5. (6 points) Two identical air capacitors with capacitance $C$ each are charged to a voltage $U$. One of them is submerged in a dielectric liquid with permittivity $\varepsilon$, after which the capacitors are connected in parallel. Determine the amount of heat released upon connecting the capacitors.

Possible solution. When the first capacitor is submerged in the liquid, its charge does not change, but its capacitance increases by a factor of $\varepsilon$. The voltage across the capacitor decreases by a factor of $\varepsilon$. When the capacitors are connected in parallel, the voltages across them become equal due to the transfer of charge from one capacitor to the other, while the total charge of the capacitors remains constant. The energy of the capacitor system before connection is

$$
W_{1}=\frac{C U^{2}}{2 \varepsilon}+\frac{C U^{2}}{2}=\frac{C U^{2}}{2} \cdot \frac{\varepsilon+1}{\varepsilon}
$$

The voltage across the parallel-connected capacitors is

$$
U^{\prime}=\frac{2 U}{\varepsilon+1}
$$

The energy of the parallel-connected capacitors is

$$
W_{2}=\frac{2 C U^{2} \varepsilon}{(\varepsilon+1)^{2}}+\frac{2 C U^{2}}{(\varepsilon+1)^{2}}=\frac{2 C U^{2}}{\varepsilon+1}
$$

The amount of heat released:

$$
Q=\Delta W=\frac{C U^{2}(\varepsilon-1)^{2}}{2 \varepsilon(\varepsilon+1)}
$$
Answer: $Q=\frac{C U^{2}(\varepsilon-1)^{2}}{2 \varepsilon(\varepsilon+1)}$.

## Evaluation Criteria

| Performance | Score |
| :--- | :---: |
| Participant did not start the task or performed it incorrectly from the beginning | $\mathbf{0}$ |
| Expression for the energy of the system before connection is written | $\mathbf{2}$ |
| Expression for the energy of the system after connection is written | $\mathbf{2}$ |
| Necessary transformations are performed, final answer is obtained | $\mathbf{2}$ |
| Total points | $\mathbf{6}$ |