Answer
(a) $Q_t = 24.2 \times 10^{3} \,\text{nC} $
(b) $Q_{35} = 7.7 \times 10^{3} \,\text{nC}$ and $Q_{75} = 16.5 \times 10^{3} \,\text{nC}$
(c) $U_{Parallel} = 2.6 \times 10^{-3} \,\text{J}$
(d) $ U_1 = 8.47 \times 10^{-3} \,\text{J} $ and $ U_2 = 18.15 \times 10^{-3} \,\text{J} $
(e) $V_1 = V_2 = 220 \,\text{V}$
Work Step by Step
(a) In parallel, the equivalent capacitance of both capacitors is given by
$$C_{eq} = 35 \,\text{nF} + 75 \,\text{nF} = 110 \,\text{nF} $$
Therefore, the total charge of the network is given by
$$Q_{t} = C_{eq}V = 110 \,\text{nF} \times 220 \,\text{V} = \boxed{24.2 \times 10^{3} \,\text{nC}} $$
(b) In parallel, the potential is the same therefore we could find the charge for each capacitor as next
$$Q_{35}= C_{35} V = 35 \,\text{nF} \times 220 \,\text{V} = \boxed{7.7 \times 10^{3} \,\text{nC}} $$
$$Q_{75}= C_{75}V = 75\,\text{nF} \times 220 \,\text{V} = \boxed{16.5 \times 10^{3} \,\text{nC}} $$
(c) Their stored energy in parallel is given by
$$U_{Parallel} = \dfrac{1}{2} C_{eq}V^2 = \dfrac{1}{2} (110 \times 10^{-9} \,\text{C})(220 \,\text{V})^2 = \boxed{2.6 \times 10^{-3} \,\text{J}}$$
(d) Their stored energy stored in each capacitor is given by:
$$U_1 = \dfrac{1}{2} C_{1}V^2 = \dfrac{1}{2} (35 \times 10^{-9} \,\text{F})(220\,\text{V})^2 = \boxed{8.47 \times 10^{-3} \,\text{J}}$$
$$U_2 = \dfrac{1}{2} C_{2}V^2 = \dfrac{1}{2} (75 \times 10^{-9} \,\text{F})(220 \,\text{V})^2 = \boxed{18.15 \times 10^{-3} \,\text{J}}$$
(e) In parallel, both cacapictiors have the same potential of the total comibination
$$\boxed{V_1 = V_2 = 220 \,\text{V}}$$
