# A charged capacitor is connected to another capacitor with identical capacitance. Where is energy stored at equilibrium?

Mar 4, 2018

Assuming that the charged capacitor having charge ${Q}_{i}$ and capacitance $C$ is connected to an identical uncharged capacitor in parallel.

We know that in the initial condition

${Q}_{i} / C = {V}_{i}$ ...........(1)
Where ${V}_{i}$ is the initial voltage across the capacitor.

Also Energy stored on the capacitor

${E}_{i} = \frac{1}{2} {Q}_{i}^{2} / C$ .........(2)

After the capacitors are connected, the charged capacitor acts a source of voltage and charges the other capacitor. In the equilibrium condition, due to the Law of Conservation of Charge half of the charge moves to other capacitor. From (2) we get energy stored on one capacitor as

${E}_{f} = \frac{1}{2} {\left({Q}_{i} / 2\right)}^{2} / C = \frac{1}{8} {Q}_{i}^{2} / C$

Similarly energy stored on the other capacitor is also same. Total energy stored on both capacitors

${E}_{f \setminus \text{total}} = 2 \left(\frac{1}{8} {Q}_{i}^{2} / C\right) = \frac{1}{4} {Q}_{i}^{2} / C$

The question is where has the remaining half of original energy gone.

And the answer is: this is expanded in charging the uncharged capacitor through $R C$ time constant of the circuit. With the flow of charging current in the circuit, this half energy is lost as heat across the resistance $R$.