Answer:
Ea = 3.2 10⁵ N / C
Explanation:
To calculate the electric field of each plane we will use Gauss's law, we create a Gaussian surface that is a cylinder that has the axis perpendicular to the plane, in this case the flow line between the cylinder walls and the surface is zero and all the flow is perpendicular to the base of the cylinder.
We apply the law of gauss flow to each side is the value of the electric field (E) for the area of the cylinder (A); whereby the flow in the two directions is 2 E A
Φ = 2E A = ρₙt / εo
Where ρₙ is the charge inside the cylinder, as the charge density gives us, σ = Q / A
ρₙ = σ A
By which we can clear the electric field
E = σ A / 2εo
where it is worth 8.85 10⁻¹² C² / N² m². Let's calculate with this equation in the field for each plane
1 plane σ= -2.0 pC / m²
E1 = -2. 10⁻¹² / 2 8.85 10⁻¹²
E1 = -0.113 N / C
The field line is directed to the plane
2 Plane σ = 5.8 mC / m2
E2 = 5.8 10-6 / 2 8.85 10-12
E2 = 3,277 10 5 N / A
Field lines leave the plane
As we have the values of each field in the whole space. Let's calculate in the field at the point x = 1 m
To do this we must add the fields at the selected point vectorally, for this distance the point is between the two planes, so the field of plane 1 points to the left and the point of plane 2 also points to the left, consequently field adds
Ea = E1 + E2
Ea = 0.113 + 3.277 10⁵
Ea = 3.2 10⁵ N / C
At point X = -1 m in this case the point is on plane 1, so this plane does not generate any field, it is an equipotential surface, the total field is equal to field 2
Ea = E1 = 3.2 10⁵ N / C
Note that there is no difference in numbers values by the difference between the load between each plane
