next up previous
Next: Charge Sheets and Dipole Up: Electrostatic Fields Previous: Electrostatic Energy


Electric Dipoles

Consider a charge $ q$ located at position vector $ {\bf r}'$ , and a charge $ -q$ located at position vector $ {\bf r}'-{\bf d}$ . In the limit that $ \vert{\bf d}\vert\rightarrow 0$ , but $ \vert q\vert\,\vert{\bf d}\vert$ remains finite, this combination of charges constitutes an electric dipole, of dipole moment

$\displaystyle {\bf p} = q\,{\bf d},$ (195)

located at position vector $ {\bf r}'$ . We have seen that the electric field generated at point $ {\bf r}$ by an electric charge $ q$ located at point $ {\bf r}'$ is

$\displaystyle {\bf E}({\bf r}) = -\nabla\left(\frac{q}{4\pi\,\epsilon_0}\,\frac{1}{\vert{\bf r}-{\bf r}'\vert}\right).$ (196)

Hence, the electric field generated at point $ {\bf r}$ by an electric dipole of moment $ {\bf p}$ located at point $ {\bf r}'$ is

$\displaystyle {\bf E}({\bf r}) = -\nabla\left(\frac{q}{4\pi\,\epsilon_0}\,\frac...
...frac{q}{4\pi\,\epsilon_0}\,\frac{1}{\vert{\bf r}-{\bf r}'+{\bf d}\vert}\right).$ (197)

However, in the limit that $ \vert{\bf d}\vert\rightarrow 0$ ,

$\displaystyle \frac{1}{\vert{\bf r}-{\bf r}'+{\bf d}\vert} \simeq \frac{1}{\ver...
...vert}-\frac{{\bf d}\cdot ({\bf r}-{\bf r}')}{\vert{\bf r}-{\bf r}'\vert^{\,3}}.$ (198)

Thus, the electric field due to the dipole becomes

$\displaystyle {\bf E}({\bf r}) =-\nabla\left(\frac{1}{4\pi\,\epsilon_0}\,\frac{{\bf p}\cdot({\bf r}-{\bf r}')}{\vert{\bf r}-{\bf r}'\vert^{\,3}}\right).$ (199)

It follows from Equation (154) that the scalar electric potential due to the dipole is

$\displaystyle \phi({\bf r}) = \frac{1}{4\pi\,\epsilon_0}\,\frac{{\bf p}\cdot({\...
...silon_0}\,{\bf p}\cdot\nabla'\left(\frac{1}{\vert{\bf r}-{\bf r}'\vert}\right).$ (200)

(Here, $ \nabla'$ is a gradient operator expressed in terms of the components of $ {\bf r}'$ , but independent of the components of $ {\bf r}$ .) Finally, because electric fields are superposable, the electric potential due to a volume distribution of electric dipoles is

$\displaystyle \phi({\bf r}) = \frac{1}{4\pi\,\epsilon_0}\int {\bf P}({\bf r'})\cdot\nabla'\left(\frac{1}{\vert{\bf r}-{\bf r}'\vert}\right) dV',$ (201)

where $ {\bf P}({\bf r})$ is the electric polarization (i.e., the electric dipole moment per unit volume), and the integral is over all space.


next up previous
Next: Charge Sheets and Dipole Up: Electrostatic Fields Previous: Electrostatic Energy
Richard Fitzpatrick 2014-06-27