# M.E. Irizarry-Gelpí

Physics impostor. Mathematics interloper. Husband. Father.

# Dipole Interactions

The scalar potential due to a static electric monopole is given by

\begin{equation*} \phi = \frac{1}{4 \pi \epsilon_{0}} \frac{e}{r} \end{equation*}

Here $$e$$ is the electric monopole moment. The electric field due to a static electric monopole is given by

\begin{equation*} \mathbf{E} \equiv -\nabla \phi = \frac{1}{4 \pi \epsilon_{0}} \frac{e \mathbf{r}}{r^{3}} \end{equation*}

The scalar potential due to a static electric dipole is given by

\begin{equation*} \phi = \frac{1}{4 \pi \epsilon_{0}} \frac{(\mathbf{f} \cdot \mathbf{r})}{r^{3}} \end{equation*}

Here $$\mathbf{f}$$ is the electric dipole moment. The electric field due to a static electric dipole is given by

\begin{equation*} \mathbf{E} \equiv - \nabla \phi = \frac{1}{4 \pi \epsilon_{0}} \frac{3 (\mathbf{f} \cdot \mathbf{r}) \mathbf{r}}{r^{5}} - \frac{1}{4 \pi \epsilon_{0}} \frac{\mathbf{f}}{r^{3}} - \frac{1}{3 \epsilon_{0}} \mathbf{f} \delta(\mathbf{r}) \end{equation*}

The energy associated to the interaction of an electric monopole with an external scalar potential is given by

\begin{equation*} U = - e \phi \end{equation*}

Similarly, the energy associated to the interaction of an electric dipole with an external electric field is given by

\begin{equation*} U = - (\mathbf{f} \cdot \mathbf{E}) \end{equation*}

Consider three cases.

## Monopole-Monopole

The interaction energy between two electric monopoles is given by

\begin{equation*} U_{12} = -e_{1} \phi_{2} = -e_{2} \phi_{1} = - \frac{1}{4 \pi \epsilon_{0}} \frac{e_{1} e_{2}}{r} \end{equation*}

This is the familiar Coulomb potential energy. Note that this energy is invariant under $$e_{1} \rightarrow -e_{1}$$ and $$e_{2} \rightarrow -e_{2}$$.

## Monopole-Dipole

The interaction energy between an electric monopole and an electric dipole is given by

\begin{equation*} U_{12} = -e_{1} \phi_{2} = - (\mathbf{f}_{2} \cdot \mathbf{E}_{1}) = - \frac{1}{4 \pi \epsilon_{0}} \frac{e_{1} (\mathbf{f}_{2} \cdot \mathbf{r})}{r^{3}} \end{equation*}

Note that this energy is invariant under $$e_{1} \rightarrow -e_{1}$$ and $$\mathbf{f}_{2} \rightarrow -\mathbf{f}_{2}$$.

## Dipole-Dipole

The interaction energy between two electric dipoles is given by

\begin{equation*} U_{12} = - (\mathbf{f}_{1} \cdot \mathbf{E}_{2}) = - (\mathbf{f}_{2} \cdot \mathbf{E}_{1}) = \frac{1}{4 \pi \epsilon_{0}} \frac{3 (\mathbf{f}_{1} \cdot \mathbf{r}) (\mathbf{f}_{2} \cdot \mathbf{r})}{r^{5}} - \frac{1}{4 \pi \epsilon_{0}} \frac{(\mathbf{f}_{1} \cdot \mathbf{f}_{2})}{r^{3}} - \frac{1}{3 \epsilon_{0}} (\mathbf{f}_{1} \cdot \mathbf{f}_{2}) \delta(\mathbf{r}) \end{equation*}

This energy is symmetric in both electric dipole moments. Note that this energy is invariant under $$\mathbf{f}_{1} \rightarrow -\mathbf{f}_{1}$$ and $$\mathbf{f}_{2} \rightarrow -\mathbf{f}_{2}$$.