Electric field inside a dielectric sphere placed in a uniform electric field I am working through Griffith's chapter 4. Example 4.7 describes a situation in which you place a sphere of homogenous linear dielectric material inside a uniform field. You are looking for the field inside the dielectric sphere. The example provides its own solution (using boundary values).
However, I approached the problem in a different way. First, I observed that the free charge density is zero. Thus, the bound charge density is also zero (free charge density is proportional to bound charge density in a homogenous linear dielectric). Thus, all excess charge must be found in the surface charge density.
I then computed the electric field inside the dielectric sphere due to the surface charge density, Pcos(\theta) (where P is polarization). This field has been computed in an earlier exercise in Griffith's and turns out to be -P/(3\epsilon_0) (in the -z direction).
Finally, I thought that the ultimate electric field produced within the dielectric material would simply be the superposition of the uniform field and the field produced by the surface charge density. Thus, I added the fields together. I then substituted P via the constitutive relationship between polarization and total electric field, collected like terms (total electric field) and came to the correct answer.
However, I am thinking that this is a much simpler solution to exercise 4.7. I am thus skeptical of it being a proper analysis (for it it were, wouldn't it be the solution itself?). I was wondering if any of the observations I made in the course of this analysis were improper.

 A: The solution presented in Griffiths follows a fully systematic method. At the very end, Griffiths makes the comment that "the field inside the sphere is (surprisingly) uniform".
Now, the sphere is made of a linear dielectric, so the polarization is proportional to the field inside the sphere. In your solution, you seem to have assumed that the sphere has a uniform polarization (allowing you to use the result from Exercise 4.2; btw, I use the 4th edition, and I'm not sure which edition you use since you didn't specify), which is equivalent to assuming that the field inside the sphere is uniform. The problem is that, like Griffiths says, this isn't actually obvious from the start. Sure, the applied field is uniform, but at the beginning, you don't have a way to immediately rule out that the polarization, and the field due to the polarization (or equivalently, due to the bound surface charge) could be non-uniform.
In effect, you made an ansatz: you assumed the form that the solution would take, and then you solved for the constants and showed that your solution is self-consistent. This is a valid solution method, but it's not the kind of fully systematic solution that the text is trying to illustrate.
To make this more precise, observe that inside the sphere, we have:
\begin{align}
E &= E_a + E_i \\
P &= \epsilon_0 \chi_e E \\
  &= \epsilon_0 \chi_e (E_a + E_i)
\end{align}
where $E_a$ is the (linear) applied field, $E_i$ is the part of the field due to the polarization (equivalently, due to the bound surface charge), and $P$ is the polarization.
If you make the assumption that $P$ sources a field inside the sphere that is proportional to itself: that is, $E_i = kP$, then we obtain $P = \epsilon_0 \chi_e (E_a + kP)$, and, by isolating $P$ on one side, we see that it is proportional to $E_a$ (and thus uniform). Finally, we need to check that a uniform $P$ actually does give rise to a uniform $E_i$ according to the laws of electrostatics, validating our initial assumption.
It's just that it wasn't obvious that the assumption was going to lead to the right answer; in general, if you start with some arbitrary polarization $P(x)$ with support in a region $R$ and calculate the field it produces in that region using the bound charge and Coulomb's law, you won't get something that's identically a scalar multiple of $P(x)$ inside $R$.
