Asolid uniformly charged insulating sphere has uniform volume charge density p and radius r. apply gauss's law to determine an expression for the magnitude of the electric field at an arbitrary distance r from the center of the sphere, such that r < r, in terms of rho and r

electric field E = (1 /3 e₀) ρ r

Explanation:

For the application of the law of Gauss we must build a surface with a simple symmetry, in this case we build a spherical surface within the charged sphere and analyze the amount of charge by this surface.

The charge within our surface is

     ρ = Q / V

     Q ’= ρ V '

The volume of the sphere is V = 4/3 π r³

     Q ’= ρ 4/3 π r³

The symmetry of the sphere gives us which field is perpendicular to the surface, so the integral is reduced to the value of the electric field by the area

      I E da = Q ’/ ε₀

      E A = E 4 πi r² = Q ’/ ε₀

      E = (1/4 π ε₀) Q ’/ r²

Now you relate the fraction of load Q ’with the total load, for this we use that the density is constant

      R = Q ’/ V’ = Q / V

How you want the solution depending on the density (ρ) and the inner radius  (r)

      Q ’= R V’

      Q ’= ρ 4/3 π r³

      E = (1 /4π ε₀) (1 /r²) ρ 4/3 π r³

     E = (1 /3 e₀) ρ r

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