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Electric Dipoles

Consider a pair of charges $\pm e$ a small distance ${\bf s}$ apart (${\bf s}$ is the vector joining the negative charge to the positive charge and is illustrated in Figure 3.5) such that we can define

Figure 3.5: A pair of opposite charges are a distance $s$ apart. A dipole is formed by letting $s \rightarrow 0$.
\includegraphics [scale=0.7]{fundfig33a.eps}


\begin{displaymath}
{\bf m} = e{\bf s},
\end{displaymath} (3.14)

which is called the dipole moment. The resulting dipole potential is

\begin{displaymath}
4\pi \epsilon F = -{e\over r} + {e\over \vert{\bf r} - {\bf...
...r (x^{2} + y^{2})^{1/2}} + {e\over ((x-s)^{2} + y^{2})^{1/2}}.
\end{displaymath}

If we assume that $s$ tends to zero, we may expand the second term in a Taylor series for small $s$. Thus, we obtain

\begin{displaymath}
4\pi \epsilon F \approx -{e\over r} + {e\over (x^{2} +
y^...
...1/2}} = {e\over r}\left
[-1 + (1 + {xs\over r^{2}})\right ].
\end{displaymath}

Hence,
\begin{displaymath}
4\pi \epsilon F = {xes\over r^{3}} = {es\cos \theta\over r^{2}} =
{m\cos \theta\over r^{2}}.
\end{displaymath} (3.15)

The resulting electric field components are given by evaluating ${\bf E} = - \nabla F$ in spherical coordinates. Hence
$\displaystyle E_{r}$ $\textstyle =$ $\displaystyle {2m\cos \theta\over 4\pi \epsilon r^{3}}$  
$\displaystyle E_{\theta}$ $\textstyle =$ $\displaystyle {m\sin \theta \over 4\pi \epsilon r^{3}}$ (3.16)

Thus, ${\bf E}$ falls off in magnitude like $r^{-3}$ and $E_{\theta}$ vanishes when $\theta = 0, \pi$. The electric field for a dipole is shown in Figure 3.6.

Figure 3.6: The electric field lines for an electric dipole.
\includegraphics [scale=0.7]{fundfig34.eps}


next up previous
Next: Magnetic Fields Up: Electrostatics Previous: Continuous distribution of charge,
Prof. Alan Hood
2000-11-06