Physics - Capacitor Question with Solution | TestHub

When ${\mathrm{C}}_1$ is connected to voltage source, it is charged to a potential $V$ and this will be stored as a potential energy in the capacitor given by $$\mathrm{U}=\frac{1}{2}{\mathrm{CV}}^2$$
When key is disconnected from battery and band c are connected, the charge will be transformed from the capacitor ${\mathrm{C}}_1$ to capacitor ${\mathrm{C}}_2$, then
The loss of energy due to redistribution of charge is given by $$\begin{array}{rl}& \begin{array}{rl}\Delta \mathrm{U}& =\frac{{\mathrm{C}}_1{\mathrm{C}}_2}{2\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)}{\left({\mathrm{V}}_1-{\mathrm{V}}_2\right)}^2\\ & =\frac{\mathrm{C}\times \mathrm{C}}{2(\mathrm{C}+\mathrm{C})}(\mathrm{V}-0{)}^2=\frac{1}{4}{\mathrm{CV}}^2\left[\because {\mathrm{C}}_1={\mathrm{C}}_2\right]\end{array}\\ & \therefore \text{ Percentage loss }=\frac{\Delta \mathrm{U}}{\mathrm{U}}\times 100=\frac{\frac{1}{4}{\mathrm{CV}}^2}{\frac{1}{2}{\mathrm{CV}}^2}\times 100=50\%\end{array}$$