#### Answer

(a) $C = 1.25\times 10^{-11}~F$
(b) $Q = 1.25\times 10^{-9}~C$

#### Work Step by Step

(a) We can find the capacitance:
$C = \frac{\epsilon_0~A}{d}$
$C = \frac{\epsilon_0~\pi~r^2}{d}$
$C = \frac{(8.854\times 10^{-12}~F/m)(\pi)~(0.015~m)^2}{0.50\times 10^{-3}~m}$
$C = 1.25\times 10^{-11}~F$
(b) We can find the magnitude of the charge $Q$ on each electrode:
$Q = C~\Delta V$
$Q = (1.25\times 10^{-11}~F)(100~V)$
$Q = 1.25\times 10^{-9}~C$