Final answer:
The electric field on the y-axis at y = 0.50 m is determined by the vector sum of fields due to the 2.0-µC charges, and the force on a -3.0 µC charge at this location can be calculated using the net electric field multiplied by the charge.
Step-by-step explanation:
Electric Field and Force on a Charge
To determine the electric field on the y-axis at y = 0.50 m due to two 2.0-µC charges located on the x-axis at x = 1.0 m and x = -1.0 m, we can use the principle of superposition. The electric fields from each charge at the given point on the y-axis will be directed along the line joining the charge and the point. These individual fields can be calculated and their vector sum will give the net electric field at y = 0.50 m.
The electric force on a -3.0 µC charge placed at this point on the y-axis can then be found using Coulomb's law, which states that the force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. The net force is the vector sum of the forces exerted by each 2.0 µC charge on the -3.0 µC charge.
To answer part (a), we calculate the electric field from each 2.0 µC charge at the point (0, 0.50 m) and then add them vectorially. For part (b), we use the net electric field calculated in part (a) and multiply it by the charge of -3.0 µC to obtain the force on this charge.