Final answer:
The magnitude of the net electric field at point y₂ = 0.180 m on the y-axis is 0.417 N/C. The electric field produced by the positive line charge is 0.225 N/C, and the electric field produced by the negative line charge is -0.192 N/C. When we subtract the magnitudes of these two electric fields, we get the net electric field.
Step-by-step explanation:
To find the magnitude of the net electric field at point y₂ = 0.180 m on the y-axis, we need to consider the electric fields produced by both the positive and negative line charges. The electric field produced by a line charge is given by the equation E = kλ/r, where E is the electric field, k is the Coulomb constant (9.0 x 10^9 N m^2/C^2), λ is the charge per unit length, and r is the distance from the line charge.
First, let's calculate the electric field produced by the positive line charge at y₂ = 0.180 m. The distance from the positive line charge to point y₂ is the perpendicular distance from the y-axis, which is also equal to y₂. Using the given charge per unit length of 4.50 μC/m, the electric field produced by the positive line charge is:
E1 = kλ/r = (9.0 x 10^9 N m^2/C^2) x (4.50 x 10^-6 C/m) / 0.180 m = 0.225 N/C
Next, let's calculate the electric field produced by the negative line charge at y₂ = 0.180 m. Since the negative line charge is parallel to the x-axis, the distance from the negative line charge to point y₂ is the y-coordinate of the negative line charge, which is given as y₁ = 0.382 m. Using the given charge per unit length of -2.56 μC/m, the electric field produced by the negative line charge is:
E2 = kλ/r = (9.0 x 10^9 N m^2/C^2) x (-2.56 x 10^-6 C/m) / 0.382 m = -0.192 N/C
To calculate the net electric field, we take the vector sum of E1 and E2. Since the electric fields produced by the positive and negative line charges have opposite directions, we subtract the magnitudes:
|Enet| = |E1| - |E2| = 0.225 N/C - |-0.192 N/C| = 0.417 N/C
Therefore, the magnitude of the net electric field at point y₂ = 0.180 m on the y-axis is 0.417 N/C.