What is the magnitude and direction of the electric field atradiaConsider a coaxial conducting cable consisting of a conductingrod of radius R1 inside of a thin-walled conducting shell of radius 2(both are infinite length). Suppose the inner rod hasradiusR1= 1.3 mm and outer shell has radiusR2= 10R1Ifthe net charge density on the center rod isq1= 3.4×10−12C/mand the outer shell isq2=−2q1,a.)What is the magnitude and direction of the electric field atradial distancer= 5R1from the center rod

 E = 9.4 10⁶ N / C ,     The field goes from the inner cylinder to the outside

Explanation:

The best way to work this problem is with Gauss's law

             Ф = E. dA = qint / ε₀

We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.

The flow on the faces is zero, since the field goes in the radial direction of the cylinders.

The area of the cylinder is the length of the circle along the length of the cable

         dA = 2π dr L

          A = 2π r L

They indicate that the distance at which we must calculate the field is

         r = 5 R₁

         r = 5 1.3

         r = 6.5 mm

The radius of the outer shell is

         r₂ = 10 R₁

         r₂ = 10 1.3

         r₂ = 13 mm

         r₂ > r

When comparing these two values we see that the field must be calculated between the two housings.

Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is

         λ = q / L

         Qint = λ L

Let's replace

      E 2π r L = λ L /ε₀

       E = 1 / 2piε₀  λ / r

Let's calculate

         E = 1 / 2pi 8.85 10⁻¹²  3.4 10-12 / 6.5 10-3

         E = 9.4 10⁶ N / C

The field goes from the inner cylinder to the outside

the components of a vector are the parts of the vectors that are to each other.

b perpendicular

snell's law

refractve index = sin incidence / sin refraction