\begin{problem*}{19.62}
Two infinite, nonconducting sheets of charge are parallel to each
other as shown in Figure P19.62.  The sheet on the left has a uniform
surface charge density $\sigma$, and the one on the right has a
uniform charge density $-\sigma$.  Calculate the electric field at
points
 \Part{a} to the left of,
 \Part{b} in between, and
 \Part{c} to the right of
the two sheets.
\end{problem*} % problem 19.62

\empaddtoprelude{
  numeric r, rf, mf, dy, height, depth, thick, theta;
  r := 1cm;
  height := 2cm;
  depth := 1cm;
  thick := 3pt;
  theta := 45;
  def draw_plates = 
    draw_plate((-r,0), height, depth, thick, theta, .5(white+red), (.7white+.3red), black);
    draw_plate(( r,0), height, depth, thick, theta, .5(white+blue), (.7white+.3blue), black);
  enddef;
}

\begin{nosolution}
\begin{center}
\begin{empfile}[6]
\begin{emp}(0, 0)
  draw_plates;
\end{emp}
\end{empfile}
\end{center}
\end{nosolution}

\begin{solution}
\hspace{\stretch{1}}
\begin{empfile}[6a]
\begin{emp}(0, 0)
  numeric rE, mE;
  rE := 2r/3;  mE := 2r/3;
  draw_ihat((2r/3,r), 0, 2r/3);
  draw_gauss_half_cyl((-thick/2,0),rE-thick/2,height/6,0.5,180); 
  draw_gauss_half_cyl((-thick/2,0),         0,height/6,0.5,180); 
  draw_plate(origin, height, depth, thick, theta, .5(white+red), (.7white+.3red), black);
  draw_gauss_half_cyl(( thick/2,0),         0,height/6,0.5,180); 
  draw_gauss_half_cyl(( thick/2,0),rE-thick/2,height/6,0.5,0); 
  label.top(btex $E_+$ etex, draw_arrow(origin,(-rE, 0), mE, 2pt, red));
  label.top(btex $E_+$ etex, draw_arrow(origin,( rE, 0), mE, 2pt, red));
\end{emp}
\end{empfile}
\hspace{\stretch{1}}
\begin{empfile}[6b]
\begin{emp}(0, 0)
  numeric rE, mE, dy;
  rE := 2r/3;  mE := 2r/3;  dy := 3pt;
  draw_plates;
  draw_ihat((-r/3,r), 0, 2r/3);
  label.top(btex $E_+$ etex, draw_arrow((-r, dy),(-r-rE, dy), mE, 2pt, red));
  label.top(btex $E_+$ etex, draw_arrow((-r, dy),(-r+rE, dy), mE, 2pt, red));
  label.top(btex $E_+$ etex, draw_arrow((-r, dy),( r+rE, dy), mE, 2pt, red));
  label.bot(btex $E_-$ etex, draw_arrow(( r,-dy),(-r-rE-mE,-dy),-mE, 2pt, blue));
  label.bot(btex $E_-$ etex, draw_arrow(( r,-dy),(-r+rE,   -dy),-mE, 2pt, blue));
  label.bot(btex $E_-$ etex, draw_arrow(( r,-dy),( r+rE+mE,-dy),-mE, 2pt, blue));
\end{emp}
\end{empfile}
\hspace{\stretch{1}}

Let \ihat\ be the direction to the right perpendicular to the sheets.
Because the problem has is symmetric to translations in the plane of
the sheets and reflections through planes perpendicular to the sheets,
the electric field must be of the form $\vect{E}(\vect{r}) =
E(x)\ihat$.

Using Gauss's law to find the electric field due to a single plate, we
imagine a cylinder that extends through the plate a length $L$ out
either side.  $\vect{E} = E\ihat$, so no flux passes through the side
walls of the cylinder.  The single sheet is symmetric to reflection in
it's plane, so (defining $x=0$ to be the $x$ value of the plane)
$\vect{E}(x) = -\vect{E}(x)$ (positive charges are repelled from both
sides).  So, letting the area of a single end cap be $A$, the charge
enclosed by the cylinder is $\sigma A$ and the flux through the
end-caps of the cylinder is given by
\begin{align}
 \Phi_E = 2EA &= \frac{q_{in}}{\epsilon_0} = \frac{\sigma A}{\epsilon_0} \\ 
 E &= \frac{\sigma}{2\epsilon_0} \label{eqn.plane_E}
\end{align}
A constant!  (See Example 19.12 on page 629 for the book's version)

\Part{a}
Using Equation \ref{eqn.plane_E} and superposition, we see
\begin{equation}
 E_L = \frac{\sigma}{2\epsilon_0} + \frac{-\sigma}{2\epsilon_0} = \ans{0}
\end{equation}

\Part{b}
\begin{equation}
 \vect{E}_B =   \frac{\sigma \ihat}{2\epsilon_0}
              + \frac{-\sigma \cdot (-\ihat)}{2\epsilon_0}
            = \frac{\sigma}{\epsilon_0}\ihat
\end{equation}

\Part{c}
Identically to \Part{a}, $E_R = 0$.
\end{solution}
