Question

A solid insulating cylinder of radius R has a positive uniform volume charge density rho. The cylinder’s axis is along the z-direction, and it can be considered infinitely long in this direction. a) Determine the electric field at a point outside the cylinder r > R, where r is the distance from the axis of the cylinder. Give both the magnitude and direction of the electric field.

Answer 1

The electric field outside of a uniformly charged infinite cylindrical conductor with radius R, at a distance r > R, is given by E = ρR^2 / (2ε_0r), directed radially outward from the axis of the cylinder.

Step-by-step explanation:

To determine the electric field outside a uniformly charged infinite cylindrical conductor, we can apply Gauss's law. Gauss's law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity of free space (ε0). For a point outside the cylinder, r > R, we can consider a cylindrical Gaussian surface co-axial with the charged cylinder. The electric field is perpendicular to the surface and has the same magnitude at every point on the Gaussian surface.

By Gauss's law, we integrate the electric field E over the curved surface area of the Gaussian surface. This gives us E(2πrl), where l is the length of the cylinder, which is equal to the enclosed charge (ρπR2l) over ε0. Solving for E:

E = ρR2 / (2ε0r)

The direction of the electric field is radially outward from the axis of the cylinder since the cylinder has a positive charge.

Answer 2

E = ρ r / 2ε₀ r>R

Step-by-step 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 =
`q_(int)` / 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 =
`q_(int)` / V

V = (π r²) L

`q_(int)` = ρ (π r² L)

Field lines are protruding to the cylinder and therefore the scalar product is reduced to the ordinary product

∫ E dA =
`q_(int)` / ε₀

E (2π r L) = ρ (π r² L) / ε₀

E = ρ r / 2ε₀ r>R