Potential of an electric dipole

Question

I'm currently working my way through Griffith's Introduction to Electrodynamics (4th ed).

In chapter 3 section 4, he shows that we can take the equation for the electric potential of a continuous charge distribution $$V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\int\frac{1}{\mathscr{R}}\rho(\mathbf{r}')\mathrm{d}\tau'$$ where $\mathscr{R} = |\mathbf{r} - \mathbf{r}'|$ and expand it out as $$V(\mathbf{r}) = V_\text{mon}(\mathbf{r}) + V_\text{dip}(\mathbf{r}) + V_\text{quad}(\mathbf{r})+\ldots$$ where $$V_\text{dip}(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\frac{\mathbf{p}\cdot\hat{\mathbf{r}}}{r^2} \\ \mathbf{p} = \int\mathbf{r}'\rho(\mathbf{r}')\mathrm{d}\tau' \tag{3.99}$$

I understand this derivation. What confuses me is what he does in the following chapter. In chapter 4 section 2, he writes the potential of a perfect dipole as $$V_\text{dip}(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\frac{\mathbf{p}\cdot\hat{\mathscr{R}}}{\mathscr{R}^2}\tag{4.8}$$

How does he switch from $\mathbf{r}$ to $\mathscr{R}$? I was under the impression that equation 3.99 already captured charges not centered at the origin.

Answer 1

The multipole expansion depends on the choice of origin. Only the lowest nonzero term is independent of the origin. In the case of a single dipole not at the origin, the higher terms will be nonzero in order to "correct" the centering of the dipole term from the origin to the dipole's actual location.

Let $\mathbf{r}'$ be the position of the dipole and $\mathbf{d}$ be the separation. Then the two charges $\pm q$ are located at $\mathbf{r}'_\pm = \mathbf{r}' \pm \mathbf{d}/2$.

The multipole expansion is $$V(\mathbf{r})=\frac{1}{4\pi\varepsilon_0} \sum_{n=0}^\infty \int\frac{{r'}^n}{r^{n+1}} P_n(\hat{\mathbf{r}}\cdot\hat{\mathbf{r}}')\rho(\mathbf{r}')\mathrm{d}^3\mathbf{r}'.$$ The dipole comprises two delta functions of $\pm q$ located at $\mathbf{r}'_\pm$ so $$V(\mathbf{r})=\frac{q}{4\pi\varepsilon_0} \sum_{n=0}^\infty \left(\frac{{r'_+}^n}{r^{n+1}} P_n(\hat{\mathbf{r}}\cdot\hat{\mathbf{r}}'_+) - \frac{{r'_-}^n}{r^{n+1}} P_n(\hat{\mathbf{r}}\cdot\hat{\mathbf{r}}'_-)\right).$$ Using the identity $$\frac{1}{|\mathbf{r}-\mathbf{r}'|} = \sum_{n=0}^\infty \frac{{r'}^n}{r^{n+1}} P_n(\hat{\mathbf{r}}\cdot\hat{\mathbf{r}}') \tag{3.94}$$ this becomes $$V(\mathbf{r}) = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{|\mathbf{r}-\mathbf{r}'_+|}-\frac{1}{|\mathbf{r}-\mathbf{r}'_-|}\right) \\ = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{|\mathscr{R}-\mathbf{d}/2|}-\frac{1}{|\mathscr{R}+\mathbf{d}/2|}\right) \\ \approx\frac{q}{4\pi\varepsilon_0}\left(\frac{1}{\sqrt{|\mathscr{R}|^2-\mathscr{R}\cdot\mathbf{d}}}-\frac{1}{\sqrt{|\mathscr{R}|^2+\mathscr{R}\cdot\mathbf{d}}}\right) \\ = \frac{q}{4\pi\varepsilon_0}\left(\frac{\hat{\mathscr{R}}\cdot\mathbf{d}}{|\mathscr{R}|^2}\right) = \frac{1}{4\pi\varepsilon_0}\frac{\hat{\mathscr{R}}\cdot\mathbf{p}}{|\mathscr{R}|^2}.$$

A much more straightforward way, without using any of the algebra above, is to observe that the formula $$V_\text{dip}(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\frac{\mathbf{p}\cdot\hat{\mathbf{r}}}{r^2}\tag{3.99}$$ is the potential of a dipole at the origin. So, in order to obtain the potential of a dipole at an arbitrary position $\mathbf{r}'$, one simply performs a translation, replacing $\mathbf{r}$ with $\mathbf{r}-\mathbf{r}'=\mathscr{R}$.

Answer 2

In chapter 4, Griffiths prepares to discuss the polarization of materials. So, in Equation 4.8, the dipole under consideration may not be at the origin. To get the potential then, we use $\mathscr{R}$, which essentially translates the dipole field to the right position.

Note that Equation 3.99, while always correct for $V_{\text{dip}}$, gives the full potential of an ideal dipole only if that dipole is at the origin - otherwise, there are higher order terms in the multipole expansion.