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Far-field expansions in monopole, dipole and quadrupole terms, dipole potentials and fields, and the torque, energy and force on dipoles in fields.
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By the end of this lesson you will be able to compute dipole moments, write the leading far-field potential and field of a localized charge distribution, and find the torque and energy of a dipole in a field.
You can compute potentials of charge distributions and fields as gradients, and you know Taylor series from calculus. From chemistry you may know that water is a polar molecule. This lesson asks what a localized charge distribution looks like from far away, and finds that its total charge, then its dipole moment, then finer details appear in order.
| Term | What it means |
|---|---|
| Multipole expansion | The series $V = \sum_n V_n$ with $V_n \propto 1/r^{n+1}$ describing a distribution from far away. |
| Monopole moment | The total charge $Q$. |
| Dipole moment | $\vec{p} = \sum q_i\vec{r}_i$ or $\int\vec{r}\,\rho\,d\tau$; for $\pm q$ separated by $\vec{d}$, $\vec{p} = q\vec{d}$. |
| Quadrupole | The next term, which dominates when both charge and dipole moment vanish. |
| Pure dipole | The limit $d \to 0$ with $p = qd$ fixed. |
| Polar molecule | A molecule with a permanent dipole moment, such as water. |
| Debye | A unit of molecular dipole moment, $3.336 \times 10^{-30}$ C m; water's is $1.85$ D. |
Expand $1/\mathscr{r}$ in powers of $r'/r$, where $r'$ locates a source charge and $r$ the distant field point. The potential becomes
$$V(\vec{r}) = k\left[\frac{Q}{r} + \frac{\vec{p} \cdot \hat{r}}{r^2} + \frac{1}{r^3}(\text{quadrupole}) + \cdots\right],$$
with total charge $Q = \sum q_i$ and dipole moment $\vec{p} = \sum q_i\vec{r}_i$. Each term falls one power of $r$ faster than the last, so from far away the first nonzero term dominates. For a neutral distribution that is the dipole term,
$$V_{\text{dip}} = \frac{kp\cos\theta}{r^2}, \qquad \vec{E}_{\text{dip}} = \frac{kp}{r^3}\left(2\cos\theta\,\hat{r} + \sin\theta\,\hat{\theta}\right).$$
When the total charge is zero, $\vec{p}$ does not depend on where the origin is. In a uniform external field, a dipole feels no net force but a torque $\vec{\tau} = \vec{p} \times \vec{E}$, and its energy is $U = -\vec{p} \cdot \vec{E}$, lowest when aligned. In a nonuniform field it also feels a force $\vec{F} = (\vec{p} \cdot \nabla)\vec{E}$, pulling it toward stronger field when aligned.
Another way: picture
Picture a distant charge distribution as a blurry blob. From very far away, you see only its total charge. Come closer and you notice that its positive and negative charges are slightly separated, a dipole, which makes the field stronger on one side. Closer still, finer lumpiness appears — quadrupole, octupole — each washed out more quickly with distance. The expansion lists the blob's features in the order you would notice them.
Another way: steps
Checks. A symmetric distribution has zero dipole moment: a uniform sphere, a pair of equal charges, a charge centered in a ring. A neutral distribution's dipole moment must not depend on the origin. The dipole field on the axis must be twice that on the bisector at the same distance. And the multipole potential must approach the exact potential as $r$ grows.
Each term of the expansion measures a finer detail of how charge is arranged. The monopole asks only how much charge there is. The dipole asks how far the average positive charge is from the average negative charge. The quadrupole asks how the charge is spread about those centers. From a distance $r$, a detail of size $d$ subtends an angle $d/r$, and each finer level of detail brings another factor of $d/r$.
This is why a neutral atom barely affects a distant charge, and why molecules interact at short range. It also explains why distant antennas can be modeled as dipoles: the radiation from any small, time-varying charge distribution is dominated by its oscillating dipole moment, the subject of the radiation lessons at the end of the course.
Differentiating $V = kp\cos\theta/r^2$ gives $E_r = 2kp\cos\theta/r^3$ and $E_\theta = kp\sin\theta/r^3$. On the axis the field points along $\vec{p}$ with strength $2kp/r^3$; on the bisector it points opposite to $\vec{p}$ with strength $kp/r^3$. Field lines leave the positive end, loop around, and return to the negative end.
Earth's magnetic field is, to a first approximation, a dipole field of exactly this shape, twice as strong at the magnetic poles as at the equator. The same geometry governs the electric fields of molecules, of the heart during each beat, and of the antennas in phones. Recognizing the $2:1$ ratio and the $1/r^3$ falloff is a quick way to tell when a field is dipolar.
In a uniform field the two ends of a dipole feel equal and opposite forces, so there is no net force but a torque $pE\sin\theta$ that turns the dipole toward the field. Its potential energy, $-pE\cos\theta$, is lowest when aligned and highest when anti-aligned, a difference of $2pE$.
In a nonuniform field the ends feel different forces, and an aligned dipole is pulled toward the stronger field. That is why a charged comb attracts neutral bits of paper: the comb's field polarizes the paper, creating dipoles aligned with the field, which are then pulled into the stronger field near the comb. Dielectrophoresis uses the same force to sort living cells in microfluidic chips, because different cell types polarize differently.
Many molecules have permanent dipole moments because their bonds share electrons unequally. Water's is $6.2 \times 10^{-30}$ C m, equivalent to a full electron charge separated by $0.039$ nm, a tenth of the molecule's size. Carbon dioxide, symmetric and straight, has none; ammonia has $1.47$ D.
Dipole moments shape chemistry and biology. They make water an excellent solvent for ions, which it surrounds with oriented molecules; they hold together the hydrogen bonds that fold proteins and pair the bases of DNA. And they let microwave radiation heat food: the oscillating field twists water's dipoles back and forth, and friction with neighbors converts that motion to heat.
At room temperature, $k_BT = 4.1 \times 10^{-21}$ J. The orientation energy of a water molecule in a field of $1$ MV/m is $pE = 6.2 \times 10^{-24}$ J, a thousand times smaller. So even strong fields align water molecules only slightly; the average alignment is about $pE/3k_BT$, a fraction of a percent. Yet because there are so many molecules, that slight alignment produces the large polarization behind water's dielectric constant of about $80$.
The balance between field and temperature gives the Curie law for dielectrics: the polarization of polar molecules falls as $1/T$. Only in fields approaching $10^9$ V/m, such as those very close to an ion, does alignment become nearly complete, a condition called dielectric saturation that chemists must account for when modeling ions in water.
When both charge and dipole moment vanish, the quadrupole term leads. Two dipoles placed tail to tail, or a linear arrangement $+q, -2q, +q$, make a quadrupole whose potential falls as $1/r^3$. Carbon dioxide, with no dipole moment, has a strong quadrupole moment, which governs how its molecules pack and interact.
Nuclei, too, have quadrupole moments when their charge is not spherical, and measuring them reveals nuclear shapes: many heavy nuclei are elongated like footballs. The electric quadrupole field inside an ion trap, the lens elements of an electron microscope and the mass filters of quadrupole mass spectrometers used in environmental and drug-testing labs all rely on the quadrupole's distinctive $x^2 - y^2$ shape.
Atoms and nonpolar molecules have no permanent dipole moment, but a field pushes their electron clouds one way and their nuclei the other, creating an induced dipole proportional to the field, $\vec{p} = \alpha\vec{E}$. The constant $\alpha$, the polarizability, measures how easily the cloud deforms; it grows with the size of the atom, because outer electrons in large atoms are loosely held. For hydrogen, $\alpha/4\pi\varepsilon_0$ is about $0.67 \times 10^{-30}$ m³, roughly the volume of the atom.
Induced dipoles explain why neutral atoms attract ions, and why even two neutral atoms attract each other weakly: fluctuations in one atom's cloud induce dipoles in the other, producing the van der Waals force that holds noble gases together as liquids at low temperature and lets geckos cling to walls. The next lesson builds the theory of dielectrics on exactly this response of matter to a field.
Water's dielectric constant of about $80$ means that the attraction between two ions in water is eighty times weaker than in vacuum. That is why table salt, held together by strong ionic bonds, falls apart in water. The reason is dipoles: water molecules rotate so that their negative oxygen ends face positive sodium ions and their positive hydrogen ends face chloride ions, and their aligned dipole fields largely cancel the ions' fields.
Each molecule's alignment is slight, because the orientation energy $pE$ is small compared with $k_BT$ except very close to an ion. But in the first layer around an ion, where fields exceed $10^9$ V/m, the molecules are strongly oriented, forming a hydration shell. Chemists at national laboratories model these shells with molecular dynamics simulations to design battery electrolytes and understand how minerals dissolve in groundwater.
A microwave oven fills its cavity with a $2.45$ GHz electromagnetic field. The oscillating electric field exerts torques $\vec{p} \times \vec{E}$ on water molecules, trying to turn them back and forth five billion times a second. In liquid water, hydrogen bonds to neighbors resist the turning, and the molecules lag behind the field; the work done by the field against that friction becomes heat.
The frequency is a compromise. Much higher frequencies are absorbed so strongly that they heat only the surface; much lower frequencies pass through with little absorption. At $2.45$ GHz microwaves penetrate a few centimeters into food, heating it from within. Ice, whose molecules are locked in a crystal and cannot rotate freely, absorbs far less, which is why frozen food thaws unevenly and why defrost settings cycle the power on and off.
Zero total charge removes only the leading term of the multipole expansion. If the positive and negative charges are separated, a dipole field remains, falling as $1/r^3$. Molecules, atoms in electric fields, the beating heart and radio antennas are all neutral and all produce measurable fields. Only when the charges coincide exactly, or are arranged with perfect symmetry, does the field vanish.
A second error is to apply the dipole approximation close to the distribution. The expansion is in powers of $d/r$; when the field point is as close as the size of the distribution, all terms matter, and the exact potential must be used.
Place $+q$ at $z = d/2$ and $-q$ at $z = -d/2$.
$Q = 0, \qquad p = q\frac{d}{2} + (-q)\left(-\frac{d}{2}\right) = qd$
Neutral, with dipole moment $qd$ along $z$.
Write the exact potential.
$V = kq\left(\dfrac{1}{\mathscr{r}_+} - \dfrac{1}{\mathscr{r}_-}\right)$
Two point charges.
Approximate the distances for $r \gg d$.
$\dfrac{1}{\mathscr{r}_\pm} \approx \dfrac{1}{r}\left(1 \pm \dfrac{d}{2r}\cos\theta\right)$
First-order expansion.
Subtract the two terms.
$V \approx \dfrac{kq\,d\cos\theta}{r^2}$
The $1/r$ terms cancel.
Write it with the dipole moment.
$V_{\text{dip}} = \dfrac{kp\cos\theta}{r^2}$
Only the product $qd$ matters far away.
Take the radial derivative.
$E_r = -\dfrac{\partial V}{\partial r} = \dfrac{2kp\cos\theta}{r^3}$
$V \propto r^{-2}$.
Take the angular derivative.
$E_\theta = -\dfrac{1}{r}\dfrac{\partial V}{\partial\theta} = \dfrac{kp\sin\theta}{r^3}$
The spherical gradient.
Evaluate on the axis.
$\theta = 0: \quad \vec{E} = \dfrac{2kp}{r^3}\hat{z}$
Along the dipole.
Evaluate on the bisector.
$\theta = 90°: \quad \vec{E} = \dfrac{kp}{r^3}\hat{\theta} = -\dfrac{kp}{r^3}\hat{z}$
Opposite to the dipole.
Evaluate for water at $1$ nm on the axis.
$E = \dfrac{2 \times 8.99 \times 10^9 \times 6.2 \times 10^{-30}}{(10^{-9})^3} = 1.1 \times 10^{8}\ \text{V/m}$
Large at molecular distances.
Compare with a single electron at the same distance.
$\dfrac{ke}{r^2} = 1.4 \times 10^{9}\ \text{V/m}$
The dipole's field is an order of magnitude weaker, and falls faster.
Place $+3q$ at $z = a$, $-2q$ at the origin, $-q$ at $z = -a$. Find the total charge.
$Q = 3q - 2q - q = 0$
No monopole term.
Find the dipole moment.
$p = 3qa + 0 + (-q)(-a) = 4qa$
Along $+z$.
Write the leading potential.
$V \approx \dfrac{4kqa\cos\theta}{r^2}$
The dipole term.
Check independence of origin.
$\text{shift by } c: \ p' = \sum q_i(z_i - c) = p - cQ = p$
Because $Q = 0$.
Evaluate on the axis for $q = 1$ nC, $a = 1$ mm, $z = 5$ cm.
$V \approx \dfrac{4 \times 8.99 \times 10^9 \times 10^{-9} \times 10^{-3}}{(0.05)^2} = 14.4\ \text{V}$
The dipole approximation.
Compute the exact potential.
$V = 8.99\left(\dfrac{3}{0.049} - \dfrac{2}{0.050} - \dfrac{1}{0.051}\right) = 14.5\ \text{V}$
Summing the three point charges.
Compare the two.
$\dfrac{14.5 - 14.4}{14.4} \approx 1\%$
The quadrupole correction, of order $a/z$ times a small factor.
Identify the quadrupole's source.
$\text{the unequal charges at } \pm a$
They are not a pure dipole pair.
Write the definition.
$p = \sum q_ix_i$
Charge times position, summed.
Substitute the values.
$p = 4 \times 3 + (-4) \times (-2)$
In nC cm.
Evaluate the moment.
Far from a small electric dipole, the field at some point is $5456$ N/C. What is the field at twice the distance, in the same direction?
Complete the worked solution: $+9$ nC sits at $x = 4$ cm and $-9$ nC at $x = -6$ cm. Find each charge's contribution $q_ix_i$ and the dipole moment, in nC cm.
Multiply the positive charge by its position.
$q_1x_1 = 9 \times 4 =$ a
Positive charge at positive $x$.
Multiply the negative charge by its position.
$q_2x_2 = (-9) \times (-6) =$ b
Two negatives make a positive contribution.
Add the contributions.
$p =$ c
The charge times the separation, $9 \times 10$.
Match each term or effect to its dependence.
| $\propto 1/r$ | $\propto 1/r^2$ | $\propto 1/r^3$ | $\vec{p} \times \vec{E}$ | |
|---|---|---|---|---|
| monopole potential | ||||
| dipole potential | ||||
| quadrupole potential | ||||
| torque on a dipole |
A small dipole has $kp/r^3 = 448$ N/C at a distance $r$. Fill in the field, in N/C, on its axis and on its perpendicular bisector at distances $r$ and $2r$.
| on the axis (N/C) | on the bisector (N/C) | |
|---|---|---|
| distance $r$ | ||
| distance $2r$ |
A water molecule, dipole moment $p = 6.2 \times 10^{-30}$ C m, sits in a field of $9$ MV/m with its dipole at $30°$ to the field. What torque acts on it, in units of $10^{-24}$ N m?
Answer: × 10⁻²⁴ N m
Charges $+3q$ at $z = a$, $-2q$ at the origin and $-q$ at $z = -a$, with $q = 7$ nC and $a = 5$ mm, sit on the $z$ axis. Estimate the potential at $z = 2$ cm on the axis, far from the charges, in volts. Use $k = 8.99 \times 10^9$ N m²/C².
Answer: V
Water's large dielectric constant comes from its molecules partly aligning with an applied field. In a field of $30$ MV/m at $300$ K, what is the ratio of the largest orientation energy of a water molecule ($p = 6.2 \times 10^{-30}$ C m) to the thermal energy $k_BT$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A small dipole has $kp/r^3 = 162$ N/C at a distance $r$. Fill in the field, in N/C, on its axis and on its perpendicular bisector at distances $r$ and $2r$.
| on the axis (N/C) | on the bisector (N/C) | |
|---|---|---|
| distance $r$ | ||
| distance $2r$ |
You can describe distributions by their multipoles. Explain to someone why a neutral molecule can still attract a nearby ion.
19. Your turn: what is the dipole moment of $+4$ nC at $x = 3$ cm and $-4$ nC at $x = -2$ cm?, step 3
$p = 20\ \text{nC cm}$
Charge times separation, $4 \times 5$.