The intepretation of $d$ in $p=qd$

Question

In Griffiths Example 4.1, the author derived the polarizability of a uniformly charged spherical cloud with radius $a$. For an external field $E$, the electron cloud will shift to the left with a length of $d$. When the $E_e$ field electron cloud induces matches the external field $E$, we have $$ E = \frac{1}{4\pi\epsilon_0} \frac{qd}{a^3}$$ with $$ p = qd = 4\pi\epsilon_0a^3E $$ thus deriving the polarizability $\alpha$. My question is: What exactly is $d$ in general in a dipole moment $p$ like this? In this problem, the $d$ is the length the center of the electron cloud shifts. But in general, $p$ is defined via the integral (in chapter 3.4) $$ \int r' \rho(r') d\tau'$$ where $q$ is just (EDIT: q should not be the total charge)

$$ q= \int \rho(r') d\tau' $$ is wrong. The total charge is $0$.

and $d$ is also defined in this problem. What is $q$ in general, and why exactly does $p = qd$ hold according to these definitions?

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I have an idea: $p=qd$ is for two point electrons, but in this example, we need to regard the electron cloud as a whole $-q$ in its center- and $p=qd$ still holds, for the integral $$\int r' \rho(r') d\tau'$$ does not change if you regard the electron cloud as that.

Is my interpretation correct?

Answer 1

Yes it is correct. Basically, the dipole moment: $$ P=\int d^3r \rho r $$ depends on the origin when the total charge: $$ Q=\int d^3r \rho $$ is does not vanish $Q\neq0$.

Typically, in this case you take the origin to be the center of charge: $$ R=\frac{P}{Q} $$ Taking the origin to be $R$ you automatically cancel the dipole moment.

When $Q=0$, the dipole moment becomes origin independent. One way to formalize your intuition is to set $\rho_\pm$ for the positive/negative charges defined by: $$ \rho=\rho_+-\rho_- \\ |\rho|=\rho_++\rho_- $$ note that they are both positive. You can then define: $$ q=\int d^3r \rho_\pm=\frac{1}{2}\int d^3r |\rho| $$ which accounts for only the positive/negative charges. You can then define: $$ R_\pm=\frac{1}{q}\int d^3r \rho_\pm r $$ which is the center of charge of the posirive/negative charges. Setting: $$ d=R_+-R_- $$ You get: $$ P=qd $$

Hope this helps.