Finding the Electric Field Outside of a Spherical Shell Using Gauss's Law

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The discussion revolves around calculating the electric field outside a uniformly charged spherical shell using Gauss's Law. The spherical shell has an inner radius of 1 cm and an outer radius of 2 cm, with a uniform charge density of 10^-3 N/m³, leading to a total charge of approximately 2.93 x 10^-8 C. The electric field inside the shell is derived using the formula EA = ρ(4π/3)(r³ - Ri³)/ε₀, resulting in a field magnitude of 3.76 x 10^7 N/Cm. The main challenge discussed is finding the electric field outside the shell, with clarification that the field behaves as if all charge were concentrated at the center of the shell. Understanding the distinction between the electric field inside the shell and the field outside is crucial for solving the problem correctly.
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Homework Statement



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.

Homework Equations



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

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?
 
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skate_nerd said:
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.
How did you get that? There should not be a field inside an empty uniformly charged spherical shell.
 
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.
 
skate_nerd said:
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.
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.
I don't feel like using that same value would make sense.
Can you explain that some more? What same value?
 
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