Answer
$3.39\times10^{-8}C$.
Work Step by Step
Combine equations 17–7 and 17–8 and find the charge.
$Q=CV$ and $C=\frac{\epsilon_oA}{d}$
$$ Q=\frac{\epsilon_oA}{d}V$$
The desired electric field is V/d for the capacitor.
$$=\epsilon_o(A)(E)$$
$$=(8.85\times10^{-12}C^2/(N\cdot m^2))(45.0\times10^{-4}m^2)(8.50\times10^5V/m)$$
$$=3.39\times10^{-8}C$$