1 \begin{problem*}{19.16}
2 Consider the electric dipole shown in Figure P19.16. Show that the
3 electric field at a distant point on the $+x$ axis is $E_x \approx
5 \end{problem*} % problem 19.16
15 drawarrow (A-(a,0))--(C+(a,0)) withpen pencircle scaled 0pt;
18 label.top("0", draw_ltic(origin, 90, 0, 3pt, 0pt, black));
20 label.bot("a", draw_length(A, origin, 10pt));
21 label.bot("a", draw_length(origin, B, 10pt));
42 label.top(btex $E_{q}$ etex, draw_Efield(B, C, 18pt));
43 label.top(btex $E_{-q}$ etex, draw_Efield(A, C, -16pt));
48 Let us assume the point in question has a positive $x$ value (just
49 reverse the sign if $x < 0$).
51 \vect{E} = k_e \sum_i \frac{q_i}{r_i^2}\rhat_i
52 = k_e \left[\frac{q}{(x-a)^2}\ihat + \frac{-q}{(x+a)^2}\ihat\right]
56 (x+c)^n = x^n \left(1+\frac{c}{x}\right)^n
57 = x^n \left[1 + n\frac{c}{x} + \frac{n(n-1)}{2}\cdot\left(\frac{c}{x}\right)^2 + \ldots\right]
58 \approx x^n (1 + n\frac{c}{x}) \;,
60 because $(c/x)^2$ is very, very small. (We are Taylor expanding
61 $(x+c)^n$ as a function of $c/x$, and keeping only the first two
62 terms.) In our case, $n = -2$ and $c = \mp a$
64 \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
65 = k_e \frac{q}{x^2}\left(1+2\frac{a}{x} - 1+2\frac{a}{x}\right)\ihat
66 = \ans{\frac{4 k_e q a}{x^3}\ihat}