Final answer:
The magnitude of the electric field is 3.00 × 10^2 N/C, and its direction is eastward. The force on a proton in this field would be westward. The correct answr is d) (3.20 X 10¹1 N/C) westward.
Step-by-step explanation:
The electric field is defined as the force per unit charge experienced by a test charge placed in the field.
The equation to calculate the magnitude of the electric field is given by:
E = F/q
where E is the electric field, F is the force exerted on the charge, and q is the charge.
In this case, the magnitude of the force on the electron is given as 4.80 × 10-17 N, and the charge of an electron is 1.60 × 10-19 C. Plugging these values into the equation, we get:
E = (4.80 × 10-17 N) / (1.60 × 10-19 C) = 3.00 × 102 N/C
Therefore, the magnitude of the electric field is 3.00 × 102 N/C.
Since the force is westward, the direction of the electric field will be in the opposite direction, which is eastward, option a.
For a proton, which has a positive charge, the force would be in the opposite direction to that experienced by the electron.
Therefore, the direction of the force on a proton would be westward.
Therefore, the correct answr is d) (3.20 X 10¹1 N/C) westward.