Final answer:
The question concerns the electric field between two oppositely charged metal plates and the resultant force on a small positive charge. The electric field in this scenario is uniform and creates a force that would accelerate a positive charge from the positively charged plate to the negatively charged one. The strength of the electric field can be calculated based on the surface charge density and the separation between the plates.
Step-by-step explanation:
The student's question relates to the electric field created between two oppositely charged metal plates and the behavior of a small positive charge when placed within this field. When the two metal plates are connected to opposite terminals of a battery, an electric field is created between them. This field is highly uniform except near the edges of the plates. The behavior of a positive charge in such a field would be to experience a force that accelerates it from the positive plate to the negative plate due to the attractive and repulsive forces exerted by the plates.
The electric field is represented by the equation E = σ/ε0, where σ is the surface charge density and ε0 is the permittivity of free space. The potential difference (potential difference) between the plates can be calculated using the relationship V = Ed, where V is the potential difference, E is the electric field strength, and d is the separation between the plates. If a positive charge were released near the positive plate, it would move towards the negative plate due to the uniform electric field.