#### (a) The capacitance of a parallel-plate capacitor is given by $$C = \varepsilon_0A/d$$, where $$A$$ is the area of each plate and $$d$$ is the plate separation. Since the plates are circular, the plate area is $$A = πR^2$$, where R is the radius of a plate. Thus,

$$ C=\dfrac{ \varepsilon_0 \pi R^2}{d}=\dfrac{(8.85\times 10^{-12}F/m)\pi(8.2\times10^{-2}m)^2}{1.3\times10^{-3}m}=1.44\times10^{-10}F=144pF$$

(b) The charge on the positive plate is given by $$q$$ = $$CV$$, where V is the potential difference across the plates. Thus,

$$q=(1.44\times10^{-10}F)(120V)=1.73\times10^{-8}F=144pF.$$