Answer
$2\times10^{-7}C$
Work Step by Step
Equation 17–4a tells us that the voltage difference V between the plates is the electric field strength, multiplied by the separation distance, so V = Ed. Use that with equation 17–8 for the capacitance of a parallel-plate capacitor.
$$Q=CV=\epsilon_o \frac{A}{d}Ed=\epsilon_o AE$$
$$=(8.85\times10^{-12}C^2/(N\cdot m^2))(65\times10^{-4}m^2)(3\times 10^6V/m) $$
$$=1.7\times10^{-7}C\approx2\times10^{-7}C$$
