# Electric field of a non-conducting sphere

## 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=Qencl / ε0
Charge density = Q/V
Volume of a sphere = 4/3ϖR3

## The Attempt at a Solution

For R/4: ∫EdA=Qencl / ε0 = 1/(4ϖε0)(Q/R3) (R/4) = E0
for 2R: ∫EdA=Qencl / 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?

Related Introductory Physics Homework Help News on Phys.org
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}##.

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.

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