E field outside of two non conductive shells?

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SUMMARY

The discussion focuses on calculating the electric fields in a system of two non-conductive shells with radii r1 and r2, and charges q1 and q2. According to Gauss's Law, the electric field inside the inner shell (E(r < r1)) is zero. For the region between the shells (E(r1 < r < r2)), the electric field is given by E = q1 / (4 * π * ε₀ * r²). The confusion arises regarding the electric field outside the outer shell (E(r > r2)), where the total enclosed charge is indeed the sum of both charges (q1 + q2), regardless of the shells being conductive or non-conductive.

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Homework Statement



Two non conductive shells surround each other, with radius r1, r2 and charges q1, q2 respectively.
Find electric fields for E(r < r1), E(r1 < r < r2), E(r > r2)

r1 is the inner shell, r2 is the outer shell (r1 < r2)

Homework Equations



I'm using gauss' law

flux = E * A = q_enclosed / epsilon_not

The Attempt at a Solution



E(r < r1) = 0
E(r1 < r < r2) = q1 / 4 * pi * episolon_not * r^2

I'm confused on E(r > r2), what is the charged enclosed for two surrounding non conductive shells?
Would I add up the charge from both (q1 + q2)? or would it only be the charge on the outer shell q2?

I don't know what is significant about the shells being "non conductive" vs "conductive" what is the difference?

please respond.
 
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q1 + q2

No difference with conductive shells.
 

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