Fix V_B typos in Serway and Jewett v8's 25.16 solution.

[course.git] / latex / problems / Serway_and_Jewett_8 / problem25.16.tex

diff --git a/latex/problems/Serway_and_Jewett_8/problem25.16.tex b/latex/problems/Serway_and_Jewett_8/problem25.16.tex
index bdf7ea852843f52ad6718e89e595e800ad53cd55..c20016d044194dcb403902df1f2d9ee888bcc114 100644 (file)
--- 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_B = k\frac{Q}{d\sqrt{2}} + k\frac{2Q}{d}
+    = k\frac{Q}{d}\p({2+\frac{1}{\sqrt{2}}})
+    = \ans{6.08\E{3}\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}