Update MetaPost to Asymptote in Serway and Jewett v4's problem 19.19.

diff --git a/latex/problems/Serway_and_Jewett_4/problem19.19.tex b/latex/problems/Serway_and_Jewett_4/problem19.19.tex
index dbccac187e7956b891f51eda4be455b9bff40081..b2daa7f2d467e677f5ada56b65c61753b0068982 100644 (file)
--- a/latex/problems/Serway_and_Jewett_4/problem19.19.tex
+++ b/latex/problems/Serway_and_Jewett_4/problem19.19.tex
@@ -10,26 +10,33 @@ ring at
 
 \begin{solution}
 \begin{center}
-\begin{empfile}[6]
-\begin{emp}(0cm,0cm)
-  pair A, B, C;
-  numeric a;
-  a := 0.75cm;
-  A := (0,a);
-  B := (0,-a);
-  C := (1cm,0);
-  draw_ijhats((-1cm,a/3), 0, a/3);
-  draw_ring(origin, a, 0, 3cm, 1cm, red, "q", "x");
-  label.bot("0", draw_ltic(origin, -90, 0, 3pt, 0pt, black));
-  label.top("A", A);
-  label.bot("B", B);
-  draw A--C; label.urt(btex $d_A$ etex, (A+C)/2);
-  draw B--C; label.lrt(btex $d_B$ etex, (B+C)/2);
-  label.lrt("E", draw_Efield(origin, C, 18pt));
-  label.top(btex $E_B$ etex, draw_Efield(B, C, 15pt));
-  label.bot(btex $E_A$ etex, draw_Efield(A, C, 15pt));
-\end{emp}
-\end{empfile}
+\begin{asy}
+import three;
+import ElectroMag;
+currentprojection = TopView;
+
+real u = 2cm;
+Ring r = Ring(normal=(1,0,0.1), radius=u,  // cheat with z component
+    axis_post=2u, outline=pChargePen, fill=pChargePen, axis=black,
+    L=Label("$q$"),
+    axis_label=Label("$x$", position=EndPoint, align=RightSide));
+r.draw();
+pair a = (0, u);
+pair b = (0, -u);
+pair c = (1.3u, 0);
+Distance da = Distance(a, c, "$r_1$");  da.draw();
+Distance db = Distance(b, c, "$r_2$");  db.draw();
+Vector Ea = EField(
+    c, dir=degrees(c-a), L=Label("$E_1$", position=EndPoint, align=RightSide));
+Ea.draw();
+Vector Eb = EField(
+    c, dir=degrees(c-b), L=Label("$E_2$", position=EndPoint, align=LeftSide));
+Eb.draw();
+Vector E = Ea + Eb;
+E.label = Label("$E$", position=EndPoint, align=RightSide);
+E.draw();
+dot(c);
+\end{asy}
 \end{center}
 
 From Example 19.5 (p.~616) we see the electric field along the axis
@@ -38,7 +45,7 @@ From Example 19.5 (p.~616) we see the electric field along the axis
  E = \frac{k_e x q}{(x^2 + r^2)^{3/2}} \ihat
 \end{equation}
 
-So applying this to our 4 distances (rembering to convert the
+So applying this to our four distances (rembering to convert the
 distances to meters), we have
 \begin{align}
  E_a &= \ans{6.64\E{6}\U{N/C}\;\ihat} \\