Calculating the Electric Field inside an insulating shell

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Homework Help Overview

The discussion revolves around calculating the electric field inside a spherical insulating shell with specified dimensions and charge density, while also considering a point charge at the center. The subject area includes electrostatics and Gauss's Law.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate the charge within the shell and apply Gauss's Law, but expresses uncertainty in the next steps. Some participants question the behavior of the electric field within the insulating shell compared to a solid sphere.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem, including the implications of the shell's geometry on the electric field and the application of Gauss's Law. There is no explicit consensus yet, but guidance on drawing Gaussian surfaces and considering charge distribution has been provided.

Contextual Notes

Participants are navigating the complexities of the problem, including the effects of the insulating shell and the central charge on the electric field, while adhering to the constraints of the homework context.

ryankunzzz
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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]
 
Last edited:
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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.
 

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