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

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Homework Help Overview

The problem involves finding the electric field outside a uniformly charged spherical shell with specified inner and outer radii and charge density. Participants are discussing the application of Gauss's Law in this context.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply Gauss's Law to find the electric field outside the shell but expresses confusion regarding the charge density outside the shell. Other participants question the validity of the electric field expression provided for inside the shell and clarify that there should be no electric field within a uniformly charged shell.

Discussion Status

The discussion is ongoing, with participants exploring the implications of Gauss's Law and clarifying concepts related to electric fields in and around spherical shells. Some guidance has been offered regarding the nature of the electric field outside the shell, but there is no explicit consensus on the original poster's approach.

Contextual Notes

Participants are navigating the complexities of applying Gauss's Law and the assumptions about charge distribution, particularly in relation to the electric field inside and outside the shell. There is a noted distinction between the field in the volume of the shell and the field outside it.

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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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