1 \begin{problem*}{19.19}
2 A uniformly charged ring of radius $r = 10.0\U{cm}$ has a total charge
3 of $q = 75.0\U{$\mu$C}$. Find the electric field on the axis of the
5 \Part{a} $x_a = 1.00\U{cm}$,
6 \Part{b} $x_b = 5.00\U{cm}$,
7 \Part{c} $x_c = 30.0\U{cm}$, and
8 \Part{d} $x_d = 100\U{cm}$ from the center of the ring.
9 \end{problem*} % problem 19.19
16 currentprojection = TopView;
19 Ring r = Ring(normal=(1,0,0.1), radius=u, // cheat with z component
20 axis_post=2u, outline=pChargePen, fill=pChargePen, axis=black,
22 axis_label=Label("$x$", position=EndPoint, align=RightSide));
27 Distance da = Distance(a, c, "$r_1$"); da.draw();
28 Distance db = Distance(b, c, "$r_2$"); db.draw();
30 c, dir=degrees(c-a), L=Label("$E_1$", position=EndPoint, align=RightSide));
33 c, dir=degrees(c-b), L=Label("$E_2$", position=EndPoint, align=LeftSide));
36 E.label = Label("$E$", position=EndPoint, align=RightSide);
42 From Example 19.5 (p.~616) we see the electric field along the axis
43 (\ihat) of a uniformly charged ring is given by
45 E = \frac{k_e x q}{(x^2 + r^2)^{3/2}} \ihat
48 So applying this to our four distances (rembering to convert the
49 distances to meters), we have
51 E_a &= \ans{6.64\E{6}\U{N/C}\;\ihat} \\
52 E_b &= \ans{24.1\E{6}\U{N/C}\;\ihat} \\
53 E_c &= \ans{6.40\E{6}\U{N/C}\;\ihat} \\
54 E_d &= \ans{0.664\E{6}\U{N/C}\;\ihat}