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

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

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).

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?

• I think you are right about the text assuming that the dipole does not change. You want the field at a point; the gradient of the potential (-kp/$r^2$) at that point when q is at that location. For the gradient you are changing r in the potential but not the distance to q. Commented May 26, 2021 at 15:20
• @R.W. Bird I think I understand. The gradient of the potential reflects the change in E when a test charge moves around, while the point charge q remains where it is (therefore dipole moment does not change)! Thanks! P.S. maybe you should put your comment in your answer too... Commented May 27, 2021 at 11:45

• The force on the dipole (or the charge q) is: F = $(kQq/{r ^2)}[(1 + a/r)^{-2} – (1 – a/r)^{-2}]$. Where the dipole has charges +Q and -Q separated by 2a, and k = 1/(4π$ε_o$). Using $(r + a)^{-2} = [r(1 + a/r)]^{-2} = (1/r^2)(1 – 2a/r)$ from the binomial expansion, Then: F = $(kQq/{r^2})(4a/r) = 2kqp/{r^3} = (2kq/{r^3})(αkq/{r^2})$. Commented May 26, 2021 at 19:02