##### Approximate Potential at Large Distance
ems.ca.me.a

Let's consider the spatial function in the potential for a single point source charge: \[ \frac{1}{|{\bf r} - {\bf r}_s|} \] How does this look when we're at large distances \(|{\bf r}| \gg |{\bf r}_s|\)? We can formally expand this in powers of \(|{\bf r}_s|/|{\bf r}|\). For simplicity, let's start by putting \({\bf r} = r ~\hat{\bf z}, r > 0\) and \({\bf r}_s = r_s \hat{\bf z}\), with \(|r_s| < r\). By Taylor expanding, we get \[ \frac{1}{|{\bf r} - {\bf r}_s|} = \frac{1}{r - r_s} = \frac{1}{r} \sum_{l=0}^{\infty} \left(\frac{r_s}{r}\right)^l \] Formally, we could do this for any vector \({\bf r}_s\) such that \(|{\bf r}_s| < |{\bf r}|\) by Taylor expanding with the \({\boldsymbol \nabla}\) operator,

\[ \frac{1}{|{\bf r} - {\bf r}_s|} = \sum_{l=0}^{\infty} \frac{1}{l!} \left( - {\bf r}_s \cdot {\boldsymbol \nabla} \right)^l \frac{1}{r} = \frac{1}{r} - {\bf r}_s \cdot {\boldsymbol \nabla} \frac{1}{r} + \frac{1}{2} \left({\bf r}_s \cdot {\boldsymbol \nabla} \right)^2 \frac{1}{r} + ... \tag{1or_grad}\label{1or_grad} \] However, it is more practical to exploit the fact that in the configuration above, the problem has azimuthal symmetry (since everything is on the \(\hat{\bf z}\) axis), and therefore the potential takes the form of the general solution of Laplace's equation Lap_sph_az_sol with \(\theta = 0\). Reading the parameters, we get \(A_l = 0\), \(B_l = r_s^l\). Putting back a generic angle (the coefficients remain the same), we thus get

- Gr(3.94)

\begin{equation} \frac{1}{|{\bf r} - {\bf r}_s|} = \sum_{l=0}^{\infty} \frac{r_s^l}{r^{l+1}} P_l (\cos \theta), \hspace{1cm} \cos \theta \equiv \hat{\bf r} \cdot \hat{\bf r}_s, \hspace{1cm} r_s < r. \tag{1or_Leg}\label{1or_Leg} \end{equation}

We thus see that our beloved Legendre polynomials are quite handy beasts indeed. Considering an arbitrary charge distribution over a volume \({\cal V}\), we can expand the potential at a point \({\bf r}\) outside \({\cal V}\) according to (here, we put the origin of our coordinate system closer to all points in \({\cal V}\) than to \({\bf r}\) to ensure convergence)

- Gr (3.95)

\[ \phi({\bf r}) = \frac{1}{4\pi \varepsilon_0} \sum_{l=0}^{\infty} \frac{1}{|{\bf r}|^{l+1}} \int_{\cal V} d\tau_s |{\bf r}_s|^l P_l (\hat{\bf r} \cdot \hat{\bf r}_s) ~\rho({\bf r}_s), \hspace{1cm} |{\bf r}| > |{\bf r}_s| ~\forall~ {\bf r}_s \in {\cal V}. \tag{p_Leg}\label{p_Leg} \] Keeping all the terms up to infinite order, this is an exact expansion of our potential. The real power of this however comes from the fact that truncating the series to only its first (few) terms gives a very good approximation to the original potential, as long as we are sufficiently far away from the source charges.

As a first very basic example, let us revisit the configuration we had in equation p_di_z which represented a (physical) electric dipole, namely two equal and opposite charges \(\pm q\) separated by a distance \(d\).

To be more general, we here put \(q\) at position \({\bf d}/2\) and \(-q\) at \(-{\bf d}/2\). The potential is then

\[ \phi({\bf r}) = \frac{q}{4\pi \varepsilon_0} \left( \frac{1}{|{\bf r} - {\bf d}/2|} - \frac{1}{|{\bf r} + {\bf d}/2|} \right). \]

We can write

\begin{equation*} |{\bf r} \pm {\bf d}/2|^2 = r^2 \pm {\bf r} \cdot {\bf d} + (d/2)^2 = r^2 \left( 1 \pm \frac{{\bf d} \cdot \hat{\bf r}}{r} + \frac{d^2}{4r^2}\right) \end{equation*}

For \(r \gg d\), we can expand (immediately dropping terms of order \(d^2/r^2\))

\[ \frac{1}{|{\bf r} \pm {\bf d}/2|} \simeq \frac{1}{r} \left( 1 \pm \frac{{\bf d} \cdot \hat{\bf r}}{r} \right)^{-1/2} \simeq \frac{1}{r} \left( 1 \mp \frac{{\bf d} \cdot \hat{\bf r}}{2r} \right). \]

Putting things together, the leading term in the expansion p_Leg for the potential of the physical dipole is given by

- Gr (3.90)

\[ \phi({\bf r}) \simeq \frac{1}{4\pi \varepsilon_0} \frac{q~ {\bf d} \cdot \hat{\bf r}}{r^2} \tag{p_physdi}\label{p_physdi} \]

Thus, although the dipole is electrically neutral overall, its influence does not vanish. Compared to a point charge, it displays two important differences: i) it falls off with one more (inverse) power of distance, and ii) it carries a vector direction, so its influence is not isotropic.

In the next section, we will revisit such dipoles, but now coming from more generic charge configurations.

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42