# Electric Dipole

An electric dipole has two equal and opposite charges separated by a small distance $d$. Its dipole moment is defined as
\begin{align}
\vec{p}=q\vec{d}.
\end{align}

## Potential of an electric dipole

Let an electric dipole $\vec{p}$ is placed at the orgin. Its potential at a point ($r,\theta$) is given by
\begin{align}
V=\frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^2}.
\end{align}

## Field of an electric dipole

Let an electric dipole $\vec{p}$ is placed at the orgin. The radial and angular components of its electric field at a point ($r,\theta$) are given by
\begin{align}
E_r & =\frac{1}{4\pi\epsilon_0}\frac{2p\cos\theta}{r^3},\nonumber\\
E_\theta&=\frac{1}{4\pi\epsilon_0}\frac{p\sin\theta}{r^3}
\end{align}

## Torque on an electric dipole placed in an electric field

The torque on an electric dipole $\vec{p}$ placed in an electric field $\vec{E}$ is given by
\begin{align}
\vec{\tau}=\vec{p}\times\vec{E}
\end{align}

## Potential energy of an electric dipole placed in an electric field

The potential energy of an electric dipole $\vec{p}$ placed in an electric field $\vec{E}$ is given by
\begin{align}
U=-\vec{p}\cdot\vec{E}
\end{align}

## Problems from IIT JEE

**Problem (IIT JEE 2003):**
A positive point charge $q$ is fixed at origin. A dipole with a dipole moment $\vec{p}\;$ is placed along the $x$-axis far away from the origin with $\vec{p}$ pointing along positive $x$-axis. Find (a) the kinetic energy of the dipole when it reaches a distance $r$ from the origin and (b) force experienced by the charge $q$ at this moment.

**Solution:**
Total energy of a dipole $\vec{p}=p\,\hat\imath$ when it is far away from the charge $q$, is zero. Now, this dipole is placed in the electric field of charge $q$. The electric field of charge $q$ and the potential energy of the dipole are given by
\begin{align}
\vec{E}_q & =\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\,\hat\imath,\nonumber\\
U &=-\vec{p}\cdot\vec{E}_q \nonumber\\
&=-\frac{qp}{4\pi\epsilon_0r^2}. \nonumber
\end{align}
The conservation of energy, $K+U=0$, gives dipole kinetic energy as $K=-U={qp}/{(4\pi\epsilon_0r^2)}$. The electric field at the origin by the dipole and force on charge $q$ are
\begin{align}
\vec{E}_p&=\frac{2p}{4\pi\epsilon_0 r^3}\,\hat\imath, \nonumber\\
\vec{F}_q &=q\vec{E}_p \nonumber\\
&=\frac{2pq}{4\pi\epsilon_0 r^3}\,\hat\imath.\nonumber
\end{align}

## Related

- Electric field
- Electric potential