-- ELECTRIC FIELDS --
-- GAUSS'S LAW --
-- ELECTRIC POTENTIAL --
-- CAPACITORS --
-- DC CIRCUITS --
-- MIDTERM 1 - STUDY GUIDE --
-- MAGNETISM --
-- INDUCTION --
-- ELECTROMAGNETIC WAVES --
-- OPTICS --
-- MIDTERM 2 - STUDY GUIDE --
-- INTERFERENCE & DIFFRACTION --
-- QUANTUM MECHANICS --
-- RELATIVITY --
-- STUDY GUIDE - FINAL --

P19L2013 – Two Electric Dipoles

Two Electric Dipoles

Two electric dipoles with dipole moments $\overrightarrow{p_1}$ and $\overrightarrow{p_2}$ are inline with one another as shown below. Assume the distance $r$ between the dipoles is much larger than the length $d$ of either dipole.

1. Find the electric field due to an electric dipole at any position on the axis of the dipole.

View answer

We consider an electric dipole centered at the origin and pointing in the $\overrightarrow{x}$ direction.

The electric field created by the dipole at a location $r\gg d$ on the $x$-axis is derived as follows

\begin{aligned}
\overrightarrow{E}&=\left[\frac{kq}{\displaystyle{{\left(r-\frac{d}{2}\right)}^2}-\frac{kq}{{\left(r+\frac{d}{2}\right)}}^2}\right]\ \hat{x} \\
\\
&=\frac{kq}{r^2}\left[\frac{1}{\displaystyle{{\left(1-\frac{d}{2r}\right)}^2}-\frac{1}{{\left(1+\frac{d}{2r}\right)}}^2}\right]\ \hat{x} \\
\\
&=\frac{kq}{r^2}\left[{\left(1-\frac{d}{2r}\right)}^{-2}-{\left(1+\frac{d}{2r}\right)}^{-2}\right]\ \hat{x} \\
\\
&\approx \frac{kq}{r^2}\left[1+\frac{d}{r}-1+\frac{d}{r}\right]\ \hat{x} \\
\\
&\approx \frac{2kqd}{r^2}\ \hat{x}
\end{aligned}

where we assumed $d/2r\ll 1$ and used the first-order Taylor expansion ${\left(1+\varepsilon \right)}^{\alpha }\approx 1+\alpha \varepsilon $. Thus, the electric field due to an electric dipole at any position on the axis of the dipole is equal to

\boxed{\overrightarrow{E}=\frac{2kqd}{r^3}\ \hat{x}=\frac{2kp}{r^3}\ \hat{x}}

as long as $d/2r\ll 1$.

2. Find the potential energy $U$ of one dipole in the presence of the other dipole (i.e. their interaction energy).

View answer

The interaction energy of the two dipoles is given by

U=-\overrightarrow{p_2}\cdot \overrightarrow{E_1}=-\left(p_2\hat{x}\right)\cdot \left(\frac{2kp_1}{r^3}\hat{x}\right)=-\frac{2kp_1p_2}{r^3}

The interaction energy of the two dipoles is equal to

\boxed{U=-\frac{2kp_1p_2}{r^3}}

Note: it is also equal to $U=-\overrightarrow{p_1}\cdot \overrightarrow{E_2}$. Indeed,

-\overrightarrow{p_1}\cdot \overrightarrow{E_2}=-\left(p_1\hat{x}\right)\cdot \left(\frac{2kp_2}{r^3}\ \hat{x}\right)=- \frac{2kp_1p_2}{r^2}

3. If the dipoles are anti-aligned with one another (see below), what is $U$?

View answer

The interaction energy of the two dipoles is given by

U=-\overrightarrow{p_2}\cdot \overrightarrow{E_1}=-\left(-p_2\hat{x}\right)\cdot \left(\frac{2kp_1}{r^3}\hat{x}\right)=\frac{2kp_1p_2}{r^3}

Thus, the interaction energy of the two dipoles is equal to

\boxed{U=\frac{2kp_1p_2}{r^3}}

4. Now assume we turn on an electric field $\overrightarrow{E_0}$ as shown below. Describe what happens to each dipole and find the maximum torque that the field can exert on each dipole.

View answer

Since each dipole is horizontal and the electric field $\overrightarrow{E_0}$ is horizontal as well, nothing happens to the dipoles. Indeed, they are both in equilibrium and experience no torque: the left dipole is at a stable equilibrium position ($\overrightarrow{p_1}$ is parallel to $\overrightarrow{E_0}$) and the right dipole is at an unstable equilibrium position ($\overrightarrow{p_2}$ is antiparallel to $\overrightarrow{E_0}$).

The maximum torque that the electric field can exert on each dipole is achieved when the dipoles are perpendicular to the electric field $\overrightarrow{E_0}$ as shown in the figure below

The net torque experienced by the dipoles is equal to

\boxed{{\tau }_{max}={\tau }_++{\tau }_-=\frac{qd}{2}E_0{\mathrm{sin} \left(90\right)\ }+\frac{qd}{2}E_0{\mathrm{sin} \left(90\right)\ }=qdE_0}