Answer:
E = ρ r / 2ε₀ r>R
Explanation:
To find the field we can use Gauss's law that says that the flow is equal to the charge of the surface divided by electric permittivity (eo)
Φ = ∫ E .dA = [tex]q_{int}[/tex] / ε₀
To use this law we must define a Gaussian surface that takes advantage of the symmetry of the problem, we use a cylindrical surface parallel to the charge cylinder
The charge density for r> R, the entire charge is internal
ρ = [tex]q_{int}[/tex] / V
V = (π r²) L
[tex]q_{int}[/tex] = ρ (π r² L)
Field lines are protruding to the cylinder and therefore the scalar product is reduced to the ordinary product
∫ E dA = [tex]q_{int}[/tex] / ε₀
E (2π r L) = ρ (π r² L) / ε₀
E = ρ r / 2ε₀ r>R