\begin{problem*}{20.54}
\Part{a} How much charge $Q_c$ can be placed on a capacitor with air
between the plates before it breaks down if the area of each plate is
$A=5.00\U{cm}^2$?
\Part{b} Find the maximum charge assuming polystyrene is used between
the plates instead of air.
\end{problem*} % problem 20.54

\begin{solution}
From Chapter 19, the voltage difference due to a constant electric
field \vect{E} over a displacement \vect{d} is given by $\Delta V =
\vect{E} \cdot \vect{d}$.  So for two plates a distance $d$ apart, the
breakdown voltage is given by
\begin{equation}
 V_c = E_c d \label{eqn.Vc_Ec}
\end{equation}
where $E_c$ is the dielectric strength of the material.

The capacitance of a parallel-plate capacitor is given by
\begin{equation}
 C = \frac{\kappa \epsilon_0 A}{d} \label{eqn.pp_cap}
\end{equation}

Combining these two formula with the definition of capacitance we have
\begin{equation}
 E_c d = V = \frac{Q}{C} = \frac{Q d}{\kappa \epsilon_0 A} \\
 Q = \kappa E_c \epsilon_0 A
\end{equation}

Looking up the values for air and polystyrene in Table 20.1 on page
699 of the text we see:
\begin{center}
 \begin{tabular}{l r r}
  Name & Dielectric constant $\kappa$ & Dielectric strength $E_c$ \\
  \hline
  \Tstrut Air & $1.00059$ & $3\E{6}\U{V/m}$ \\
  Polystyrene & $2.56$ & $24\E{6}\U{V/m}$ \\
 \end{tabular}
\end{center}

So plugging into our formula for the charge
\begin{align}
 Q_a &= 1.00 \cdot (3\E{6}\U{V/m}) \cdot (8.85\E{-12}\U{C$^2$/N m$^2$}) \cdot 5\E{-4}\U{m$^2$} = \ans{1.33\E{-8}\U{C}} \\
 Q_b &= 2.56 \cdot (24\E{6}\U{V/m}) \cdot (8.85\E{-12}\U{C$^2$/N m$^2$}) \cdot 5\E{-4}\U{m$^2$} = \ans{2.72\E{-7}\U{C}}
\end{align}
\end{solution}
