Two capacitors 3 $\mathrm{\mu }$F and 4 $\mathrm{\mu }$F, are individually charged across a 6 V battery. After being disconnected from the battery, they are connected together with the negative plate of one attached to the positive plate of the other. What is the final total energy stored? 

Store energy in capacitor of 3 $\mathrm{\mu }$F,

                      ${\mathrm{U}}_1=\frac{1}{2}\times {\mathrm{C}}_1{\mathrm{V}}^2$

                           $=\frac{1}{2}\times 3\times (6{)}^2\times {10}^{-6}$

                           $=54\times {10}^{-6} \mathrm{J}$

Store energy in capacitor of 4 $\mathrm{\mu }$F,

                      ${\mathrm{U}}_2=\frac{1}{2}{\mathrm{C}}_2{\mathrm{V}}^2$

                           $=\frac{1}{2}\times 4\times (6{)}^2\times {10}^{-6}$

                           $=72\times {10}^{-6} \mathrm{J}$

When both capacitors are connected in series,

                       ${\mathrm{C}}_{\mathrm{eq}}=\frac{{\mathrm{C}}_1{\mathrm{C}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}$

                             $=\frac{3\times 4}{3+4}=\frac{12}{7}\mathrm{\mu F}$

Energy lost, $\mathrm{U}=\frac{1}{2}{\mathrm{C}}_{\mathrm{eq}}{\left({\mathrm{V}}_1-{\mathrm{V}}_2\right)}^2$

                       $=\frac{1}{2}\times \frac{12}{7}\times (0{)}^2\times {10}^{-6}=0$

Total energy = U₁ + U₂

                    $=1.26\times {10}^{-4} \mathrm{J}$