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core34 min studyLesson 12 in Physics

Gauss's law and multipole expansion

Separate exact flux constraints from symmetry arguments and estimate distant fields with controlled multipole approximations.

Before you begin
On this page

The question for today

What you’ll understand

  • Identify when Gauss's law determines a local field.
  • Compute monopole and dipole contributions to a distant potential.
  • Estimate the error and angular limitations of a multipole truncation.

A closed surface counts enclosed charge through electric flux, but it does not usually tell you the field at every point on that surface. Symmetry supplies the missing information in special cases. For a complicated localized source seen from far away, a different simplification becomes useful: organize its field by progressively finer features of the charge distribution.

Flux is an integral statement

Electric flux measures the normal field integrated over an oriented surface. A charge outside a closed surface can produce a large field on it while contributing zero net flux: field entering one side leaves another. Conversely, zero enclosed charge does not imply that every normal field component vanishes. Gauss's law constrains the sum, not each contribution separately.

For a spherically symmetric isolated charge distribution, rotations require the field to be radial and to have the same magnitude everywhere on a centered sphere. Only then may that magnitude be removed from the surface integral. Cylindrical and planar symmetry support other useful surfaces, but a rectangular charged object does not become symmetric merely because one draws a spherical surface around it.

∮SE⋅dA=Qencϵ0,Er(r)=Qenc(r)4πϵ0r2(spherical symmetry)\oint_S\mathbf E\cdot d\mathbf A=\frac{Q_{\mathrm{enc}}}{\epsilon_0},\qquad E_r(r)=\frac{Q_{\mathrm{enc}}(r)}{4\pi\epsilon_0r^2}\quad\text{(spherical symmetry)}
The flux equation is general in electrostatics in vacuum; the expression for a pointwise radial field requires the stated symmetry. Charge is measured in coulombs, and the resulting field in N/C.↔ Scroll wide equations horizontally.

A distant observer resolves moments

Let all source charges lie within a radius a of the chosen origin and observe at distance r greater than a. Expanding the Coulomb kernel in powers of source position divided by r gives a convergent exterior multipole series. The monopole depends only on total charge. The dipole records the first spatial moment, so its potential has angular structure and decreases one additional power of distance.

For a neutral source the dipole moment does not depend on a translation of origin; for a charged source it does. This is not a physical ambiguity because the whole expansion transforms together. Truncating after the dipole is controlled by a/ra/r, provided one compares nonvanishing terms. On a nodal direction where a leading term is zero, relative error can be large even when the absolute omitted field is small.

ϕ(r)=14πϵ0[Qr+p⋅r^r2+⋯ ],p=∫r′ρ(r′) d3r′\phi(\mathbf r)=\frac{1}{4\pi\epsilon_0}\left[\frac Qr+\frac{\mathbf p\cdot\hat{\mathbf r}}{r^2}+\cdots\right],\qquad \mathbf p=\int\mathbf r'\rho(\mathbf r')\,d^3r'
This expansion assumes a localized source and observation outside its enclosing sphere. Dipole moment has units C·m. The dots contain quadrupole and higher angular moments, not a second independent monopole.↔ Scroll wide equations horizontally.

Differentiate before interpreting a zero

For a dipole along the vertical axis, potential is proportional to cosine of polar angle. Its equatorial potential vanishes, but its electric field does not: a spatial derivative of a function can be nonzero where the function itself is zero. Along the axis the field points with the dipole moment; in the equatorial plane it points opposite to it. A scalar contour and a vector field answer different questions.

An ideal point dipole is an exact mathematical source with a singular center. Two finite separated charges reproduce its leading distant field, with corrections near the pair. A dipole approximation is therefore a statement about observational scale, not a claim that the actual charges occupy the same point. Always compare the observation distance with the charge separation before using its inverse-cube field.

Edip=3(p⋅r^)r^−p4πϵ0r3\mathbf E_{\mathrm{dip}}=\frac{3(\mathbf p\cdot\hat{\mathbf r})\hat{\mathbf r}-\mathbf p}{4\pi\epsilon_0r^3}
The formula is exact away from the origin for an ideal electrostatic point dipole, and leading order for a small neutral pair at large distance. The singular contact contribution at the origin is outside this exterior formula.↔ Scroll wide equations horizontally.

Work it through

How accurate is a small dipole?

Charges ±2.0 nC\pm2.0\,\mathrm{nC} are separated by 1.0 cm. Find the potential 10 cm from their midpoint along the positive-charge axis, using 1/(4πϵ0)=9.0×109 N m2/C21/(4\pi\epsilon_0)=9.0\times10^9\,\mathrm{N\,m^2/C^2}.

  1. The dipole moment magnitude is p=qd=(2.0×10−9)(0.010)=2.0×10−11 C mp=qd=(2.0\times10^{-9})(0.010)=2.0\times10^{-11}\,\mathrm{C\,m}. The distance ratio is d/r=0.10d/r=0.10, making a distant approximation plausible but worth checking.

  2. On the positive axis, p⋅r^=p\mathbf p\cdot\hat{\mathbf r}=p. The dipole estimate is ϕ=kp/r2=18.0 V\phi=kp/r^2=18.0\,\mathrm V, and the axial field estimate is 2kp/r3=360 N/C2kp/r^3=360\,\mathrm{N/C}.

  3. The exact potential is kq[1/(0.100−0.005)−1/(0.100+0.005)]=18.0451 Vkq[1/(0.100-0.005)-1/(0.100+0.005)]=18.0451\,\mathrm V. The approximate potential is about 0.25% below the exact result; symmetry removes the next even-power charge moment for this pair.

The result

The potential is approximately 18 V, with a directly checked 0.25% relative error in this geometry. A generic source need not share the pair's cancellation of its quadrupole moment.

The living notebook

Explore the model

Interactive lab

Interactive lab · fields

Flux-like source structure and circulation

Adjust source strength q and rotation Ω in the softened two-dimensional field q r/(r²+ε) + Ω(−y,x). Setting Ω to zero exposes radial structure; adding rotation shows why flux and circulation differ. This is not a Coulomb dipole, a three-dimensional Gaussian surface, or a multipole solver.

Flux-like source structure and circulationThe softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Current readings: Fx at (1, 0): 0.833; Fy at (1, 0): 0.5; Field magnitude there: 0.972. All parameters are adjustable with the labeled controls below.−3.2−1.601.63.2−1.5−0.7500.751.5xyq = 1
Local field directionSource strength
Fx at (1, 0)
0.833
Fy at (1, 0)
0.5
Field magnitude there
0.972
t = 0 s

The softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA picture of arrows must be interpreted with its actual dimensionality and source law.

Pause & reason

Check your understanding

An electrostatic dipole has zero potential on its equatorial plane. What follows?

Your turn

Try it yourself

For the pair in the worked example, estimate the electric field at a point 10 cm from the midpoint in the equatorial plane. State its direction and the status of the approximation.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The dot product of the dipole moment with the equatorial radial unit vector vanishes.

Compare with the solution

The dipole formula becomes E=−p/(4πϵ0r3)\mathbf E=-\mathbf p/(4\pi\epsilon_0r^3). Its magnitude is 9.0×109(2.0×10−11)/(0.10)3=180 N/C9.0\times10^9(2.0\times10^{-11})/(0.10)^3=180\,\mathrm{N/C}, directed opposite to the dipole moment, from the positive-charge side toward the negative-charge side. This is a far-field estimate for finite separated charges; its error must be checked separately rather than copied from the axial potential calculation.

Keep the idea

Take it with you

  • Gauss's law becomes a pointwise shortcut only with sufficient symmetry.
  • Multipoles organize source information by angular structure and distance scaling.
  • A zero leading potential does not automatically imply a zero electric field.

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