27 May 2025
Electric Dipole and Quadrupole Fields
Far-field multipole expansion and the potentials and electric fields of electric dipoles and axial quadrupoles.
For localized point charges $q_a$ at positions $\mathbf r^{\prime}_a$,
\[V(\mathbf r)=\frac{1}{4\pi\epsilon_0} \sum_a\frac{q_a}{\lvert\mathbf r-\mathbf r^{\prime}_a\rvert}.\]When $r\gg r^{\prime}_a$, put $u=r^{\prime}_a/r$ and let $\gamma_a$ be the angle between $\mathbf r$ and $\mathbf r^{\prime}_a$. Then
\[\frac{1}{\lvert\mathbf r-\mathbf r^{\prime}_a\rvert} =\frac{1}{r}\left(1-2u\cos\gamma_a+u^2\right)^{-1/2}.\]Keeping terms through $u^2$,
\[\frac{1}{\lvert\mathbf r-\mathbf r^{\prime}_a\rvert} =\frac{1}{r} +\frac{r^{\prime}_a\cos\gamma_a}{r^2} +\frac{r_a^{\prime 2}}{2r^3}\left(3\cos^2\gamma_a-1\right) +O(r^{-4}).\]Thus the first three contributions are the monopole, dipole, and quadrupole potentials:
\[V(\mathbf r)=\frac{1}{4\pi\epsilon_0} \left[ \frac{Q}{r} +\frac{\mathbf p\cdot\hat{\mathbf r}}{r^2} +\frac{1}{2r^3}\sum_{i,j}Q_{ij}\hat r_i\hat r_j +\cdots \right],\]where
\[Q=\sum_aq_a,\qquad \mathbf p=\sum_aq_a\mathbf r^{\prime}_a,\]and, with the traceless convention,
\[Q_{ij}=\sum_aq_a \left(3x^{\prime}_{a,i}x^{\prime}_{a,j}-r_a^{\prime 2}\delta_{ij}\right).\]The dipole moment has SI unit $\mathrm{C\,m}$, and $Q_{ij}$ has unit $\mathrm{C\,m^2}$.
Electric dipole
Place $-q$ and $+q$ a distance $d$ apart, and let $\mathbf d$ point from $-q$ to $+q$. The dipole moment is
\[\mathbf p=q\mathbf d.\]The total charge is zero, so the leading far-field potential is
\[\boxed{V_{\mathrm{dip}}(\mathbf r) =\frac{1}{4\pi\epsilon_0} \frac{\mathbf p\cdot\hat{\mathbf r}}{r^2}}, \qquad r\gg d.\]Choose $\mathbf p=p\hat{\mathbf z}$. Then
\[V_{\mathrm{dip}}(r,\theta) =\frac{p\cos\theta}{4\pi\epsilon_0r^2}.\]In spherical coordinates,
\[E_r=-\frac{\partial V}{\partial r} =\frac{2p\cos\theta}{4\pi\epsilon_0r^3},\] \[E_\theta=-\frac{1}{r}\frac{\partial V}{\partial\theta} =\frac{p\sin\theta}{4\pi\epsilon_0r^3}, \qquad E_\phi=0.\]Since $\mathbf p=p(\cos\theta\,\hat{\mathbf r}-\sin\theta\,\hat{\boldsymbol\theta})$, the two components combine to
\[\boxed{\mathbf E_{\mathrm{dip}}(\mathbf r) =\frac{1}{4\pi\epsilon_0r^3} \left[3(\mathbf p\cdot\hat{\mathbf r})\hat{\mathbf r}-\mathbf p\right]}.\]On the axis $(\theta=0)$, $E=2p/(4\pi\epsilon_0r^3)$; on the equatorial plane $(\theta=\pi/2)$, the field has magnitude $p/(4\pi\epsilon_0r^3)$ and points opposite to $\mathbf p$.
Axial electric quadrupole
Take $+q$ at $z=\pm a$ and $-2q$ at the origin. Both $Q$ and $\mathbf p$ vanish. The second-order term from each outer charge is
\[\frac{1}{4\pi\epsilon_0} \frac{qa^2}{2r^3}\left(3\cos^2\theta-1\right).\]Adding the two equal terms gives
\[\boxed{V_{\mathrm{quad}}(r,\theta) =\frac{qa^2}{4\pi\epsilon_0r^3} \left(3\cos^2\theta-1\right)}, \qquad r\gg a.\]Let $A=qa^2/(4\pi\epsilon_0)$. Direct differentiation gives
\[E_r=-\frac{\partial}{\partial r} \left[\frac{A}{r^3}(3\cos^2\theta-1)\right] =\frac{3A}{r^4}(3\cos^2\theta-1),\]and
\[E_\theta=-\frac{1}{r}\frac{\partial}{\partial\theta} \left[\frac{A}{r^3}(3\cos^2\theta-1)\right] =\frac{6A}{r^4}\sin\theta\cos\theta.\]Therefore
\[\boxed{\mathbf E_{\mathrm{quad}} =\frac{qa^2}{4\pi\epsilon_0r^4} \left[ 3(3\cos^2\theta-1)\hat{\mathbf r} +6\sin\theta\cos\theta\,\hat{\boldsymbol\theta} \right]}.\]The quadrupole potential decreases as $r^{-3}$, one power faster than the dipole potential.
Solved Problems
1. Dipole field at an off-axis point
An electric dipole $\mathbf p=p\hat{\mathbf z}$ is at the origin. Find its field at the point $(s,0,s)$.
Solution. At this point,
\[r=\sqrt2s,\qquad \hat{\mathbf r}=\frac{\hat{\mathbf x}+\hat{\mathbf z}}{\sqrt2}, \qquad \mathbf p\cdot\hat{\mathbf r}=\frac{p}{\sqrt2}.\]Insert these in the invariant dipole field:
\[\mathbf E=\frac{1}{4\pi\epsilon_0r^3} \left[3(\mathbf p\cdot\hat{\mathbf r})\hat{\mathbf r}-\mathbf p\right].\]The vector in brackets becomes
\[\frac{3p}{2}(\hat{\mathbf x}+\hat{\mathbf z})-p\hat{\mathbf z} =\frac{p}{2}(3\hat{\mathbf x}+\hat{\mathbf z}).\]Since $r^3=2\sqrt2s^3$,
\[\boxed{\mathbf E(s,0,s)= \frac{p}{16\sqrt2\pi\epsilon_0s^3} (3\hat{\mathbf x}+\hat{\mathbf z})}.\]Differentiating $V=pz/[4\pi\epsilon_0(x^2+z^2)^{3/2}]$ gives the same two components.
2. Nodal cone of an axial quadrupole
For $V=A(3\cos^2\theta-1)/r^3$, find the cone on which $V=0$ and determine the field there.
Solution. The nodal condition is
\[3\cos^2\theta_0-1=0 \quad\Longrightarrow\quad \boxed{\cos\theta_0=\pm\frac{1}{\sqrt3}}.\]On either cone the radial component
\[E_r=\frac{3A}{r^4}(3\cos^2\theta-1)\]vanishes. In the upper hemisphere, $\sin\theta_0=\sqrt{2/3}$, so
\[E_\theta=\frac{6A}{r^4}\sin\theta_0\cos\theta_0 =\boxed{\frac{2\sqrt2A}{r^4}}.\]Thus zero potential on the cone does not mean zero electric field; the field is tangent to the sphere there.
Descriptive Questions
- Why does the leading nonzero multipole term depend on the lower moments that vanish?
- How do the radial dependences of monopole, dipole, and quadrupole potentials differ?
- Why is the dipole field on the equatorial plane opposite to the dipole moment?
- What physical information is encoded in the traceless quadrupole tensor?
Numerical Problems
1. Axial dipole field
A dipole has $p=2.00\times10^{-8}\,\mathrm{C\,m}$. Find its field on the axis at $r=0.300\,\mathrm m$.
Answer: $E=1.33\times10^4\,\mathrm{N\,C^{-1}}$, along $\mathbf p$.
2. Equatorial dipole field
For the same dipole, find the field magnitude on its equatorial plane at $r=0.200\,\mathrm m$.
Answer: $E=2.25\times10^4\,\mathrm{N\,C^{-1}}$, opposite to $\mathbf p$.
3. Axial quadrupole potential
An axial quadrupole has outer charges $q=5.00\,\mathrm{nC}$ at $z=\pm a$, with $a=1.00\,\mathrm{cm}$. Find the far-field potential at $r=0.200\,\mathrm m$ on its axis.
Answer: $V=1.12\,\mathrm V$.
The symbolic solutions and all printed numerical answers are verified in the Unit I multipoles worksheet.
References
- Multipole expansion: Wikipedia
- David J. Griffiths, Introduction to Electrodynamics, 4th ed., Cambridge University Press, 2017.
- John D. Jackson, Classical Electrodynamics, 3rd ed., Wiley, 1998.
- Edward M. Purcell and David J. Morin, Electricity and Magnetism, 3rd ed., Cambridge University Press, 2013.
Discussion