Answer:
same value in R and 2R E = E₀ = σ / 2ε₀
Explanation:
For this exercise we use Gauss's law
Ф = E. dA = [tex]q_{int}[/tex] /ε₀
We define a Gaussian surface with a cylinder with the base being parallel to the load sheet, so the electic field line and the normal line to the base are parallel and the scalar product is reduced to the algebraic product, in the parts the angle is 90º and the dot product is zero
As the sheet has two faces
2E A = q_{int} /ε₀
The charge inside the cylinder is
σ = q_{int} / A
q_{int} = σ A
We substitute
E = σ / 2ε₀
We see that this expression is independent of the distance, so it has the same value in R and 2R
E = E₀ = σ / 2ε₀