Calculating the Electric Field inside an insulating shell

1. Feb 14, 2015

ryankunzzz

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

2. Relevant equations

Gauss's Law

∫EdS=Qin0

and Q=ρV

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

3. 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 Guass'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.

Last edited: Feb 14, 2015
2. Feb 14, 2015

Staff: Mentor

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)?

3. Feb 14, 2015

ryankunzzz

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?

4. Feb 14, 2015

Staff: Mentor

When you're within the shell, effectively you have a shell exterior to your position, and another interior to your position.