Answer
a) $Q = 120 \times 10^{-6} \,\text{C}$\\
b) $Q = 60 \times 10^{-6} \,\text{C}$\\
c)$Q = 480\times 10^{-6} \,\text{C}$\\
Work Step by Step
a) The charge $Q$ is related to the capacitance and the voltage of the capacitor by equation 24.2
\begin{equation}
Q = CV_{ab}
\end{equation}
By substitute $C$ and $V_{ab}$ values into this equation we can get $Q$
\begin{align*}
Q &= CV_{ab} = (10 \times 10^{-6} \,\text{F})(12 \,\text{V}) = 120 \times 10^{-6} \,\text{C}
\end{align*}
b) The capacitance $C$ is inversely proportional to the separated distance $d$ as shown by equation 24.2, so as $d$ is doubled, the capacitance is halved. Therefore, the charge $Q$ is halved and becomes
\begin{align*}
Q &=120 \times 10^{-6} \,\text{C}/2 = 60\times 10^{-6} \,\text{C}
\end{align*}
c) The capacitance is directly proportional to the area of the sphere hence, the capacitance is directly proportional to the square radius. As the radius is doubled, the capacitance increases by a factor 4. Therefore the charge $Q$ increases four times and becomes
\begin{align*}
Q &=4(120 \times 10^{-6} \,\text{C}) = 480\times 10^{-6} \,\text{C}
\end{align*}
Note: The voltage is constant for the three cases.
