Answer:
To determine the magnitude of the electric field at a point with z = 1.0 m, we can use the principle of superposition. The electric field due to each charge distribution will be calculated separately and then added together.
Let's start with the charge distribution on the xy plane with a uniform density of 8.0 nC/m^2. The electric field due to this charge distribution can be calculated using the formula:
E1 = σ / (2ε₀)
Where E1 is the electric field, σ is the charge density, and ε₀ is the permittivity of free space.
Substituting the given values:
E1 = (8.0 nC/m^2) / (2ε₀)
Next, let's consider the charge distribution on the parallel plane defined by z = 2.0 m with a uniform density of 5.0 nC/m^2. The electric field due to this charge distribution can also be calculated using the same formula:
E2 = σ / (2ε₀)
Substituting the given values:
E2 = (5.0 nC/m^2) / (2ε₀)
Now, we need to add the electric fields due to both charge distributions to find the total electric field at the point with z = 1.0 m. Since the electric fields are vectors, we need to consider their magnitudes and directions.
Since the charge distributions are parallel to the xy plane, the electric fields will be perpendicular to the xy plane and will only have a z-component. Therefore, we can simply add the magnitudes of the electric fields.
E_total = |E1| + |E2|
Substituting the calculated values:
E_total = |(8.0 nC/m^2) / (2ε₀)| + |(5.0 nC/m^2) / (2ε₀)|
Now, we need to convert the charge densities from nC/m^2 to C/m^2 by multiplying by 10^(-9):
E_total = |(8.0 × 10^(-9) C/m^2) / (2ε₀)| + |(5.0 × 10^(-9) C/m^2) / (2ε₀)|
Finally, we can simplify the expression by substituting the value of ε₀, which is approximately 8.854 × 10^(-12) C^2/(N·m^2):
E_total = |(8.0 × 10^(-9) C/m^2) / (2 × 8.854 × 10^(-12) C^2/(N·m^2))| + |(5.0 × 10^(-9) C/m^2) / (2 × 8.854 × 10^(-12) C^2/(N·m^2))|
Evaluating this expression will give us the magnitude of the electric field at the point with z = 1.0 m.