From 4e9676fc85bf0ba0ae326a4bbdcc76d9d387bb8d Mon Sep 17 00:00:00 2001
From: "W. Trevor King" <wking@tremily.us>
Date: Fri, 20 Apr 2012 17:48:23 -0400
Subject: [PATCH] Add solution text for Serway and Jewett v8's 25.7, .12, and
 .16.

---
 .../Serway_and_Jewett_8/problem25.07.tex      | 12 +++++++++
 .../Serway_and_Jewett_8/problem25.12.tex      | 18 +++++++++++++
 .../Serway_and_Jewett_8/problem25.16.tex      | 26 +++++++++++++++++++
 3 files changed, 56 insertions(+)

diff --git a/latex/problems/Serway_and_Jewett_8/problem25.07.tex b/latex/problems/Serway_and_Jewett_8/problem25.07.tex
index 238876f..315852d 100644
--- a/latex/problems/Serway_and_Jewett_8/problem25.07.tex
+++ b/latex/problems/Serway_and_Jewett_8/problem25.07.tex
@@ -35,4 +35,16 @@ label("Top view", (u/2, 0), align=S);
 \end{problem*}
 
 \begin{solution}
+This is just like a gravitational pendulum, except that the external
+force is $qE$ from the electric field instead of the usual $mg$ from
+the gravitational field.  We can solve the problem by conserving
+energy between the two snapshots shown in the figure.  The
+gravitational potential energy $mgh$ is now $qEh$, so
+\begin{align}
+  qEL(1-\cos(\theta)) &= \frac{1}{2} mv^2 \\
+  v &= \sqrt{\frac{2qEL(1-\cos(\theta))}{m}}
+    = \sqrt{\frac{2\cdot(2.00\E{-6}\U{C})\cdot(300\U{V/m})
+                  \cdot(1.50\U{m})\cdot(1-\cos(60.0\dg))}{0.0100\U{kg}}}
+    = \ans{0.300\U{m/s}} \;.
+\end{align}
 \end{solution}
diff --git a/latex/problems/Serway_and_Jewett_8/problem25.12.tex b/latex/problems/Serway_and_Jewett_8/problem25.12.tex
index b3a8dc2..9eda325 100644
--- a/latex/problems/Serway_and_Jewett_8/problem25.12.tex
+++ b/latex/problems/Serway_and_Jewett_8/problem25.12.tex
@@ -23,4 +23,22 @@ label("$d$", (u/2, 0), align=S);
 \end{problem*}
 
 \begin{solution}
+\Part{a}
+Using the formula for electric potential due to a point charge, we have
+\begin{equation}
+  V_A = k\frac{-15.0\U{nC}}{d} + k\frac{27.0\U{nC}}{d}
+    = \frac{8.99\E{-9}\U{Nm$^2$/C$^2$}}{2.00\E{-2}\U{m}}
+      \cdot(-15.0+27.0)\E{-9}\U{C}
+    = \ans{5.39\E{3}\U{V}} = \ans{5.39\U{kV}} \;.
+\end{equation}
+Because electric potential is a scalar quantity, there's no need to
+mess around with any pesky vectors.
+
+\Part{b}
+The same formula holds, but now there is half the distance between the
+charges and the point of interest.
+\begin{equation}
+  V_B = \frac{k}{d/2}\cdot(-15.0+27.0)\E{-9}\U{C} = 2V_A
+      = \ans{10.8\U{kV}} \;.
+\end{equation}
 \end{solution}
diff --git a/latex/problems/Serway_and_Jewett_8/problem25.16.tex b/latex/problems/Serway_and_Jewett_8/problem25.16.tex
index bdf7ea8..f6e2fa7 100644
--- a/latex/problems/Serway_and_Jewett_8/problem25.16.tex
+++ b/latex/problems/Serway_and_Jewett_8/problem25.16.tex
@@ -22,4 +22,30 @@ label("$d$", (u/2, 0), align=S);
 \end{problem*}
 
 \begin{solution}
+\Part{a}
+Using the formula for electric potential due to a point charge, we have
+\begin{equation}
+  V_A = k\frac{Q}{d} + k\frac{2Q}{d\sqrt{2}}
+    = k\frac{Q}{d}\p({1+\sqrt{2}})
+    = \ans{5.43\E{3}\U{V}} = \ans{5.43\U{kV}} \;,
+\end{equation}
+where we used the pythagorean theorem to find the distance from the
+$2Q$ charge to $A$ ($\sqrt{d^2 + d^2} = \sqrt{2d^2} = d\sqrt{2}$).
+
+\Part{b}
+The geometry is flipped for $B$, so we have
+\begin{equation}
+  V_A = k\frac{Q}{d\sqrt{2}} + k\frac{2Q}{d}
+    = k\frac{Q}{d}\p({2+\frac{1}{\sqrt{2}}})
+    = \ans{6.08\E{-15}\U{V}} = \ans{6.08\U{kV}} \;,
+\end{equation}
+
+\Part{c}
+\begin{equation}
+  \Delta V_{AB} = V_B - V_A
+    = k\frac{Q}{d}\p({2+\frac{1}{\sqrt{2}}}) - k\frac{Q}{d}\p({1+\sqrt{2}})
+    = k\frac{Q}{d}\p({1 + \frac{1}{\sqrt{2}} - \frac{2}{\sqrt{2}}})
+    = k\frac{Q}{d}\p({1 - \frac{1}{\sqrt{2}}})
+    = \ans{658\U{V}} \;.
+\end{equation}
 \end{solution}
-- 
2.26.2