Answer
See answers.
Work Step by Step
a. The capacitance is given in Problem 94.
Use the equation for the capacitance of a parallel-plate capacitor with a dielectric.
$$C=\kappa\epsilon_o \frac{A}{d} $$
Solve for the area.
$$A= \frac{Cd}{\kappa\epsilon_o } $$
$$A= \frac{(35\times10^{-15}F)( 2.0\times10^{-9}m)}{25(8.85\times10^{-12}C^2/(N\cdot m^2)) } $$
$$A=3.164\times10^{-13}m^2=0.3164\mu m^2$$
b. The plate area A is half of the cell’s area. $1.5\;cm^2$ is used for capacitance. One capacitor is one bit.
$$1.5cm^2(\frac{10^6 \mu m}{10^2 cm})^2(\frac{1bit}{0.32 \mu m^2})^2(\frac{1byte}{8bits})\approx 59MB$$
