# Peeter Joot's (OLD) Blog.

• 324,835

## Dipole Moment from constant electric field

Posted by peeterjoot on December 27, 2012

## Question: Dipole Moment from constant electric field

In [1] it is stated that the force per unit angle on a dipole system as illustrated in Fig.1 is

\begin{aligned}F_\theta = -p \mathcal{E} \sin\theta,\end{aligned} \hspace{\stretch{1}}(1.0.1)

where $\mathbf{p} = q \mathbf{r}$. The text was also referring to torques, and it wasn’t clear to me if the result was the torque or the force. Derive the result to resolve any doubt (in retrospect dimensional analysis would also have worked).

Fig1: Dipole moment coordinate

Let’s put the electric field in the $\hat{\mathbf{x}}$ direction ($\theta = 0$), so that the potential acting on charge $i$ is given implicitly by

\begin{aligned}\mathbf{F}_i = q_i \mathcal{E} \hat{\mathbf{x}} = -\nabla \phi_i = -\hat{\mathbf{x}} \frac{d{{\phi_i}}}{dx}\end{aligned} \hspace{\stretch{1}}(1.0.2)

or

\begin{aligned}\phi_i = -q_i (x_i - x_0).\end{aligned} \hspace{\stretch{1}}(1.0.3)

Our positions, and velocities are

\begin{aligned}\mathbf{r}_{1,2} = \pm \frac{r}{2} \hat{\mathbf{x}} e^{\hat{\mathbf{x}} \hat{\mathbf{y}} \theta}\end{aligned} \hspace{\stretch{1}}(1.0.4a)

\begin{aligned}\frac{d{{\mathbf{r}_{1,2}}}}{dt} = \pm \frac{r}{2} \dot{\theta} \hat{\mathbf{y}} e^{\hat{\mathbf{x}} \hat{\mathbf{y}} \theta}.\end{aligned} \hspace{\stretch{1}}(1.0.4b)

Our kinetic energy is

\begin{aligned}T = \frac{1}{{2}} \sum_i m_i \left( \frac{d{{\mathbf{r}_i}}}{dt} \right)^2 = \frac{1}{{2}} \sum_i m_i \left( \frac{r}{2} \right)^2 \dot{\theta}^2= \frac{1}{{2}} (m_1 + m_2) \left( \frac{r}{2} \right)^2 \dot{\theta}^2.\end{aligned} \hspace{\stretch{1}}(1.0.5)

For our potential energies we require the $x$ component of the position vectors, which are

\begin{aligned}x_i = \mathbf{r}_i \cdot \hat{\mathbf{x}}=\pm \left\langle{{ \frac{r}{2} \hat{\mathbf{x}} e^{\hat{\mathbf{x}} \hat{\mathbf{y}} \theta} \hat{\mathbf{x}}}}\right\rangle=\pm \frac{r}{2} \cos\theta\end{aligned} \hspace{\stretch{1}}(1.0.6)

Our potentials are

\begin{aligned}\phi_1 = -q_1 \mathcal{E} \frac{r}{2} \cos\theta + \phi_0\end{aligned} \hspace{\stretch{1}}(1.0.7a)

\begin{aligned}\phi_2 = q_2 \mathcal{E} \frac{r}{2} \cos\theta + \phi_0\end{aligned} \hspace{\stretch{1}}(1.0.7b)

Our system Lagrangian, after dropping the constant reference potential that doesn’t effect the dynamics is

\begin{aligned}\mathcal{L} = \frac{1}{{2}} (m_1 + m_2) \left( \frac{r}{2} \right)^2 \dot{\theta}^2+q_1 \mathcal{E} \frac{r}{2} \cos\theta-q_2 \mathcal{E} \frac{r}{2} \cos\theta\end{aligned} \hspace{\stretch{1}}(1.0.8)

For this problem we had two equal masses and equal magnitude charges $m = m_1 = m_2$ and $q = q_1 = -q_2$

\begin{aligned}\mathcal{L} = \frac{1}{{4}} m r^2 \dot{\theta}^2 + q r \mathcal{E} \cos\theta\end{aligned} \hspace{\stretch{1}}(1.0.9)

\begin{aligned}p_\theta = \frac{\partial {\mathcal{L}}}{\partial {\dot{\theta}}} = \frac{1}{{2}} m r^2 \dot{\theta}\end{aligned} \hspace{\stretch{1}}(1.0.10)

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\theta}} = -q r \mathcal{E} \sin\theta=\frac{d{{p_\theta}}}{dt}=\frac{1}{{2}} m r^2 \dot{d}{\theta}\end{aligned} \hspace{\stretch{1}}(1.0.11)

Putting these together, with $p = q r$, we have the result stated in the text

\begin{aligned}F_\theta = \frac{d{{p_\theta}}}{dt} = -p \mathcal{E} \sin\theta.\end{aligned} \hspace{\stretch{1}}(1.0.12)

# References

[1] E.A. Jackson. Equilibrium statistical mechanics. Dover Pubns, 2000.