# Dipole in an Electric Field

In the last post, we defined the electric dipole moment  $\bold{p}$ associated with a charge distribution of density $\rho$:

$\displaystyle\bold{p}=\int \rho(\bold{r})\,\bold{r}\,d^3\bold{r}$

We might like to know how an idealised dipole responds to an electric field. To do so, we can evaluate the dipole’s interaction energy with said field.

The electrostatic potential energy of a small element of charge $\rho\,d$3$\bold{r}$ in a potential field $\varphi$ is

$dU=\rho(\bold{r})\varphi(\bold{r})\,d^3\bold{r}$

To find the energy associated with a macroscopic charge distribution, we integrate over space:

$\displaystyle U=\int \rho(\bold{r})\varphi(\bold{r})\,d^3\bold{r}$

As in the previous post, we’re going to Taylor expand part of the integrand to get an analytic approximation. Let’s expand the potential field $\varphi$ about some point $\bold{r}$0 that represents the ‘centre’ of the charge distribution. Doing so gives

$\varphi(\bold{r})\approx\varphi(\bold{r}_0)+\nabla\varphi(\bold{r}_0)\cdot(\bold{r}-\bold{r}_0)$

where $\nabla$ is the gradient operator. Substituting this into the integrand gives

$\displaystyle U=\int\rho(\bold{r})(\varphi(\bold{r}_0)+\nabla\varphi(\bold{r}_0)\cdot(\bold{r}-\bold{r}_0))\,d^3\bold{r}$

$\displaystyle U=\left(\varphi(\bold{r}_0)-\nabla\varphi(\bold{r}_0)\cdot\bold{r}_0\right)\int\rho(\bold{r})\,d^3\bold{r}+\nabla\varphi(\bold{r}_0)\cdot\int\rho(\bold{r})\,\bold{r}\,d^3\bold{r}$

Fear not! This looks a mess, but if we demand that the charge distribution be electrically neutral, the first term vanishes entirely, leaving

$\displaystyle U=-\left(-\nabla\varphi(\bold{r}_0)\right)\cdot\int\rho(\bold{r})\,\bold{r}\,d^3\bold{r}$

We can simplify this expression noticing that i) the first term is equal to the electric field at the ‘location of the dipole’ $\bold{r}$0 and ii) the integral is equal to the charge distribution’s dipole moment. Hence the electrostatic energy of a neutral dipole in an electric field is

$U=-\bold{p}\cdot\bold{E}$

For this expression to be valid, we have to assume the dipole is ‘small’ compared to the length scale over which the electric field varies. If the electric field varied dramatically over the spatial extent of the charge distribution, we would have to consider higher-order terms in the expansion of $\varphi$.

What does the equation tell us? It says the configuration with the least energy is that in which the dipole moment $\bold{p}$ is parallel to the electric field $\bold{E}$, since its energy is $-pE$; the dipole will naturally attempt to align itself with the electric field to which it is subjected. Conversely, the least energetically favourable configuration is that in which the dipole is antiparallel to the field, since its energy is $+pE$. In-between these two limits, the interaction energy varies continuously with the cosine of the angle made between $\bold{p}$ and $\bold{E}$.

Why should a dipole behave like this? Consider again the definition:

$\bold{p}=\int\rho(\bold{r})\,\bold{r}\,d^3\bold{r}$

For convenience, suppose the dipole comprises a point charge $q$ and another point charge $-q$ separated by a neutral stick, a bit like a twirling baton. The electric dipole moment associated with this electret is

$\bold{p}=q\bold{r}_{q}+(-q)\bold{r}_{-q}$

The two vectors represent the positions of the charges. The expression above can be factorised:

$\bold{p}=q\left(\bold{r}_q-\bold{r}_{-q}\right)$

The vector $\bold{p}$ is therefore parallel to the line joining the charge $-q$ to the charge $q$. That is, the dipole moment points from negative to positive. We’ve chosen a convenient charge distribution here, but this is true in general: the dipole moment joins the ‘more negative’ bit of the distribution to the ‘more positive’ bit.

That the dipole tends to align with the electric field should therefore not be surprising. The positive part of the dipole wants to move with the electric field, and the negative part wants to move against it. In doing so, the vector we call the dipole moment naturally spins around, until it becomes aligned with the field. It’s nice to see that the outcome of a load of Taylor expansions and integrals has a simple physical interpretation.

$\,$

For fun (!), let’s do a little Lagrangian mechanics. Consider a dipole $\bold{p}$ with moment of inertia $I$ and mass $m$ subject to a uniform electric field $\bold{E}$. The two coordinates with which we’ll describe the system are $x$, the position of the dipole’s centre of mass measured along the direction of the field, and $\theta$, the angle the dipole moment makes with the field.

The dipole’s kinetic energy is

$T=\frac{1}{2}m\dot{x}^2+\frac{1}{2}I\dot{\theta}^2$

and its potential energy, as we’ve established, is

$U=-\bold{p}\cdot\bold{E}=-pE\cos\theta$

The system’s Lagrangian is therefore

$L=\frac{1}{2}m\dot{x}^2+\frac{1}{2}I\dot{\theta}^2+pE\cos\theta$

We’ll use the Euler-Lagrange equations to determine the differential equations governing the evolution of $x$ and $\theta$. Let’s first do the easy one:

$\displaystyle \frac{\partial L}{\partial x}=\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{x}}\right)\implies 0=m\ddot{x}$

The motion of the dipole as a whole is therefore inertial; the dipole’s centre of mass moves almost as if the electric field wasn’t there at all! This makes sense, given that the dipole is assumed to be electrically neutral, and each part of the dipole experiences the same electric field; the forces on different parts of the dipole cancel one another perfectly to give a net force of zero.

The evolution of the angle $\theta$ is determined by the equation

$\displaystyle \frac{\partial L}{\partial\theta}=\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{\theta}}\right)\implies \ddot{\theta}+\frac{pE}{I}\sin\theta=0$

Hence if we assume the angle made by the dipole with the electric field is small, the dipole undergoes simple harmonic motion at angular frequency

$\displaystyle \omega=\left(\frac{pE}{I}\right)^{\frac{1}{2}}$

In this simple model, the dipole wiggles back and forth, attempting to align itself with the field and consistently overshooting. Note that, if we had assumed that $\bold{E}$ were inhomogeneous (that is, dependent on the coordinate $x$), the dipole’s motion would not have been inertial! The dipole would therefore have experienced both a torque and a force. Fun indeed!

$\,$

In summary, the energy of a neutral, small electric dipole $\bold{p}$ in an electric field $\bold{E}$ is

$U=-\bold{p}\cdot\bold{E}$

By ‘small’, we mean that the electric field does not vary significantly over the spatial extent of the dipole. The energy of interaction provides a means of deriving the equation of motion of an idealised dipole. In a future post, we’ll combine the results of this and the previous post to model dipole-dipole interactions.

Advertisements