\begin{problem*}{20.51}
Show that the force between two plates of a parallel-plate capacitor
each have an attractive force given by
\begin{equation}
 F = \frac{Q^2}{2\epsilon_0 A}
\end{equation}
\end{problem*} % problem 20.51

\begin{solution}
The electric field generated by the plate $A$ is given by $E_A = Q/2
\epsilon_0 A$ (which we derived for P19.62, along with $\sigma =
Q/A$).  So the force on plate $B$ due to plate $A$ is given by
\begin{equation}
 F = QE_A = \frac{Q^2}{2 \epsilon_0 A}
\end{equation}
\end{solution}
