Electric Dipoles and Multipole Expansions

1. The Simplest Case: The Physical Dipole

Let's start not with the general theory, but with the most fundamental example that motivates it: two equal and opposite point charges, +q and -q, separated by a small distance vector $\mathbf{s}$. This is called a physical dipole.

Our goal is to find the potential $\varphi (\mathbf{r})$ at a point $\mathbf{r}$ that is very far away from the dipole, i.e., $r\gg s$.

The potential is simply the sum of the potentials from each point charge:
$\varphi (\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\left(\frac{q}{|\mathbf{r}-{\mathbf{r}}_{+}|}+\frac{-q}{|\mathbf{r}-{\mathbf{r}}_{-}|}\right)$
where ${\mathbf{r}}_{+}$ and ${\mathbf{r}}_{-}$ are the positions of the positive and negative charges, respectively.

When we are far away, this expression can be simplified dramatically. The key insight is to define a vector quantity called the electric dipole moment:
$\mathbf{p}=q\mathbf{s}$
where $\mathbf{s}={\mathbf{r}}_{+}-{\mathbf{r}}_{-}$ is the vector pointing from the negative to the positive charge. For a continuous charge distribution, this becomes an integral:
$\mathbf{p}=\int {\mathbf{r}}^{\prime }\rho ({\mathbf{r}}^{\prime })d^3{r}^{\prime }$

After a mathematical approximation (a Taylor expansion), the potential of the dipole at a large distance $r$ is found to be:
${\varphi }_{\text{dipole}}(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\frac{\mathbf{p}\cdot \mathbf{r}}{r^3}=\frac{1}{4\pi {ϵ}_0}\frac{p\cos \theta }{r^2}$
where $\theta$ is the angle between the dipole moment vector $\mathbf{p}$ and the position vector $\mathbf{r}$. Notice that the dipole potential falls off as $1/r^2$, faster than the $1/r$ potential of a single point charge. This makes sense: from far away, the two opposite charges nearly cancel each other out.

The following reading provides a rigorous derivation of this result and formally defines the dipole moment.

2. The Electric Field of a Dipole

Once we have the potential, we can find the electric field using our fundamental relation $\mathbf{E}=-\nabla \varphi$. Applying the gradient in spherical coordinates to the dipole potential ${\varphi }_d=\frac{p\cos \theta }{4\pi {ϵ}_0{r}^2}$ is a great exercise for your vector calculus skills.

The gradient in spherical coordinates is:
$\nabla =\hat{\mathbf{r}}\frac{\partial }{\partial r}+\hat{\mathbf{\theta }}\frac{1}{r}\frac{\partial }{\partial \theta }+\hat{\mathbf{\varphi }}\frac{1}{r\sin \theta }\frac{\partial }{\partial \varphi }$

Applying this to ${\varphi }_d$ (which has no $\varphi$ dependence), we get:
${\mathbf{E}}_d=-\left(\hat{\mathbf{r}}\frac{\partial }{\partial r}\left(\frac{p\cos \theta }{4\pi {ϵ}_0{r}^2}\right)+\hat{\mathbf{\theta }}\frac{1}{r}\frac{\partial }{\partial \theta }\left(\frac{p\cos \theta }{4\pi {ϵ}_0{r}^2}\right)\right)$
${\mathbf{E}}_d=-\left(\hat{\mathbf{r}}\left(\frac{-2p\cos \theta }{4\pi {ϵ}_0{r}^3}\right)+\hat{\mathbf{\theta }}\frac{1}{r}\left(\frac{-p\sin \theta }{4\pi {ϵ}_0{r}^2}\right)\right)$
${\mathbf{E}}_d=\frac{p}{4\pi {ϵ}_0{r}^3}(2\cos \theta \hat{\mathbf{r}}+\sin \theta \hat{\mathbf{\theta }})$

This expression gives the field components in the radial and polar directions. While useful, there is an even more elegant, coordinate-free form. It can be shown that the above expression is equivalent to:
${\mathbf{E}}_{\text{dipole}}(\mathbf{r})=\frac{1}{4\pi {ϵ}_0{r}^3}\left[3(\mathbf{p}\cdot \hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{p}\right]$
This powerful formula gives the dipole field at any point in space, given the vectors $\mathbf{p}$ and $\mathbf{r}$. Notice the field strength falls off as $1/r^3$, even faster than the potential.

For a clear derivation of this coordinate-free form and a step-by-step calculation of the field on the axis and equator, the following video is excellent.

3. The Multipole Expansion: A General Framework

We started with a simple dipole. Now let's generalize. The multipole expansion is a mathematical technique for approximating the potential of any localized charge distribution $\rho ({\mathbf{r}}^{\prime })$ at a point $\mathbf{r}$ far from the distribution ($r\gg r^{\prime }$).

The starting point is the exact formula for the potential:
$\varphi (\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\int \frac{\rho ({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}{d}^3{r}^{\prime }$

The core of the method is to expand the term $1/|\mathbf{r}-{\mathbf{r}}^{\prime }|$ in a power series of $r^{\prime }/r$. This can be done using the law of cosines and the Taylor expansion, which results in a series involving Legendre Polynomials, $P_n(\cos \theta )$.

The following video provides a concise and clear derivation of this expansion.

The result of the expansion is a series for the potential:
$\varphi (\mathbf{r})={\varphi }_{\text{mono}}+{\varphi }_{\text{dipole}}+{\varphi }_{\text{quad}}+\dots$
$\varphi (\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\left(\frac{Q}{r}+\frac{\mathbf{p}\cdot \hat{\mathbf{r}}}{r^2}+\frac{1}{2r^3}{\sum }_{i,j}{\hat{r}}_i{\hat{r}}_j{Q}_{ij}+\dots \right)$

Let's dissect this:

1. Monopole Term ($\propto 1/r$): The first term depends on the monopole moment, which is just the total charge of the system, $Q=\int \rho ({\mathbf{r}}^{\prime })d^3{r}^{\prime }$. If the system has a net charge, this term dominates at large distances.
2. Dipole Term ($\propto 1/r^2$): The second term is the dipole potential we already studied. Its strength depends on the dipole moment $\mathbf{p}=\int {\mathbf{r}}^{\prime }\rho ({\mathbf{r}}^{\prime })d^3{r}^{\prime }$. If the total charge $Q=0$ (e.g., a neutral atom or molecule), this is often the dominant, or "leading," term.
3. Quadrupole Term ($\propto 1/r^3$): The third term depends on the quadrupole moment tensor $Q_{ij}$. This term becomes important if both $Q=0$ and $\mathbf{p}=0$. It describes more complex charge arrangements, like two back-to-back dipoles.

This expansion is incredibly useful: instead of needing to know the entire charge distribution, we only need to calculate a few key numbers (the moments $Q$, $\mathbf{p}$, etc.) to get a very accurate approximation of the field far away.

The following reading details this expansion, defines the moments, and crucially, discusses how the choice of origin affects these moments.

4. Dipoles in an External Field

So far, we've focused on the fields created by a dipole. What happens when we place a dipole in an external field ${\mathbf{E}}_{\text{ext}}$ created by other sources?

• Torque: A uniform electric field exerts no net force on a dipole (the force on +q is equal and opposite to the force on -q). However, it does exert a torque, which tends to align the dipole moment $\mathbf{p}$ with the external field ${\mathbf{E}}_{\text{ext}}$. The torque is given by:
  $\mathbf{\tau }=\mathbf{p}\times {\mathbf{E}}_{\text{ext}}$

• Potential Energy: The work done to rotate the dipole against this torque is stored as potential energy. The potential energy of a dipole in an external field is:
  $U=-\mathbf{p}\cdot {\mathbf{E}}_{\text{ext}}$
  The energy is minimized when $\mathbf{p}$ is aligned with ${\mathbf{E}}_{\text{ext}}$ ($\theta =0$).

• Force: If the external field is non-uniform, there will be a net force on the dipole. This happens because the force on the positive end may not exactly cancel the force on the negative end. The force is given by the gradient of the potential energy:
  $\mathbf{F}=-\nabla U=\nabla (\mathbf{p}\cdot {\mathbf{E}}_{\text{ext}})$
  Which can be written in the more intuitive form:
  $\mathbf{F}=(\mathbf{p}\cdot \nabla ){\mathbf{E}}_{\text{ext}}$

These interactions are fundamental to understanding how materials behave in electric fields.

Engineering Application: Microwave Ovens

A fantastic real-world example of dipole dynamics is the microwave oven. Water molecules ($H_{2}O$) are polar molecules; they have a permanent electric dipole moment because the oxygen atom pulls electrons away from the hydrogen atoms.

The microwave oven generates a rapidly oscillating electric field. This oscillating field exerts a torque on the water molecules in food, forcing them to rotate back and forth billions of times per second. This rapid rotation creates friction and transfers kinetic energy to surrounding molecules, heating the food. This is a direct application of $\mathbf{\tau }=\mathbf{p}\times \mathbf{E}$.

Conclusion

In this lesson, we introduced the multipole expansion as a powerful tool to approximate the potential of complex charge distributions.

Key Takeaways:

• The potential of any localized charge distribution can be approximated by a series of terms (monopole, dipole, quadrupole, etc.) that fall off with increasing powers of $1/r$.
• The electric dipole is the second term in this series, with a potential falling as $1/r^2$ and a field falling as $1/r^3$.
• The dipole moment vector $\mathbf{p}$ characterizes the strength and orientation of a dipole. Its value is independent of origin choice only if the net charge is zero.
• The electric field of a dipole has a characteristic shape given by ${\mathbf{E}}_d\propto [3(\mathbf{p}\cdot \hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{p}]/r^3$.
• In an external field ${\mathbf{E}}_{\text{ext}}$, a dipole experiences a torque $\mathbf{\tau }=\mathbf{p}\times {\mathbf{E}}_{\text{ext}}$ and has potential energy $U=-\mathbf{p}\cdot {\mathbf{E}}_{\text{ext}}$. A net force arises only in a non-uniform field.

Preview of the Next Lesson:
We have now completed our foundational study of electrostatics—the physics of stationary charges. Our next major step is to explore what happens when charges start to move. We will enter the realm of magnetostatics, the study of steady currents and the magnetic fields they produce. We will begin by introducing the fundamental laws for calculating magnetic fields from currents: the Biot-Savart Law and Ampère's Law.