Σ free on two dielectric spherical surfaces

AI Thread Summary
The discussion focuses on calculating the electric field and dipole moment for a system involving two dielectric spherical surfaces. The user has determined the total dipole moment but is uncertain about the electric field, particularly for regions inside the sphere (r<R) and outside (r>2R). They believe their expression for the electric field in the region r>2R is correct but are questioning their findings for r<R due to inconsistencies with previous problems. There is confusion regarding the surface charge densities, specifically σ1 and σ2, with σ1 being interpreted as 4σcosθ and uncertainty surrounding σ2. The user seeks clarification on the analytic expressions for the surface charge densities to resolve their doubts.
guyvsdcsniper
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Homework Statement
σ 1 and σ 2 are pasted to 2 spherical dielectric surfaces w/ radius r and 2R

2.)Calc. E when r<R and r>2R
3.)Calc E when R<r<2R
Relevant Equations
Dipole Potential
I have found the total dipole moment of for this problem but am having trouble finding the electric field.

I believe my electric field when r>2R ( I mistakenly wrote it as r<2R on my work, but it is the E with a coefficient of 2/3) is correct as it fits the equation:
Screen Shot 2022-04-18 at 10.06.02 PM.png
.
I don't believe this formula applies inside the sphere though, just based off experience with other problems because with other problems, I don't get that 2cosθ +sinθ. Which is making me second guess my E for r<R. Mathematically it seems correct but I feel I may be missing something fundamental.

Do my E.F. for r<R and r>2R seem correct?
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Screen Shot 2022-04-18 at 10.02.08 PM.png
 
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Please show the complete statement of the problem as given to you. Specifically, do you have analytic expressions for ##\sigma_1## and ##\sigma_2##? The figure shows σ1=4σcosθ on the inner sphere which I can interpret as ##\sigma_1=4\sigma \cos\!\theta## but then I see on the outer sphere σ2=σdos0 which I don't know how to interpret and which reminds of covfefe.
 
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