P19L2013 – Two Electric Dipoles

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 $d/2r\ll 1$.

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}

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}}

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}