\begin{problem*}{19.55}
Four identical point charges ($q = +10.0\U{$\mu$C}$) are located on
the corners of a rectangle as shown in Figure P19.55.  The dimensions
of the rectangle are $L = 60.0\U{cm}$ and $W = 15.0\U{cm}$.  Calculate
the magnitude and direction of the resultant electric force exerted on
the charge at the lower left corner by the other three charges.
\end{problem*} % problem 19.55

\empaddtoprelude{
  numeric a;
  pair A, B, C, D;
  a := 1cm;
  A := (0, a); % q1, labeled CCW from upper left
  B := origin; % q2
  C := (4a, 0); % q3
  D := (4a, a); % q4
  def drawE = 
    label.rt("W", draw_length(C, D, 8pt));
    label.bot("L", draw_length(B, C, 8pt));
    labeloffset := 5pt;
    draw_pcharge(A, 3pt);
    label.bot(btex $q_1$ etex, A);
    draw_pcharge(B, 3pt);
    label.top(btex $q_2$ etex, B);
    draw_pcharge(C, 3pt);
    label.top(btex $q_3$ etex, C);
    draw_pcharge(D, 3pt);
    label.bot(btex $q_4$ etex, D);
  enddef;
}
\begin{nosolution}
\begin{center}
\begin{empfile}[3p]
\begin{emp}(0cm,0cm)
  drawE;
\end{emp}
\end{empfile}
\end{center}
\end{nosolution}

\begin{solution}
\begin{center}
\begin{empfile}[3]
\begin{emp}(0cm,0cm)
  label.rt(btex $F_{12}$ etex, draw_force(A, B, 2cm));
  label.lft(btex $F_{32}$ etex, draw_force(C, B, .5cm));
  label.bot(btex $F_{42}$ etex, draw_force(D, B, .48cm));
  draw A--(D+(a/2,0)) dashed evenly;
  draw A--B dashed evenly;
  draw B--D dashed evenly;
  label.top(btex $\theta$ etex, draw_lout_angle(D+(1,0),D,B, a/3));
  draw_right_angle(B, A, D, a/3);
  draw_ijhats((5.5*a, a/3), 0, a/3);
  drawE;
\end{emp}
\end{empfile}
\begin{center}
\begin{empfile}[3]
\begin{emp}(0cm,0cm)
  label.rt(btex $F_{12}$ etex, draw_force(A, B, 2cm));
  label.lft(btex $F_{32}$ etex, draw_force(C, B, .5cm));
  label.bot(btex $F_{42}$ etex, draw_force(D, B, .48cm));
  draw A--(D+(a/2,0)) dashed evenly;
  draw A--B dashed evenly;
  draw B--D dashed evenly;
  label.top(btex $\theta$ etex, draw_lout_angle(D+(1,0),D,B, a/3));
  draw_right_angle(B, A, D, a/3);
  draw_ijhats((5.5*a, a/3), 0, a/3);
  drawE;
\end{emp}
\end{empfile}
The unit vector \rhat\ diagonally across from the upper right is given by
\begin{align}
 \rhat &= \cos\theta \ihat + \sin\theta \jhat \\
 \theta &= \arctan{W/L} + 180\dg = 194\dg \\
 \cos\theta &= -0.970 \\
 \sin\theta &= -0.243
\end{align}
So the electric field in the lower left corner is given by
\begin{align}
 \vect{E} &= k_e \sum_i \frac{q_i}{r_i^2}\rhat_i
           = k_e \left(\frac{q}{L^2}(-\ihat)
                       + \frac{q}{(L^2 + W^2)}(\cos\theta\ihat + \sin\theta\jhat)
		       + \frac{q}{W^2}(-\jhat) \right) \\
          &= -k_e q \left[
               \left(\frac{1}{L^2} - \frac{\cos\theta}{L^2+W^2}\right)\ihat
               + \left(\frac{1}{W^2} - \frac{\sin\theta}{L^2+W^2}\right)\jhat
	            \right]
\end{align}

So the magnitude of \vect{E} is given by
\begin{equation}
 E = k_e q \sqrt{ \left(L^{-2} - \frac{\cos\theta}{L^2+W^2}\right)^2
                + \left(W^{-2} - \frac{\sin\theta}{L^2+W^2}\right)^2 }
  = 4.08\E{6}\U{N/C}
\end{equation}
(Remembering to convert $L$ and $W$ to meters.)  And the direction
$\phi$ (measured counter clockwise from \ihat) of \vect{E} is given by
\begin{equation}
 \phi = \arctan\left(\frac{-W^{-2}+\frac{\sin\theta}{L^2+W^2}}{-L^{-2}+\frac{\cos\theta}{L^2+W^2}}\right) + 180\dg
        = \ans{263\dg}
\end{equation}
Where the $+180\dg$ is because the tangent has a period of $180\dg$,
and the angle we want is in the backside $180\dg$.

$\vect{F} = q \vect{E}$ so the direction of \vect{F} is the same as
the direction of \vect{E}.  The magnitude of \vect{F} is given by
\begin{equation}
 F = 10.0\E{-6}\U{C} \cdot 4.08\E{6}\U{N/C} = \ans{40.8\U{N}}
\end{equation}
\end{solution}
