Answer:
E = 2k [tex]\frac{\lambda}{ r}[/tex]
Explanation:
Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.
We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.
Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀
tells us that the linear charge density is
λ = q_ {int} /l
q_ {int} = l λ
we substitute
E A = l λ /ε₀
is area of cylinder is
A = 2π r l
we substitute
E = [tex]\frac{ l \ \lambda}{ \epsilon_o \ 2\pi \ r \ l }[/tex]
E = [tex]\frac{\lambda}{ 2\pi \epsilon_o \ r}[/tex]
the amount
k = 1 / 4πε₀
E = 2k [tex]\frac{\lambda}{ r}[/tex]
