Answer:
a) E = σ / 2 ε₀ = Q / 2A ε₀, b) E = 2Q/A ε₀
Explanation:
For this exercise we can use Gauss's Law
Ф = E. dA = / ε₀
Let us define a Gaussian surface as a cylinder with the base parallel to the plane. In this case, the walls of the cylinder and the charged plate have 90 degrees whereby the scalar product is zero, the normal vector at the base of the cylinder and the plate has zero degrees whereby the product is reduced to the algebraic product
Φ = E dA = q_{int} / ε₀ (1)
As they indicate that the plate has an area A, we can use the concept of surface charge density
σ = Q / A
Q = σ A
The flow is to both sides of loaded plate
Φ = 2 E A
Let's replace in equation 1
2E A = σA / ε₀
E = σ / 2 ε₀ = Q / 2A ε₀
This is in the field at point P.
b) Now we have two plates each with a load Q and 3Q respectively and they ask for the field between them
The electric field is a vector quantity
E = E₁ + E₂
In the gap between the plates the two fields point in the same direction whereby they add
σ₁ = Q / A
E₁ = σ₁ / 2 ε₀
For the plate 2
σ₂ = -3Q / A = -3 σ₁
E₂ = σ₂ / 2 ε₀
E₂ = -3 σ₁ / 2 ε₀
The total field is
E = σ₁ / 2 ε₀ + 3 σ₁ / 2 ε₀
E = σ₁ / 2 ε₀ (1+ 3)
E = 2 σ₁ / ε₀
E = 2Q/A ε₀