
+  q_o = Q-q_i = 4.50\U{$\mu$C}
+\end{equation}
+is distributed uniformly over the outer surface, so
+\begin{equation}
+  \sigma_o = \frac{q_o}{4\pi R_o^2} = \sigma-\frac{-q}{4\pi R_o^2}
+    = \ans{5.73\U{$\mu$C}} \;.
+\end{equation}
+
+\Part{b}
+Now that we know the charge distribution, we can calculate electric
+fields with Gauss' law for spherically symmetric charge distributions.
+For any point outside the sphere, the total charged enclosed will be
+$q_e=q_o+q_i+q=q_o$, so
+\begin{equation}
+  E_o = \frac{kq_o}{R_o^2} = \ans{647\U{kN/C}} \;.
+\end{equation}
+
+\Part{c}
+At any point inside a steady-state conductor, $E=0$, so $\ans{\Phi_E=AE=0}$.
 \end{solution}
diff --git a/latex/problems/Young_and_Freedman_12/problem22.25.tex b/latex/problems/Young_and_Freedman_12/problem22.25.tex
index 4b85589..ff51065 100644
--- a/latex/problems/Young_and_Freedman_12/problem22.25.tex
+++ b/latex/problems/Young_and_Freedman_12/problem22.25.tex
@@ -8,4 +8,32 @@ from the center.
 \end{problem*}
 
 \begin{solution}
+\Part{a}
+Using Gauss' law for spherically symmetric charge distributions
+\begin{align}
+  E &= \frac{kq}{r^2} \\
+  q &= \frac{Er^2}{k}
+    = \frac{1750\U{N/C} \cdot (0.145\U{m}+0.355\U{m})^2}{8.99\E{9}\U{Nm$^2$/C$^2$}}
+    = 48.7\U{nC} \;.
+\end{align}
+For a uniform distribution, the charge density (per unit volume)
+$\rho$ is the total charge divided by the total volume of the sphere.
+\begin{equation}
+  \rho = \frac{q}{\frac{4/3}\pi R^3} = \ans{2.60\E{-7}\U{C/m$^3$}}
+\end{equation}
+
+\Part{b}
+Once you know the charge density, we can find the charge enclosed by a
+gaussian surface of radius $r=0.200\U{m}$.
+\begin{equation}
+  q_e = \frac{4/3}\pi r^3 \rho = q \frac{r^3}{R^3} = 8.70\U{nC} \;.
+\end{equation}
+Then back to our Gauss' law formula for the electric field
+\begin{equation}
+  E = \frac{kq_e}{r^2}
+    = \frac{kq\frac{r^3}{R^3}}{r^2}
+    = \frac{kqr}{R^3}
+    = \frac{8.99\U{Nm$^2$/C$^2$} \cdot 48.7\U{nC} \cdot 0.200\U{m}}{(0.355\U{m})^3}
+    = \ans{1960\U{N/C}}
+\end{equation}
 \end{solution}
diff --git a/latex/problems/Young_and_Freedman_12/problem22.37.tex b/latex/problems/Young_and_Freedman_12/problem22.37.tex
index b12c0a5..5b087c9 100644
--- a/latex/problems/Young_and_Freedman_12/problem22.37.tex
+++ b/latex/problems/Young_and_Freedman_12/problem22.37.tex
@@ -3,14 +3,83 @@ A long coaxial cable consists of an inner cylindrical conductor with a
 radius $a$ and an outer coaxial cylinder with inner radius $b$ and
 outer radius $c$.  The outer cylinder is mounted on insulating
 supports and has no net charge.  The inner cylinder has a uniform
-positive charge per unit length $\lambda$.  Caluculate the electric
+positive charge per unit length $\lambda$.  Calculate the electric
 field \Part{a} at any point between the cylinders a distance $r$ from
-the axis and \Part{b} at any point outside the cylinder.  \Part{c}
-Graph the magnitude of the electric field as a function of the
-distance $r$ from the axis of the cable, from $r=0$ to
+the axis and \Part{b} at any point outside the outer
+cylinder.  \Part{c} Graph the magnitude of the electric field as a
+function of the distance $r$ from the axis of the cable, from $r=0$ to
 $r=2c$.  \Part{d} Find the charge per unit length on the inner surface
 and on the outer surface of the outer cylinder.
 \end{problem*}
 
 \begin{solution}
+For this question, we need to apply Gauss' law to cylindrically
+symmetric charge distributions.  Take a look at Example 22.6 in the
+text to see their solution for an infinitely long, charged wire.
+Eventually you end up with
+\begin{equation}
+  E = \frac{\lambda_e}{2\pi\varepsilon_0 r}
+\end{equation}
+for the electric field $E$ a distance $r$ from a long, straight,
+cylindrically symmetric charge distribution carrying a charge density
+(per unit length) of $\lambda_e$.  That's really all we need for this
+problem.
+
+\Part{a}
+Between the two cylinders, the enclosed charge density
+$\lambda_e=\lambda$, the charge density of the inner cylinder, so
+\begin{equation}
+  E = \ans{\frac{\lambda}{2\pi\varepsilon_0 r}} \;.
+\end{equation}
+
+\Part{b}
+Outside the outer cylinder, the enclosed charge density is still given
+by $\lambda_e=\lambda$, because the outer cylinder carries no net
+charge.  Therefore
+\begin{equation}
+  E = \ans{\frac{\lambda}{2\pi\varepsilon_0 r}} \;,
+\end{equation}
+which is the same formula as in \Part{a}.
+
+\Part{c}
+We've already determined the electric field in all non-conductor
+regions.  Inside any steady-state conductor, $E=0$, otherwise charges
+would be moving, and the conductor would not be in a steady-state.
+Therefore, $E(r)$ looks like
+\begin{center}
+\begin{asy}
+import graph;
+size(5cm, 4cm, IgnoreAspect);
+
+real a=1;
+real b=4;
+real c=5;
+
+real E(real r) {
+  if (r < a) return 0;
+  else if (r > b && r < c) return 0;
+  else return 1/r;
+}
+
+draw(graph(E, 0, 2*c, n=600), red);
+
+pen thin=linewidth(0.5*linewidth())+grey;
+xaxis("$r$");
+yaxis("$E$");
+
+real dy=0.15; // >= the height of the letter "b" in graph units...
+label("$a$", align=N, Scale((a,-dy)));
+label("$b$", align=N, Scale((b,-dy)));
+label("$c$", align=N, Scale((c,-dy)));
+
+draw(graph(E, 0, 2*c, n=600), red); // overwrite the axis
+\end{asy}
+\end{center}
+
+\Part{d}
+The inner surface of the outer cylinder must carry
+$\ans{\lambda_i=-\lambda}$ so that the total charge enclosed by any
+gaussian surface embedded inside the outer cylinder is $0$ (as it must
+be, because $E=0$ inside a conductor).  Because the outer cylinder
+carries no net charge, $\ans{\lambda_o=-\lambda_i=\lambda}$.
 \end{solution}
-- 
2.26.2