Is the dipole moment of a polarized particle supposed to be a constant?

Question

Problem $4.4$ A point charge $q$ is situated a large distance $r$ from a neutral atom of polarizability $\alpha$. Find the force of attraction between them.

Source: Griffiths, Electrodynamics

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Textbook solution:

$ \text { Field of } q: \frac{1}{4 \pi \epsilon_{0}} \frac{q}{r^{2}} \hat{\mathbf{r}} \text { . Induced dipole moment of atom: } \\ \mathbf{p}=\alpha \mathbf{E}= \frac{\alpha q}{4 \pi \epsilon_{0} r^{2}} \hat{\mathbf{r}} $ Field of this dipole, at location of $q(\theta=\pi$, in Eq. $3.103): E=\frac{1}{4 \pi \epsilon_{0}} \frac{1}{r^{3}}\left(\frac{2 \alpha q}{4 \pi \epsilon_{0} r^{2}}\right)$ (to the right). Force on $q$ due to this field: $\quad F=2 \alpha\left(\frac{q}{4 \pi \epsilon_{0}}\right)^{2} \frac{1}{r^{5}}$ (attractive).

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What I have tried:

$\textbf{p} = \alpha \textbf{E}$

the potential due to the induced dipole is: $V_{\mathrm{dip}}(\mathbf{r})=\frac{1}{4 \pi \epsilon_{0}} \frac{\mathbf{p} \cdot \hat{\mathbf{r}}}{r^{2}} = (\frac{1}{4 \pi \epsilon_{0}})^2 \frac{\alpha q}{r^{4}}$

However, why isn't $E = - \frac{\partial V_{dip}}{\partial r} = \frac{\partial}{\partial r} (\frac{1}{4 \pi \epsilon_{0}})^2 \frac{\alpha q}{r^{4}} = (\frac{1}{4 \pi \epsilon_{0}})^2 \frac{4 \alpha q}{r^{5}}$, which is larger than the answer given by a factor of 2?

The author appears to take $p$ as a constant, thus it is somehow left out of the partial derivative.

Thus my questiom is

Is dipole moment, in this case, supposed to be a constant?

Answer 1

I found the net force on the dipole and it agrees with the answer from the text (using a binomial approximation). I think you have an extra factor of 2 in your expression for the potential energy. Be careful that you don't confuse the dipole charge with the external charge.

Answer 2

Your observation is correct. The force is the gradient of the potential energy, that is, the rate of change when the charge is moved. This includes the change in polarization.