91Ó°ÊÓ

To get the decrease in energy of the system we will get energy at the initial connection$U_1$,then the energy in the final connection$U_2$.

In the parallel connection, the total charge of the system equals the summation of both capacitors. To get the energy of the system we should get the equivalent capacitance for the initial state where the two capacitors are connected in parallel Where the equivalent capacitance is given by the equation.

$C_{eq}=C_1+C_2=9.0\mathrm{μ}F+4.0\mathrm{μ}F=13.0\mathrm{μ}F$ (1)

Now let us get storage energy for the system by using the equation(given below) in the form

$U_1=\frac{1}{2}{C}_{eq}{V}^2$ (2)

Now we can plug our values forand V into equation (1) to get U

$$\begin{gathered} U_1=\frac{1}{2}{C}_{eq}{V}^2 \\ {U}_1=\frac{1}{2}\left(13×{10}^{-6}\right){\left(64.0V\right)}^2 \\ {U}_1=26.3×{10}^{-3}J \end{gathered}$$

After connecting the two capacitors the opposite to each other the total charge of the system will be

$Q=Q_1-Q_2$

Where the charge of each capacitor depends on the capacitance of each capacitor so the total charge Q after connecting in the opposite will be

role="math" localid="1664263799558" $\begin{array}{l}Q=Q_1-Q_2\\ Q=C_{1}V-C_{2}V\\ Q=\left(C_1-C_2\right)V\end{array}$

The energy storage after connecting in the opposite direction will be given by equation (2) in the form

$\begin{array}{l}{U}_2=\frac{Q^2}{2C_{eq}}\\ U_2=\frac{{\left[\left(C_1-C_2\right)V\right]}^2}{2C_{eq}}\end{array}$ (3)

Now we can plug our values for $C_1,C_2,C_{eq}$ and V into equation 3 to get

$$\begin{gathered} U_2=\frac{{\left[\left(C_1-C_2\right)V\right]}^2}{2C_{eq}} \\ {U}_2=\frac{{\left[\left(9.0×{10}^{-6}F-4.0×{10}^{-6}F\right)64.0V\right]}^2}{2\left(13×{10}^{-6}\right)} \\ {U}_2=3.9×{10}^{-3}J \end{gathered}$$

The change in the energy system will be

$∆U=U_2-U_1=3.9×{10}^{-3}J-26.3×{10}^{-3}J=-22.7×{10}^{-3}J$