Question

A spherical shell has inner radius Rin and outer radius Rout. The shell contains total charge Q, uniformly distributed. The interior of the shell is empty of charge and matter. Find the electric field strength within the shell, Rin≤r≤Rout.

Answer 1

Answer:

  E = Q / 4 π ε₀  (r-R_int) / (R_put³ -R_int³)

Explanation:

For this exercise we can use Gauss's law

          Ф = ∫ E. dA = $q_{int}$ / ε₀

Where q is the charge inside the surface.

In this case the surface must be a sphere, the electric field lines and the radius of the sphere are parallel, so the scalar product is reduced to the algebraic product

           ∫ E dA = q_{int}/ ε₀

The area of ​​a sphere is

           A = 4π r²

- The electric field for a distance r < R_int

The charge inside is zero, so the electric field

          E = 0        r <R_in t

- The field for a radio inside the shell

   Let's use the concept of density

         ρ = Q / V

         q = ρ (4/3 π r³)

         dq = ρ 4π r² dr

We substitute in the Gaussian equation

     E ∫ dA = ρ 4π r² dr / ε₀

    E 4π r² = ρ 4π/ε₀   r³ / 3

     E = ρ / 3ε₀ r

We evaluate between the lower limit r = R_int, E = 0 and the upper limit r = r, E = E

      E- 0 = ρ / 3ε₀ (r –R_int)

Density  is

      ρ = q / 4/3 π (R_out³ - R_int³)

Where R <r

     E = Q / 4 π ε₀  (r-R_int) / (R_put³ -R_int³)