Electric field of a non-conducting sphere

[TishBass]

Homework Statement

A solid non-conducting sphere of radius R carries a uniform charge density. At a radial distance r₁= R/4 the electric field has a magnitude Eo. What is the magnitude of the electric field at a radial distance r₂=2R?

Homework Equations

Gauss's Law: ∫EdA=Q_encl / ε₀
Charge density = Q/V
Volume of a sphere = 4/3ϖR³

The Attempt at a Solution

For R/4: ∫EdA=Q_encl / ε₀ = 1/(4ϖε₀)(Q/R³) (R/4) = E₀
for 2R: ∫EdA=Q_encl / ε₀ = 1/(4ϖε₀)(Q/ [2R]²)

So for R/4, it simplifies to 1/(4ϖε₀)(Q/4R²). I know the answer is E₀ so i guess 2R simplifies to 1/(4ϖε₀)(Q/ 4R²) and everything cancels to leave E₀.

However, I feel like I'm lost in the math at this point. If I'm right, could someone explain the math and if I'm wrong could someone just point me in the right direction?

 

[Dewgale]

You've got the wrong answer. Here's a hint to point you in the right direction: The charge density is the same at every point. However, the enclosed charge is not the same over every enclosed surface, so you may want to write your charge in terms of this invariant charge density, you can call it $\rho$. For the case you know, try solving for rho. Then, solve for the case where you don't know E -- bearing in mind the definition of $dA$, and $Q_{encl}$.

 

[TishBass]

So, solving for Q in terms of ρ for the first case, I get Q = ρ3ϖR³
I plug this into the case I don't know, 2R, and I end up with E = (ρ3ϖR³)/(8ϖR² ε₀)
I know I'm making an easy mistake but I can't pin down what it is.

 

[Dewgale]

How much charge is enclosed at 2R vs 1R?