Gauss's law
Define signed electric flux through an oriented closed surface; derive Gauss's law from Coulomb's point-source field and solid angle; distinguish total enclosed charge from the distribution of local electric field; solve spherical, cylindrical and planar symmetry cases, including a uniformly charged ball, an infinite line and an ideal sheet; state precisely when symmetry allows E to leave the surface integral.
- Define dA as outward normal oriented area and calculate signed electric flux with the correct N·m²/C units.
- Derive the point-charge flux through a sphere, then motivate the 4π solid-angle result for arbitrary closed surfaces.
- State the superposition/integral form ∮E·dA=Qencl/ε₀ and its distributional local form ∇·E=ρ/ε₀.
- Explain why Qencl=0 implies zero NET FLUX, not zero field at each surface point.
- Choose a Gaussian surface only after identifying the actual field symmetry; reject symmetry arguments that assume what must be proved.
- Derive fields inside and outside a uniformly charged solid sphere and contrast an infinitesimally thin spherical shell.
- Derive an infinite-line field using a coaxial cylinder with noncontributing end caps.
- Derive the two-sided field of an isolated uniformly charged infinite plane with a centered pillbox.
- Distinguish arbitrary spherical Gaussian surfaces containing off-center charges from spherically symmetric source distributions.
- Solve problems by checking the enclosed charge, outward normal signs, continuity across model interfaces and limits.
Gauss's law relates an integral over a CLOSED surface to its enclosed charge. It holds whether or not the field is simple. The much stronger statement that E can be taken outside the surface integral requires symmetry of the actual source distribution and the correct orientation of E.
The dashed equator is the BACK half of a closed spherical surface, not an open circular contour. Only +q is enclosed: ∯s E·n dA = +q/ε0. An outside charge can alter E locally but not the enclosed charge.
03.1Define signed flux before using a Gaussian surface
At each regular point of a smooth closed surface S, define the oriented area element dA=n̂ dA, with n̂ the OUTWARD unit normal. The electric flux is the scalar surface integral of the normal component:
E parallel to the surface contributes zero locally; E pointing outward contributes positively; E pointing inward contributes negatively. The signed flux is not the total number of drawn field lines, nor the integral of |E|. A patch can have zero flux even when its tangential field is large.
03.2Derive the centered point-charge sphere result
Put an ideal point charge q at the center of a sphere of radius R>0. Coulomb's field is E=kq r̂/R² and the outward normal is n̂=r̂ everywhere. Both magnitude and normal alignment are constant ON THIS SPHERE, so
The cancellation of R² comes from the inverse-square field together with the sphere's area. If q is negative, the flux is negative even though the geometric outward normal remains unchanged.
03.3From solid angle and superposition to arbitrary closed surfaces
The oriented flux element of a point source can be written dΦ=kq dΩ, where dΩ is the SIGNED solid-angle element seen from the charge. The total oriented solid angle is 4π for a closed surface enclosing the source and zero if the source lies outside. Deforming the surface does not change which case holds, provided it does not cross the source.
The last equation follows by linear superposition for point charges and by the continuum limit for density ρ. The enclosed charge is the algebraic sum/integral of charges INSIDE S, not the total charge in the universe. External sources may change E point by point on S while contributing zero net flux across it.
03.4The differential form and its distributional meaning
For a smooth volume charge density ρ and regular field, the divergence theorem turns the closed-surface flux into a volume integral:
For an ideal point charge, the field has zero ordinary divergence away from the charge yet nonzero flux around it. The source is represented distributionally by ρ=qδ³(r−r₀). For the radial kernel one can write ∇·[(r−r₀)/|r−r₀|³]=4πδ³(r−r₀) in the sense of distributions; evaluating the ordinary divergence only away from r₀ and concluding zero enclosed charge would be incorrect.
03.5A zero flux does not mean a zero field
Take a closed sphere placed inside a spatially uniform externally imposed field E₀. There is no source charge inside, so Qenclosed=0. Nevertheless E₀ is nonzero at every point: the inward and outward contributions to E₀·n̂ cancel in the surface integral. The same issue arises when a Gaussian sphere contains an off-center point charge: the total flux is q/ε₀, but field magnitude and direction need not be constant on that sphere.
Only after independently establishing a suitable symmetry can one replace the flux integral by a simple product E times an area.
03.6Spherical symmetry: uniformly charged solid ball
Consider a stationary ball of radius R carrying constant volume density ρ in vacuum. Its total charge is Q=ρ(4πR³/3). Rotational symmetry implies E=E(r)r̂. For a concentric spherical Gaussian surface of radius r, E·n̂=E(r) is constant, and the enclosed charge is different inside and outside the ball:
At r=R the inside/outside expressions agree for a uniform volume density with no singular surface sheet. At r=0 the field vanishes by symmetry. For an IDEAL thin spherical shell instead, rotational symmetry and zero enclosed charge give E=0 for r<R and E=kQ/r² r̂ for r>R; a jump at the charged sheet is treated by one-sided limits.
03.7Cylindrical symmetry: infinite uniform line
For an ideal infinite line with signed linear density λ on the z axis, translation/rotation symmetry makes E radial and independent of z and polar angle. Use a closed coaxial cylinder of radius s>0 and length L. The two end caps have normals parallel to ±ẑ and therefore zero flux; only the lateral surface contributes:
The line result matches the direct Coulomb integral of EM-01.01. Its radial 1/s dependence and absence of axial field rely on the ideal infinite uniform distribution; a finite line does not have the full required symmetry.
03.8Planar symmetry: an isolated infinite charged sheet
An ideal infinitely extended uniform charged sheet at z=0 has constant surface density σ. Reflection symmetry across the plane and translation/rotation symmetry within it imply equal-magnitude opposite-normal fields on its two sides. A centered pillbox with cap area A has total flux 2EA; its side contributes zero.
For a positive σ both fields point away from the sheet. This two-sided result is for an isolated ideal sheet in vacuum, not the outside field of a conducting surface whose inside field vanishes and whose geometry/boundary conditions differ; conductors are studied in Chapter 04.
03.9A checklist for choosing a Gaussian surface
- Identify the actual source distribution and its symmetries BEFORE selecting a surface.
- Write E's permitted direction and coordinate dependence from those symmetries.
- Choose a closed surface on which the normal component is constant or zero by a justified geometric argument.
- Compute Qenclosed carefully, including signs and the appropriate line/surface/volume density.
- Write ∮E·dA first; simplify it only after steps 2–3; then solve for the unknown field.
- Check units, one-sided surface limits, and agreement with direct integration or a known asymptotic regime.
Gauss's law always constrains the total flux, but without enough symmetry it may not determine the local field. A Gaussian sphere centered at the origin does not magically make an arbitrary off-center charge distribution radially symmetric.
03.10Mastery exercises and worked solutions
- A closed surface encloses +3.00 nC and −1.00 nC. Find its net flux and say whether the shape matters.
- A sphere in a nonzero uniform external field contains no charges. Find net flux and explain whether E on the surface is zero.
- A uniform solid ball has Q=+1.00 nC and R=0.100 m. Find E at r=0.050 m and r=0.200 m; compare with an ideal spherical shell of the same Q,R.
- An ideal infinite line has λ=+1.00 nC/m. Find E at s=0.020 m from a Gaussian cylinder.
- An isolated infinite sheet has σ=+1.00 nC/m². Find the field magnitude on each side and state directions.
- A charge is inside a sphere but displaced from the sphere center. Does Gauss's law imply E=q/(4πε₀R²) everywhere on the sphere? Explain.
Solutions, units and validity checks
1. Qenc=+2.00 nC, so Φ=2.00×10⁻⁹/ε₀≈+2.26×10² N·m²/C. Surface shape does not affect the total flux if charges do not cross it.
2. Net flux is zero: the incoming and outgoing contributions cancel. E itself remains nonzero throughout the surface.
3. Inside the ball, E=kQr/R³≈4.49×10² N/C outward; outside, E=kQ/r²≈2.25×10² N/C outward. For the ideal shell E=0 at r=0.050 m, and outside at 0.200 m the same ≈2.25×10² N/C.
4. E=λ/(2πε₀s)=2kλ/s≈8.99×10² N/C, radially outward. Caps contribute zero because E is perpendicular to their outward normal.
5. |E|=σ/(2ε₀)≈56.5 N/C. The upper field points +ẑ and the lower field −ẑ.
6. No. The total flux is q/ε₀, but the off-center charge produces nonuniform E·n̂ over the sphere. E=q/(4πε₀R²) locally requires actual centered spherical symmetry.
Chapter 04 will use electrostatic equilibrium and boundary conditions to explain why fields inside ideal conductors vanish and why an isolated sheet cannot be confused with a charged conductor interface.