Answer
(a) $Q = 3200 \,\text{nC}$
(b) $Q_1 = Q_2 = 3200 \,\text{nC}$
(c) $U = 7.70\times 10^{-5} \,\text{J}$
(d) $U_1 = 0.35 \times 10^{-3} \,\text{J}$ and $U_2 = 0.28 \times 10^{-3} \,\text{J}$
(e) $V_1 = 21.33 \,\text{V}$ and $V_2 = 26.66 \,\text{V}$
Work Step by Step
(a) The capacitance in series for two capacitors is given by
\begin{align*}
C_{eq} = \dfrac{C_1 C_2}{C_1 +C_2} = \dfrac{( 150\,\text{nF}) (120\,\text{nF})}{150\,\text{nF} + 120\,\text{nF}} = 66.67 \,\text{nF}
\end{align*}
Therefore, we could get the charge in their combination by
$$Q = C_{eq} V = (66.67 \,\text{nF}) (48 \,\text{V}) = \boxed{3200 \,\text{nC}} $$
(b) In series, both capacitors have the same charge even for their combination and the charge for each capacitor is given by
$$\boxed{Q_{1} = Q_2 = 3200 \,\text{nC}}$$
(c) Their stored energy in series is given by
$$U_{series} = \dfrac{1}{2} C_{eq}V^2 = \dfrac{1}{2} (66.67\times 10^{-9} \,\text{F})(48 \,\text{V})^2 = \boxed{7.70\times 10^{-5} \,\text{J}}$$
(d) Their stored energy stored in each capacitor is given by
$$U_1 = \dfrac{1}{2} C_{1}V^2 = \dfrac{1}{2} (150 \times 10^{-9} \,\text{F})(48 \,\text{V})^2 = \boxed{0.35 \times 10^{-3} \,\text{J}}$$
$$U_2 = \dfrac{1}{2} C_{2}V^2 = \dfrac{1}{2} (120 \times 10^{-9} \,\text{F})(48 \,\text{V})^2 = \boxed{0.28 \times 10^{-3} \,\text{J}}$$
(e) The potential for each capacitor is given by
$$V_1 = \dfrac{Q_1}{C_1} = \dfrac{3200 \,\text{nC}}{ 150 \,\text{nF}} =\boxed{ 21.33 \,\text{V}} $$
$$V_2 = \dfrac{Q_2}{C_2} = \dfrac{3200 \,\text{nC}}{ 120 \,\text{nF}} = \boxed{26.66 \,\text{V}} $$
