# Gaussian sphere problem

1. Feb 28, 2013

### skate_nerd

1. The problem statement, all variables and given/known data

The problem states that you have a spherical shell with inner radius Ri=1 cm and outer radius R0=2 cm. The shell also has uniform charge density of ρ=10-3 N/m3. I found the first few answers of the question already. First was to get the charge of the shell, which is simply ρVshell, or Q=ρ(4∏/3)(R03-Ri3). This ends up being 2.93(10-8)C.
Next I found that the electric field magnitude everywhere inside the shell could be expressed in terms of r by just setting EA=ρ(4∏/3)(r3-Ri3)/ε0, where the 4∏ cancels. the answer ends up being ((1/3)ρ/ε0)(r-(Ri3/r2), or 3.76(107) N/Cm.
Now the part I'm having trouble with is finding an equation in terms of r for the electric field magnitude outside of the spherical shell.

2. Relevant equations

Gauss's Law. Surface area and volume of a sphere.

3. The attempt at a solution

I tried using Gauss's law in a similar way to the last part of the problem by having EA=Q/ε0, but I don't understand how you are supposed to solve for this if you don't know what the uniform charge density is outside of the shell. I don't feel like using that same value would make sense. Could somebody explain this to me?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 28, 2013

### haruspex

How did you get that? There should not be a field inside an empty uniformly charged spherical shell.

3. Mar 1, 2013

### skate_nerd

I said that is the E field inside the shell, as in like in the volume of the shell. Thats why I used the gaussian radius as arbitrary r minus the inner radius of the sphere.

4. Mar 1, 2013

### haruspex

OK - that's often a tricky distinction to make verbally.
The field outside a uniformly charged spherical shell happens to be exactly as though all the charge were concentrated at the sphere's centre.
Can you explain that some more? What same value?