Electric Flux Density and Gauss's Law
Counting how much field pierces a closed surface turns out to depend on nothing but the charge trapped inside. That single idea — Gauss's law — collapses the laborious integrals of Chapter 4 into one elegant line, and becomes the first of Maxwell's four equations.
- The electric flux density \(\vec{D}\) and why it is independent of the medium.
- Electric flux \(\Psi\) as the surface integral of \(\vec{D}\).
- Gauss's law in integral form — flux out equals charge enclosed.
- The point (differential) form \(\nabla\cdot\vec{D} = \rho_v\), the first Maxwell equation, via the divergence theorem.
- How to choose a Gaussian surface that matches a problem's symmetry.
- One-line derivations for the point charge, line, sheet, coaxial cable and charged sphere.
Electric Flux Density D
The field \(\vec{E}\) depends on the medium through \(\varepsilon_0\). It is often cleaner to work with a quantity that depends only on the charge, not the material: the electric flux density \(\vec{D}\), also called the electric displacement. In free space it is simply \(\vec{E}\) scaled by the permittivity:
For a point charge, this gives \(\vec{D} = \dfrac{Q}{4\pi R^2}\hat{a}_R\) — notice \(\varepsilon_0\) has vanished. \(\vec{D}\) measures charge per unit area, and its field lines (called displacement or flux lines) begin and end only on charge, regardless of what medium fills the space between. This is exactly why \(\vec{D}\) is the natural object for Gauss's law.
Electric Flux
Electric flux \(\Psi\) measures the amount of \(\vec{D}\)-field passing through a surface — the surface integral met in Chapter 3. For a small patch \(d\vec{S}\), only the component of \(\vec{D}\) normal to the patch contributes, which the dot product handles automatically:
Gauss's Law
Gauss's law states that the total electric flux out of any closed surface equals the total charge enclosed by that surface — no more, no less. Charge outside the surface contributes nothing, because whatever flux it sends in, it takes back out.
The shape of the closed surface is irrelevant to the total flux — only the enclosed charge matters. The art is choosing a surface whose shape lets you pull \(\vec{D}\) outside the integral, which is what makes the law so powerful.
The Point Form
Gauss's law as written relates a surface to a volume. Applying the divergence theorem from Chapter 3 to the left side converts that surface integral into a volume integral of \(\nabla\cdot\vec{D}\):
Since this holds for any volume, the integrands must match everywhere. That gives the point (or differential) form — the first of Maxwell's equations:
Choosing a Gaussian Surface
Gauss's law is always true, but only useful for finding \(\vec{E}\) when you can choose a closed surface — a Gaussian surface — on which \(\vec{D}\) is either constant and normal, or zero. The right surface mirrors the symmetry of the charge, so that \(\oint \vec{D}\cdot d\vec{S}\) simplifies to \(D\) times an area.
| Symmetry | Gaussian surface | Coordinate |
|---|---|---|
| Point / spherical | Concentric sphere | Spherical \(r\) |
| Line / cylindrical | Coaxial cylinder | Cylindrical \(\rho\) |
| Sheet / planar | Pillbox (box straddling sheet) | Cartesian |
On faces where \(\vec{D}\perp d\vec{S}\) the flux is zero; on faces where \(\vec{D}\parallel d\vec{S}\) with constant \(D\), the integral becomes \(D\times(\text{area})\). Match those conditions and the field falls out in one step.
Standard Applications
The power of Gauss's law is that the same three lines — enclose the charge, evaluate the flux as \(D\times\text{area}\), solve for \(\vec{D}\) — reproduce every result of Chapter 4 instantly, and adds new ones like the coaxial cable.
Outside the outer conductor of a coaxial cable the enclosed charge is zero, so \(\vec{D}=0\) — the cable confines its field entirely between the conductors. This shielding property is exactly why coaxial cable carries signals cleanly, a point we return to in Part 6.
Worked Examples
Problem. In free space \(\vec{E} = 6\,\hat{a}_x\ \text{V/m}\). Find \(\vec{D}\).
Solution. Multiply by \(\varepsilon_0\):
Problem. A \(+8\ \text{nC}\) charge sits at the centre of a sphere. Find the total flux \(\Psi\) leaving it.
Solution. By Gauss's law, flux equals enclosed charge, independent of radius:
Problem. Re-derive the field of an infinite line charge \(\rho_L\) using a Gaussian cylinder.
Solution. Flux leaves only through the curved side, area \(2\pi\rho L\):
Same answer as Chapter 4 — but in three lines instead of an integral.
Problem. A sphere of radius \(a\) has uniform \(\rho_v\). Find \(\vec{E}\) for \(r Solution. The Gaussian sphere of radius \(r\) encloses only the charge within it: Inside, the field grows linearly with \(r\); outside it falls as \(1/r^2\).
Problem. Given \(\vec{D} = 3x\,\hat{a}_x + 2y\,\hat{a}_y\ \text{C/m}^2\), find the volume charge density \(\rho_v\).
Solution. Take the divergence, \(\rho_v = \nabla\cdot\vec{D}\):
Problem. If \(\vec{D} = x\,\hat{a}_x\ \text{C/m}^2\), find the charge enclosed by the unit cube \(0\le x,y,z\le 1\).
Solution. Use the point form and integrate, or note flux leaves only the \(x=1\) face:
Chapter Summary
\(\vec{D}=\varepsilon_0\vec{E}\) in free space; measured in C/m²; independent of medium; lines end on charge.
\(\Psi=\int_S\vec{D}\cdot d\vec{S}\); only the normal component of \(\vec{D}\) counts.
\(\oint_S\vec{D}\cdot d\vec{S}=Q_{\text{enc}}\); outside charge contributes nothing.
\(\nabla\cdot\vec{D}=\rho_v\) — Maxwell's first equation, via the divergence theorem.
Match the symmetry so \(\vec{D}\) is constant-normal or zero; sphere, cylinder, or pillbox.
Point, line, sheet recovered in three lines each; coaxial field lives only between conductors.
Problems
For each item, first identify the symmetry and pick a Gaussian surface, then apply Gauss's law or its point form. Difficulty rises down the list.
- Convert \(\vec{E} = 10\,\hat{a}_r\ \text{V/m}\) (free space) to \(\vec{D}\).
- A \(+5\ \mu\text{C}\) charge is enclosed by a cube. State the total flux through the cube and through one face.
- Use Gauss's law to find \(\vec{E}\) of an infinite sheet \(\rho_S = 4\ \text{nC/m}^2\).
- For \(\vec{D} = 2x^2\,\hat{a}_x + 3y\,\hat{a}_y\), find \(\rho_v\) at \((1,1,0)\).
- A coaxial cable has \(\rho_L = 6\ \text{nC/m}\) on the inner conductor. Find \(\vec{E}\) at \(\rho = 2\ \text{mm}\) (between conductors).
- A uniformly charged sphere of radius \(a\) carries total charge \(Q\). Find \(\vec{E}\) both inside and outside, and sketch \(E\) versus \(r\).
- A spherical shell carries surface charge \(\rho_S\). Show the field is zero inside and \(1/r^2\) outside.
- Given \(\vec{D} = \dfrac{5r^2}{4}\,\hat{a}_r\ \text{C/m}^2\) (spherical), find \(\rho_v\) using the spherical divergence.
- Two concentric spheres carry \(+Q\) (inner) and \(-Q\) (outer). Find \(\vec{E}\) in all three regions.
- A slab of thickness \(2d\) carries uniform \(\rho_v\). Find \(\vec{E}\) inside and outside the slab.
- Verify the divergence theorem for \(\vec{D} = r\,\hat{a}_r\) over a sphere of radius \(a\), confirming both sides equal the enclosed charge.
- Explain why Gauss's law, though always true, fails to give \(\vec{E}\) for an off-centre or asymmetric charge unless combined with superposition.