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

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

The energy of the connected parallel 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:

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

From this expression, we get

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

## Evaluation Criteria

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