Answer
See answers.
Work Step by Step
a. Use Q=CV to find the amount of charge on each plate.
$$Q=CV$$
The negative plate has excess electrons.
$$N(-e)=-CV$$
$$N=\frac{CV}{e}=\frac{(35\times10^{-15}F)(1.5V)}{1.60\times10^{-19}C}\approx 3.3\times10^5$$
b. The stored charge is directly proportional to the potential difference. A $2.0\%$ decrease in potential difference means a $2.0\%$ decrease in charge.
$$\Delta Q=0.020Q$$
$$\Delta t=\frac{\Delta Q}{\Delta Q/\Delta t}=\frac{0.020CV}{\Delta Q/\Delta t }$$
$$\Delta t=\frac{0.020(35\times10^{-15}F)(1.5V)}{0.30\times10^{-15}C/s}=3.5s$$
