n identical capacitors are joined in parallel and the combination is charged to voltage V. The total energy stored is U. The capacitors are now disconnected and joined in series. The total ·energy stored in the series combination will be

Let C be the capacitance of each capacitor. For parallel combination, the net capacitance is C₁ = nC. Also V₁ = V. Therefore, the energy stored in the parallel combination is

$$\begin{gathered} U_1=\frac{1}{2}{C}_1{V}_1^2=\frac{1}{2}\times nC\times V^2=\frac{1}{2}nC{V}^2 \\ For series combination, we have C_2 = \raisebox{1ex}{$C$}\!\left/ \!\raisebox{-1ex}{$n$}\right. and V_2 = nV(sum of all the potentials) \\ Therefore, the energy stored in the series combination is \\ {U}_2=\frac{1}{2}{C}_2{V}_2^2=\frac{1}{2}\times \frac{C}{n}\times n^2{V}^2=\frac{1}{2}nC{V}^2 \end{gathered}$$