5. (6 points) Two identical conducting spheres are located at a large distance from each other and have positive charges $Q_{1}$ and $Q_{2}$. A neutral metal ball on a non-conductive suspension is brought close to the first sphere and touches it. Then the ball is brought close to the second sphere and touches it. After these manipulations, the charge of the ball is found to be $q_{2}$. Find the charge of the second sphere after all manipulations.

Possible solution. Let the capacitances of the spheres be $C_{1}$, and the capacitance of the ball be $C$. After the ball touches the first sphere, its charge will become

$$
q_{1}=\frac{Q_{1}}{1+\frac{C_{1}}{C}}
$$

From the law of conservation of charge for the contact of the ball with the second sphere, we get:

$$
q_{1}+Q_{2}=q_{2}+Q_{2}^{\prime}
$$

The potentials of the second sphere and the ball at the moment of contact are equal, so

$$
C q_{2}=C_{1} Q_{2}^{\prime}
$$

Then we can write:

$$
q_{1}=\frac{Q_{1}}{1+\frac{Q_{2}^{\prime}}{q_{2}}}
$$

We obtain the equation:

$$
Q_{2}^{\prime 2}-\left(Q_{2}-2 q_{2}\right) Q_{2}^{\prime}-q_{2}\left(Q_{1}+Q_{2}-q_{2}\right)=0
$$
Solution of this equation:

$$
Q_{2}^{\prime}=\frac{Q_{2}}{2}-q_{2} \pm \sqrt{\frac{Q_{2}^{2}}{4}+Q_{1} q_{2}}
$$

Answer: $Q_{2}^{\prime}=\frac{Q_{2}}{2}-q_{2} \pm \sqrt{\frac{Q_{2}^{2}}{4}+Q_{1} q_{2}}$.

## Evaluation Criteria

| Performance | Score |
| :--- | :---: |
| The participant did not start the task or performed it incorrectly from the beginning | $\mathbf{0}$ |
| Expression for the charge $q_{1}$ is written | $\mathbf{1}$ |
| Expression for the law of charge conservation is written | $\mathbf{1}$ |
| Expression for the equality of potentials upon contact of the sphere and the ball is written | $\mathbf{1}$ |
| Necessary transformations are performed, the final answer is obtained | $\mathbf{3}$ |
| Total points | $\mathbf{6}$ |