Determination of $C_1$ and $C_2$Determination of Charge on each capacitor in parallel combinationEnergy stored in a capacitor $E=\;\frac{1}{2} C V^2$In series combination $0.045\;=\;\frac{1}{2}\;\frac{C_1 C_2}{C_1+C_2}(100)^2$ $\Rightarrow\;\;\;\frac{C_1 C_2}{C_1+C_2}\;=0.09 \times 10^{-4}$ ......(i)In Parallel combination $0.25\;=\;\frac{1}{2} (C_1+C_2) (100)^2$ $\Rightarrow\;\; C_1+C_2\;=0.5 \times 10^{-4}$ .......(ii) On simplifying (i) & (ii) $C_1 C_2\;=0.045 \times 10^{-8}$ $(C_1-C_2)^2\;= (C_1+C_2)^2-4 C_1 C_2$ $\;= (0.5 \times 10^{-4})^2-4 \times 0.045 \times 10^{-8}$ $\;=0.25 \times 10^{-8}-0.180 \times 10^{-8}$ $(C_1-C_2)^2\;=0.07 \times 10^{-8}$ $(C_1-C_2) \;=2.6 \times 10^{-5}$ $\;=0.26 \times 10^{-4}$ .......(iii) From (ii) and (iii) we have $\Rightarrow C_1=0.38 \times 10^{-4}\ \mathrm{F}$ $\;\text{ and }\;\;C_2=0.12 \times 10^{-4}\ \mathrm{F}$ Charges on capacitor $C_1$ and $C_2$ in Parallel combination $Q_1\;=C_1 V= (0.38 \times 10^{-4} \times 100)$ $\;=0.38 \times 10^{-2}\ \mathrm{C}$ $Q_2\;=C_2 V= (0.12 \times 10^{-4} \times 100)$ $\;=0.12 \times 10^{-2}\ \mathrm{C}$ [Note: If the student writes the relations/ equations $E=\;\frac{1}{2} C V^2$ and $\;\; 0.045=\;\frac{1}{2} \left(\;\frac{C_1 C_2}{C_1+C_2}\right) (100)^2$ $0.25=\;\frac{1}{2} (C_1+C_2) (100)^2$But is unable to calculate $C_1$ and $C_2$, award him/her full 2 marks. Also if the student just writes$Q_1=C_1 V=C_1(100) \;\text{ and }\; Q_2=C_2 V=C_2(100)$Award him/her one mark for this part of the question.]