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
21 draw_ijhats((-1cm,a/3), 0, a/3);
22 draw_ring(origin, a, 0, 3cm, 1cm, red, "q", "x");
23 label.bot("0", draw_ltic(origin, -90, 0, 3pt, 0pt, black));
26 draw A--C; label.urt(btex $d_A$ etex, (A+C)/2);
27 draw B--C; label.lrt(btex $d_B$ etex, (B+C)/2);
28 label.lrt("E", draw_Efield(origin, C, 18pt));
29 label.top(btex $E_B$ etex, draw_Efield(B, C, 15pt));
30 label.bot(btex $E_A$ etex, draw_Efield(A, C, 15pt));
35 From Example 19.5 (p.~616) we see the electric field along the axis
36 (\ihat) of a uniformly charged ring is given by
38 E = \frac{k_e x q}{(x^2 + r^2)^{3/2}} \ihat
41 So applying this to our 4 distances (rembering to convert the
42 distances to meters), we have
44 E_a &= \ans{6.64\E{6}\U{N/C}\;\ihat} \\
45 E_b &= \ans{24.1\E{6}\U{N/C}\;\ihat} \\
46 E_c &= \ans{6.40\E{6}\U{N/C}\;\ihat} \\
47 E_d &= \ans{0.664\E{6}\U{N/C}\;\ihat}