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Electric Dipole

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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}

electric-dipole

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}

potential-of-an-electric-dipole

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}

field-of-an-electric-dipole
electric field lines of dipole

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

  1. Electric field
  2. Electric potential
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