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 $s \rightarrow 0$.

$${\bf m} = e{\bf s},$$ (3.14)

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

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

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

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

Hence,

$$4\pi \epsilon F = {xes\over r^{3}} = {es\cos \theta\over r^{2}} = {m\cos \theta\over r^{2}}.$$ (3.15)

The resulting electric field components are given by evaluating ${\bf E} = - \nabla F$ in spherical coordinates. Hence

$$E_{r} = {2m\cos \theta\over 4\pi \epsilon r^{3}}$$

$$E_{\theta} = {m\sin \theta \over 4\pi \epsilon r^{3}}$$ (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.