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While solving the electrostatic field for a dielectric in an electric field, we take the potential at origin to not blow and thus, we eliminate the inverse powers of r in the expression for general potential for azimuthal symmetry. Why is that valid :- we can have bound charge inside the sphere at r = 0 and which can contribute to infinite potential at the origin by the usual electrostatic laws. Does this depend on whether the sphere is homogeneous or inhomogeneous ? There are potentials which we do blow up in our models at the origin, is there any specific constraint in blowing up the potentials inside the materials when we are considering only limited set of ideal conditions and nothing pertaining to ionization or maximum electric field for stability of the system has been mentioned.

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The validity of setting the electrostatic potential to a certain value at a certain point comes from the fact that the electrostatic potential, like the potential, is defined up to a constant. The numerical value of this constant does not change the physics, but can sometimes make expressions nicer.

The potential should usually be finite everywhere, except in some limiting situations. The case of a point charge is one of such, because to have a point (of infinitesimal dimensions by definition) carrying a finite charge would require an infinite charge density, leading to an infinite energy and to an infinite electrostatic potential. You can avoid these pathologies defining the electrostatic potential by its value at infinity (instead that by its value at the origin where the charge is), and considering the resulting expression $-Q/r$ to be valid only outside the point charge, i.e. for $r \neq 0$. In this way we ignore the immediate proximities of the charge, meaning that we are not interested in probing its internal structure. When this is not true, and we care about how the charge is distributed in the body, the point charge model simply is no more suitable and something else should be used.

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When you have a uniformly charged dielectric sphere, there is no finite charge at the origin. Instead, you have uniform charge density, so the total charge inside of a sphere of radius $r$ is

$$q = \frac43 \pi r^3 \rho$$

where $\rho$ is the volumetric charge density, $\rho = \frac{Q}{V}$ where $Q$ is total charge and $V = \frac43\pi R^3$ is total volume.

Now you can see that as you approach the origin, the equation does not "blow up" - the force that you would experience goes down linearly to zero, instead of going asymptotically to infinity.

As long as you don't treat charge as a finite point charge (where you have a number in the numerator that remains finite, and a denominator that goes to zero) but instead as a charge density (which doesn't have to be uniform as long as it doesn't have a delta function anywhere) the above analysis (or a variation) is valid.

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