Does the Net Charge of a Dielectric Affect the External Electric Field?

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SUMMARY

The discussion centers on the impact of a dielectric's net charge on the external electric field surrounding a uniformly charged sphere with radius R, enveloped by a dielectric with a polarizability of kz in the z direction, extending to radius A. It concludes that the external electric field is determined solely by the free charge on the sphere, rendering the dielectric's net charge ineffective in altering the external field. The analysis indicates that while the total bound surface charge is zero, the volume bound charge is -4/3(A^3-R^3)*pi*k, and the external field remains non-zero due to the monopolar nature of the central sphere and the dipolar nature of the dielectric coating.

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Let's say we have a sphere, radius R, which has a uniform volume charge density. Then we wrap it around with a dielectric with a frozen in polarizability of kz in the z direction. This dielectric goes from R to a radius A. The total surface bound charge on either surface of the dielectric is zero; however, the volume bound charge total is -4/3(A^3-R^3)*pi*k. Now I would think that this net charge would change the E field outside the dielectric. Looking at it through the electric displacement vector shows that in a region outside the dielectric, the E field would be determined only by the free charge on the sphere. Which would say that the charge of the dielectric has no effect. Anybody who would know which of these scenarios is actually happening and why?
 
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Although the total bound surface charge (bsc) is zero, the bsc varies with angle.
Therefor, you cannot use Gauss's law.
The problem requires a Legendre polynomial expansion.
 
The external field will be non-zero, because the central sphere is a monopolar source and the coating is essentially a dipolar source.

As a first attempt at the answer I would add the fields from: monopole sphere (radius R) + dipole sphere (radius A) - dipole sphere (radius R)

I would guess given the geometry that this is actually correct but this would need to be proven properly.
 

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