Fix subscripting (E{600} -> E_{600}) in Serway and Jewett v8's 24.13.

diff --git a/latex/problems/Serway_and_Jewett_8/problem24.13.tex b/latex/problems/Serway_and_Jewett_8/problem24.13.tex
index 98034e84eb34e49afaa3d507ba2a347c55686008..6b19c69a2879aa9f260eefb0e7d38eeccbc2c85b 100644 (file)
--- a/latex/problems/Serway_and_Jewett_8/problem24.13.tex
+++ b/latex/problems/Serway_and_Jewett_8/problem24.13.tex
@@ -17,14 +17,14 @@ at the end caps.  If the area of the end cap is $A$, that flux is
 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 \;.
+  \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{(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}