Calculation of electrostatic energy in $12\ \mathrm{pF}$ capacitorTotal charge stored in combination Potential difference across each capacitor Energy stored, in the capacitor of capacitance 12 $\ \mathrm{pF}$ , $U\;=\;\frac{1}{2} C_1 V^2$ $\;=\;\frac{1}{2} \times 12 \times 10^{-12} \times 50 \times 50 \ \mathrm{J}$ $\;=1.5 \times 10^{-8} \ \mathrm{J}$ $C=$ Equivalent capacitance of $12\ \mathrm{pF}$ and $6\ \mathrm{pF}$, in series, is given by $\;\;\frac{1}{C_{\;\text{eq }\;}} = \;\frac{1}{12} + \;\frac{1}{6} = \;\frac{1+2}{12}$ $\therefore \;\; C_{\;\text{eq }\;}\;=4\ \mathrm{pF}$ $\therefore$ Charge stored across each capacitor$q\;=C_{e q} V$$\;=4 \times 10^{-12} \times 50\ \mathrm{C}$$\;=2 \times 10^{-10}\ \mathrm{C}$Charge on each capacitor $12\ \mathrm{pF}$ as well as $6\ \mathrm{pF}$$\therefore$ Potential difference across capacitor $C_1$$\therefore V_1=\;\frac{2 \times 10^{-10}}{12 \times 10^{-12}} \;\text{ volt }\;=\;\frac{50}{3}\ \mathrm{V}$Potential difference across capacitor $C_2$$\therefore V_2=\;\frac{2 \times 10^{-10}}{6 \times 10^{-12}} \;\text{ volt }\;=\;\frac{100}{3}\ \mathrm{V}$