6. (15 points) A capacitor with capacitance $C_{1}=10$ μF is charged to a voltage $U_{1}=15$ V. A second capacitor with capacitance $C_{2}=5$ μF is charged to a voltage $U_{2}=10$ V. The capacitors are connected with their oppositely charged plates. Determine the voltage that will establish across the plates.
Answer: 6.67 V.

Solution. The law of charge conservation $C_{1} U_{1}-C_{2} U_{2}=C_{1} U+C_{2} U$. We obtain:

$$
U=\frac{c_{1} U_{1}-C_{2} U_{2}}{C_{1}+C_{2}}=\frac{10 \cdot 15-5 \cdot 10}{10+5}=6.67 \mathrm{~V} .
$$

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