Question

What is the stored energy before and after the switch $S$ is closed?

Answer 1

The common potential is $V_c=\frac{Q_1+Q_2}{C_1+C_2}=\frac{C_1{V}_0+0}{C_1+C_2}=\frac{8\times 120}{8+4}=80V$

The initial energy stored, $U_i=\frac{1}{2}{C}_1{V}_0^2=0.5\times (8\times {10}^{-6})\times {120}^2=57.6mJ$

The final energy stored, $U_f=\frac{1}{2}(C_1+C_2)V_c^2=0.5\times (8+4)\times {10}^{-6}\times {80}^2=38.4mJ$