\begin{problem*}{19.16}
Consider the electric dipole shown in Figure P19.16.  Show that the
electric field at a distant point on the $+x$ axis is $E_x \approx
4k_eqa/x^3$.
\end{problem*} % problem 19.16

\empaddtoprelude{
  pair A, B;
  numeric a;
  a := 1cm;
  A := (-a,0);
  B := (a, 0);
  C := (6a, 0);
  def drawC = 
    drawarrow (A-(a,0))--(C+(a,0)) withpen pencircle scaled 0pt;
    draw_ncharge(A, 6pt);
    draw_pcharge(B, 6pt);
    label.top("0", draw_ltic(origin, 90, 0, 3pt, 0pt, black));
    dotlabel.bot("x", C);
    label.bot("a", draw_length(A, origin, 10pt));
    label.bot("a", draw_length(origin, B, 10pt));
    labeloffset := 8pt;
    label.top("-q", A);
    label.top("q", B);
  enddef;
}

\begin{nosolution}
\begin{center}
\begin{empfile}[3p]
\begin{emp}(0cm, 0cm)
  drawC;
\end{emp}
\end{empfile}
\end{center}
\end{nosolution}

\begin{solution}
\begin{center}
\begin{empfile}[3]
\begin{emp}(0cm, 0cm)
  label.top(btex $E_{q}$ etex, draw_Efield(B, C, 18pt));
  label.top(btex $E_{-q}$ etex, draw_Efield(A, C, -16pt));
  drawC;
\end{emp}
\end{empfile}
\end{center}
Let us assume the point in question has a positive $x$ value (just
reverse the sign if $x < 0$).
\begin{equation}
 \vect{E} = k_e \sum_i \frac{q_i}{r_i^2}\rhat_i
          = k_e \left[\frac{q}{(x-a)^2}\ihat + \frac{-q}{(x+a)^2}\ihat\right]
\end{equation}
For $|x| \gg |c|$,
\begin{equation}
 (x+c)^n = x^n \left(1+\frac{c}{x}\right)^n
         = x^n \left[1 + n\frac{c}{x} + \frac{n(n-1)}{2}\cdot\left(\frac{c}{x}\right)^2 + \ldots\right]
         \approx x^n (1 + n\frac{c}{x}) \;,
\end{equation}
because $(c/x)^2$ is very, very small.  (We are Taylor expanding
$(x+c)^n$ as a function of $c/x$, and keeping only the first two
terms.)  In our case, $n = -2$ and $c = \mp a$
\begin{equation}
 \vect{E} = k_e \left[\frac{q}{x^2}\left(1-2\frac{-a}{x}\right) + \frac{-q}{x^2}\left(1-2\frac{a}{x}\right)\right]\ihat
          = k_e \frac{q}{x^2}\left(1+2\frac{a}{x} - 1+2\frac{a}{x}\right)\ihat
          = \ans{\frac{4 k_e q a}{x^3}\ihat}
\end{equation}
\end{solution}
