Two charges \(Q\) with opposite signs are separated by a distance \(\mathbf{a}\).
The electric dipole moment is defined as \(\mathbf{p} = Q \mathbf{a}\) and points from the negative to the positive charge.
Figure 16.1: Sketch of an electrical dipole aligned with the \(z\)-axis.
According to the principle of superposition, the potential is given by \[
U(\mathbf{r}) = \frac{Q}{4 \pi \varepsilon} \left[
\frac{1}{|\mathbf{r} - \mathbf{a}|}
-
\frac{1}{|\mathbf{r}|}
\right].
\]
For \(|\mathbf{a}| \ll |\mathbf{r}|\), the reciprocal distance (first term in the brackets) can be expanded into a series \[
\begin{align}
\frac{1}{|\mathbf{r} - \mathbf{a}|} & \approx
\frac{1}{|\mathbf{r}|} \left\{
1 - 2 \frac{\mathbf{a} \cdot \mathbf{r}}{|\mathbf{r}|^2}
\right\}^{-1/2} \\
& \approx \frac{1}{|\mathbf{r}|}
\left[
1 + \frac{\mathbf{a} \cdot \mathbf{r}}{|\mathbf{r}|^2}
\right].
\end{align}
\]
The dipole potential is thus \[
U(\mathbf{r}) = \frac{Q}{4 \pi \varepsilon}
\frac{\mathbf{a} \cdot \mathbf{r}}{|\mathbf{r}|^3}.
\]
The scalar product, using \(r = |\mathbf{r}|\) and \(p = |\mathbf{p}|\), leads to the equation \[
U(r) = \frac{p}{4 \pi \varepsilon} \frac{\cos \theta}{r^2}.
\]
We see that the dipole potential can be derived from the single charge potential by differentiating with respect to the coordinates of the single potential along the dipole axis. The term \(\mathbf{p} \cdot \nabla\) is understood as a directional derivative. It calculates the projection of the gradient onto the dipole axis. \[
U(\mathbf{r}) = -\frac{1}{4 \pi \varepsilon} \mathbf{p} \cdot \nabla \frac{1}{|\mathbf{r}|}.
\]
Note
This result is an approximate solution, as higher-order terms in the Taylor series expansion are neglected.
16.2 Field components
The electric field can be obtained from the potential by applying the gradient
\[
\mathbf E = -\grad U.
\]
Show the code
import numpy as npfrom sympy import*from IPython.display import display, MathU, r, theta = symbols("U r theta")U = cos(theta) / r**2display(Math('U(r, \\theta) = \\frac{p}{4 \\pi \\varepsilon} '+ latex(U)))grad_sph = [diff(U, r), 1/ r * diff(U, theta)]E = [-grad_sph[i] for i in [0,1]]display(Math('\\mathbf E(r, \\theta) = \\frac{p}{4 \\pi \\varepsilon}'+ latex(E) +'^\\top'))
