What does "weak-field limit" mean when it refer to dipole and atomic polarization?

Question

In Griffiths' book, Introduction to Electrodynamics 3rd, problem 4.3,it requires me to find the condition on ρ(r) such that equation $p = αE$ will hold in the weak-field limit. The answer gives $ρ（r）=Ar$, and i don't understand why, either that it requires $p$ is proportional to $E$ while the answer gives $E^{1/2}$,is it because of the weak-field limit? And why the answer requires $E$ to be proportional to $r$ and $ρ$ to be zero at the origin?

$$p=αE \tag{4.1}\label{}$$ α means atomic polarization

Question: Problem 4.3 According to Eq. 4.1， the induced dipole moment of an atom is proportional to the external field. This is a"rule of thumb not a fundamental law, and it is easy to concoct exceptions--in theory. Suppose, for example, the charge density of the electron cloud were proportional to the distance from the center, out to a radius $R$. To what power of E would p be proportional in that case？ Find the condition on ρ（r） such that Eq. 4.1 will hold in the weak-field limit

Answer: ρ（r）=Ar. Electric field（ by Gauss's Law：$$∮E·da=E（4πr²）=Q/(ε₀),or E=(Ar²)/(4ε₀)$$ This “internal” field balances the external field $E$ when nucleus is “off-center” an amount $$d:(ad²)/(4ε₀)=E⇨d={\sqrt {(4ε₀E)/A}}$$So the induced dipole moment is $$p= ed= 2e·{\sqrt {ε₀/A}}E^{1/2}$$ Evidently p is proportional to $E^{1/2}$. For Eq. 4.1 to hold in the weak-field limit, E must be proportional to r, for small r, which means that ρ must go to a constant（not zero）at the origin:ρ（0）≠0（nor infinite）

Answer 1

Most systems have $p$ proportional to $E$ for small $E$. The example from Griffiths is meant to illustrate an exception to that general pattern, hence they pick a specific distribution of charge so that $p\propto E^{1/2}$, which is not proportional to $E$.

The second part of the question asks for what exactly makes some distributions $\rho$ give a linear dependence and some do not. The answer is that the density at $0$ needs to be nonzero. The distribution with $\rho\propto r$ is zero at $r=0$, so the polarization is not proportional to $E$.

Answer 2

This problem has 2 questions.

The 1st is : if ρ=Ar , the relation between E and p.

The 2ed is : if I want p=αE , how to solve ρ?

Lets solve 2ed question:

Assume ρ=B (B is a constant）.

∮E⋅da=E（4πr²）=Q/(ε₀)=∫B4πr²dr/(ε₀) .

We can solve：

E=Br/3(ε₀)，

so r=3(ε₀)E/B,

so p=qr=3q(ε₀)E/B，

then we can see p is proportional to E.

So we can see if ρ=B(the charge density is a constant），p is proportional to E.