\begin{problem*}{19}
A uniformly charged ring of radius $r = 10.0\U{cm}$ has a total charge of $q = 75.0\U{$\mu$C}$.
Find the electric field on the axis of the ring at
 \Part{a} $x_a = 1.00\U{cm}$,
 \Part{b} $x_b = 5.00\U{cm}$,
 \Part{c} $x_c = 30.0\U{cm}$, and
 \Part{d} $x_d = 100\U{cm}$ from the center of the ring.
\end{problem*}

\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}
\end{center}

From Example 19.5 (p. 616) we see the electric field along the axis (\ihat) of a uniformly charged ring is given by
\begin{equation}
 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 distances to meters), we have
\begin{align}
 E_a &= \ans{6.64\E{6}\U{N/C}\;\ihat} \\
 E_b &= \ans{24.1\E{6}\U{N/C}\;\ihat} \\
 E_c &= \ans{6.40\E{6}\U{N/C}\;\ihat} \\
 E_d &= \ans{0.664\E{6}\U{N/C}\;\ihat}
\end{align}
\end{solution}
