# Electric Potential Energy of Electric Dipole in a Uniform Electrostatic Field

## 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, && U=-\vec{p}\cdot\vec{E}_q=-\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, && \vec{F}_q=q\vec{E}_p=\frac{2pq}{4\pi\epsilon_0 r^3}\,\hat\imath.\nonumber \end{align}