Answer
See answers.
Work Step by Step
a. Let N electrons flow onto the plate. The charge on the capacitor is Q=Ne. However, we also know that Q=CV for a capacitor.
$$Ne=CV$$
$$V=\frac{Ne}{C}$$
V is directly proportional to N, which was to be shown.
b. We’re told that $\Delta V=1\;mV$. It is desired to resolve $\Delta N=1$. Solve for the required capacitance.
$$V=\frac{Ne}{C}$$
$$\Delta V=\frac{e \Delta N}{C}$$
$$C=\frac{e \Delta N}{\Delta V }$$
$$C=\frac{(1.60\times10^{-19}C) (1)}{0.001 V }\approx 2\times10^{-16}F$$
c. Use the equation for the capacitance of a parallel-plate capacitor with a dielectric.
$$C=\kappa\epsilon_o \frac{A}{d} =\kappa\epsilon_o \frac{\ell ^2}{d} $$
Solve for the side length of the square.
$$\ell= \sqrt{\frac{Cd}{\kappa\epsilon_o }} $$
$$\ell= \sqrt{\frac{(1.60\times10^{-16}F) (100\times10^{-9}m)}{3(8.85\times10^{-12}C^2/(N\cdot m^2)) }} $$
$$\ell=7.76\times10^{-7}m=0.8\mu m$$