Answer
See answers.
Work Step by Step
a. Calculate the capacitance from Equation 17–9.
$$C=\frac{\kappa \epsilon_o A}{d}=\frac{3.7(8.85\times10^{-12}F/m)(0.21m)(0.14m)}{0.11\times10^{-3}m}=8.8nF$$
b. The charge equals the capacitance multiplied by the voltage (this is simply the electric field multiplied by d, the separation of the plates).
$$Q_{max}=CV=CEd$$
When the maximum charge is stored, the electric field between the plates equals the dielectric strength of paper.
$$=(8.752\times10^{-9}F)(15\times10^6 V/m)( 0.11\times10^{-3}m)=14\mu C$$
