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

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 = $q_{int}$ / 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