\begin{problem*}{11}
In Figure P19.11, determine the point (other than infinity) at which the electric field is zero.
$q_1 = -2.50\U{$\mu$C}$ and $q_2 = 6.00\U{$\mu$C}$.
\end{problem*}

\empaddtoprelude{
  pair A, B;
  A := origin;
  B := (2cm, 0);
  C := (-3.6cm, 0);
  numeric orig_labeloffset;
  orig_labeloffset := labeloffset;
  def drawD =
    label.bot("x", draw_arrow((-2cm,0), (-1cm,0), 4cm, 0pt, black));
    label.bot("0", draw_ltic(A, -90, 0, 10pt, 0pt, black));
    label.bot("1m", draw_ltic(B, -90, 0, 10pt, 0pt, black));
    draw_ncharge(A, 4pt);
    draw_pcharge(B, 6pt);
    labeloffset := 8pt;
    label.top(btex $q_1$ etex, A);
    label.top(btex $q_2$ etex, B);
    labeloffset := orig_labeloffset;
  enddef;
}

\begin{nosolution}
\begin{center}
\begin{empfile}[4ns]
\begin{emp}(0cm,0cm)
  drawD;
\end{emp}
\end{empfile}
\end{center}
\end{nosolution}

\begin{solution}
\begin{center}
\begin{empfile}[4]
\begin{emp}(0cm,0cm)
  drawD;
  draw_tic(A, -180, 0, 4.5cm, 0pt, black); % extend x axis
  label.top(btex $r_1$ etex, draw_length(A, C, 20pt));
  label.bot(btex $r_2$ etex, draw_length(C, B, 20pt));
  label.bot(btex $E_1$ etex, draw_Efield(A, C, -10pt));
  label.bot(btex $E_2$ etex, draw_Efield(B, C, 10pt));
  dotlabel("",C);
\end{emp}
\end{empfile}
\end{center}

First, we need a coordinate system.
Let $q_1$ be the origin ($x_1 = 0$), and $q_2$ be at $x_2 = 1.00\U{m}$.

The electric field of a finite number of point charge is given by (p. 612, 19.6)
\begin{equation}
 \vect{E} = k_e \sum_i \frac{q_i}{r_i^2}\rhat_i
\end{equation}
For any point off the $x$ axis, there would be some force moving the charge in the vertical $y$ direction, so we only need to look at positions on the $x$ axis.

A positive test charge placed between the two charges would be pulled to the left by $q_1$ and pushed to the left by $q_2$.
A positive test charge placed to the right of $q_2$ would be pushed to the right by $q_2$ more strongly (because $q_2 > q_1$ and $r_2 < r_1$) than it would be pulled to the left by $q_1$.
So the only place to look for equilibrium is to the left of $q_1$,  ($x < 0$, where $r_2 = r_1 + x_2$).
\begin{align}
 \vect{E} &= k_e \left(-\frac{q_1}{r_1^2} - \frac{q_2}{(r_1 + x_2)^2}\right)
           = 0 \\
 \frac{q_1}{r_1^2} &= -\frac{q_2}{(r_1 + x_2)^2} \\
 \frac{r_1+x_2}{r_1} = 1 + \frac{x_2}{r_1} &= \pm\sqrt{\frac{-q_2}{q_1}} \\
 r_1 &= \frac{x_2}{\sqrt{\pm\frac{-q_2}{q_1}} - 1} = 1.82\U{m}, -0.392\U{m}
\end{align}
But $r_1 = -0.392\U{m}$ is between the two charges (where our assumption about the electric fields opposing each other doesn't hold), so $\vect{E} = 0$ only at a $r_1 = 1.82\U{m}$ ($x=-1.82\U{m}$).
\end{solution}
