D

To analyze the behavior of the charge +q thrown towards the charged disc, we start with the following detailed steps:
1. **Understanding the system**: The disc has a uniform surface charge density, leading to an electric field (E) directed outward from the disc. This electric field will affect the charge +q as it approaches.
2. **Kinetic energy conversion**: The charge +q is thrown with a specific kinetic energy, given by KE = \frac{1}{2} mv^2. As +q approaches the disc, this kinetic energy will convert into potential energy due to the electric field generated by the disc.
3. **Electric field of the disc**: The electric field due to a uniformly charged infinite plane sheet at a distance 'z' from the sheet is given by E = \frac{\sigma}{2\epsilon_0}, where \sigma is the surface charge density and \epsilon_0 is the permittivity of free space. For a finite disc, the electric field changes with distance, especially near the center.
4. **Potential energy and forces**: As +q approaches the disc, the electric field exerts a force on it, which can either be attractive or repulsive (since +q interacts with like charges). If the kinetic energy of +q is sufficient to overcome the potential due to the electric field, the charge will continue to approach the disc. If not, it could return.
5. **Possible outcomes**: The behavior of charge +q will depend heavily on the interplay of its initial kinetic energy, the magnitude of the electric field at that distance, and the charge distribution on the disc.
6. **Conclusion**: Therefore, depending on the initial conditions (like the speed and distance of approach), +q could either hit the center of the disc, return after touching, or just barely miss and return without touching; hence, the final conclusion is that any of these situations is possible.
Therefore, the correct answer is D: Any of the above three situations is possible depending on the magnitude of E.