Answer
(a) $Q = 120 \mathrm{~\mu C}$
(b) $Q = 60 \mathrm{~\mu C}$
(c) $Q = 480 \mathrm{~\mu C}$
Work Step by Step
(a) The capacitance between the two plates occur due to the accumulation of the charges on the plates because of the potential of the battery and it is given by
$$Q = CV_{ab}$$
Where $V_{ab} $ is the potential between the two plates due to the battery. Let us substitute the values $C$ and $V_{ab}$ to get the charge on each plate
$$Q = CV_{ab} = (10 \times 10^{-6} \,\text{F})(12 \,\text{V}) = \boxed{120 \mathrm{~\mu C}}$$
(b) The capacitance is inversely proportional to the distance $d$ between the two plates and it is given by
$$C = \dfrac{\epsilon_o A}{d}$$
Also, the capacitance is directly proportional to the charge on the plates. Therefore, we conclude that the charge is inversely proportional to the distance $d$ between the two plates
$$Q \propto \dfrac{1}{d}$$
Hence, as the distance is doubled, the charges are halved and becomes
$$Q = \dfrac{1}{2} \times 120 \mathrm{~\mu C} = \boxed{60 \mathrm{~\mu C}}$$
(c) As shown by the second equation above, the capacitance is directly proportional to the area of the plates and the radius of the plates
$$C \propto r^2$$
So, when the radius is doubled, the capacitance will increase four times. Also, the capacitance is directly proportional to the charge on the plates. therefore, the charge on the plates will increase four times when the radius of the plates is doubled
$$Q = 4 \times 120 \mathrm{~\mu C} = \boxed{480 \mathrm{~\mu C}}$$
