Calculating the Electric Field inside an insulating shell

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To calculate the electric field inside a spherical insulating shell with an inner radius of 10 cm, outer radius of 20 cm, and a charge density of 80 µC/m³, the total charge in the shell is determined to be 2.3 µC. Using Gauss's Law, the electric field inside the shell is expressed as E = (9.0 x 10^9 * 8.0 x 10^-6) / r², resulting in E = 7.2 x 10^4 / r². The discussion highlights the behavior of electric fields in spherical shells, noting that within the shell, the electric field is influenced by both the inner and outer charges. The key point is that while the formula for the electric field inside a solid sphere applies, the presence of the shell alters the calculations. Understanding these principles is crucial for solving the problem effectively.
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Homework Statement


We are asked to calculate the electric field inside of the a spherical insulating shell with an inner radius of 10cm and an outer radius of 20 cm and a charge density of 80 uC/M^3. Additionally, a +8uC charge is added to the center of the shell.

Homework Equations



Gauss's Law

∫EdS=Qin0

and Q=ρV

And the volume of a shell is V=4/3π(R3-r3)[/B]

The Attempt at a Solution



So first I want to find the amount of charge in the shell,

Q=(80 uC/M3)*(4/3π(.23-.13) = 2.3 uC

Now I need to use Gauss's Law.

I found the E field inside the shell up to the surface to be

E=(9.0*109 * 8.0*10-6)/r2

Which is 7.2*104/r2

So now my Efield in the shell is going to be that added to whatever the Efield at the point in the shell we are at is.

so the shell should be ∫Eds=(2.3*10^-6)/ε0 but this is where I get stuck.[/B]
 
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Draw your Gaussian surface and determine how much charge is inside. What do you know about how spherical shells of charge behave for an observer outside the shell (interior or exterior)?
 
Well outside the shell we know that we can treat it like a point charge, but we are withing the material of shell.

I know what the formula for an E field inside of an solid sphereical insulator is, but does it change because now the insulator is a shell?
 
When you're within the shell, effectively you have a shell exterior to your position, and another interior to your position.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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