Charge distributions and Gauss’s law

Continuous charge distributions

Narrated video. Transcript:

Charge is quantised. But a charged body carries so many electrons that it is far easier to treat the charge as smeared out smoothly, like paint, rather than as separate dots.

For charge spread along a thin wire, we use the linear charge density λ: the charge Δq on a small length Δl, divided by Δl. Its unit is coulomb per metre.

For charge spread over a surface, like a thin sheet, we use the surface charge density σ: charge ΔQ on a small area ΔS, divided by ΔS. Unit: coulomb per square metre.

For charge filling a volume, we use the volume charge density ρ: ΔQ over ΔV, in coulomb per cubic metre.

These small pieces are small on our everyday scale, but each still holds a huge number of electrons. So the density is a smooth, average quantity.

Example: a wire 50 centimetres long carries 2 microcoulombs spread evenly. λ is 2 × 10⁻⁶ over 0.5, which is 4 × 10⁻⁶ coulomb per metre.

A thin spherical shell of radius 10 centimetres carries 2 microcoulombs. Its area is 4π times 0.1 squared, about 0.126 square metre. So σ is about 1.6 × 10⁻⁵ coulomb per square metre.

To find the field of such a body, cut it into tiny pieces. Each piece, ρ ΔV, acts like a point charge. Add up all their fields by superposition. In the limit, the sum becomes an integral.

That can be hard work. But for shapes with symmetry, there is a beautiful shortcut: Gauss’s law.

Gauss’s law

Narrated video. Transcript:

Put a point charge q at the centre of a sphere of radius r. Everywhere on the sphere, E has size q over 4πε₀r², and points straight out, along the outward normal.

So the flux through the sphere is just E times the area: q over 4πε₀r², times 4πr². The r² cancels, and the 4π cancels. The flux is q over ε₀.

It does not depend on the radius at all. That makes sense: every field line from q crosses every sphere around it, big or small.

Now squash the sphere into any closed shape you like. The same lines still cross it, so the flux is still q over ε₀.

This is Gauss’s law: the total electric flux through any closed surface equals the total charge enclosed by the surface, divided by ε₀. Φ equals q over ε₀.

What about a charge outside the surface? Every line that enters the surface leaves it again somewhere else. In minus out is zero, so outside charges add nothing to the net flux.

Careful: E in Gauss’s law is the total field, due to all charges, inside and outside. Outside charges change E on the surface; they just do not change the total flux.

The closed surface we choose is called a Gaussian surface. We can pick any we like, but it must not pass through a point charge, where the field is not defined.

Example: 10 microcoulombs at the centre of a cube of side 10 centimetres. The total flux is 10⁻⁵ over 8.854 × 10⁻¹², about 1.13 × 10⁶. By symmetry, each of the six faces gets one sixth: about 1.88 × 10⁵ newton metre squared per coulomb.

And in reverse: if the net flux out of a closed box is 8.0 × 10³ newton metre squared per coulomb, the charge inside is ε₀ Φ: 8.854 × 10⁻¹² times 8 × 10³, about 7.1 × 10⁻⁸ coulomb.

Try it: a Gaussian surface

Interactive simulation. Things to try:

• The green circle stands for a Gaussian sphere. Drag its centre (+) to move it, and its rim to resize it.
• With only the +2 μC charge inside, Φ = q/ε₀ = 2 × 10⁻⁶ / 8.854 × 10⁻¹², about 2.26 × 10⁵ N m²/C. Make S bigger: Φ does not change until another charge gets inside.
• Enclose the −1 μC charge as well: the flux drops to that of +1 μC.
• Move S so that it contains no charge at all. Lines go in and come out again: Φ = 0.

Charge in a cube

A charge of 6 μC sits at the centre of a cube. What is the electric flux through one face of the cube? (ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻²)

1. 1.13 × 10⁵ N m²/C
2. 6.78 × 10⁵ N m²/C
3. Zero
4. It depends on the size of the cube

Zero flux

A closed surface S encloses +2 μC and −2 μC. A +3 μC charge sits just outside S. Which statement is correct?

1. Φ = 0, and E = 0 everywhere on S
2. Φ = 0, but E on S need not be zero
3. Φ = 3 μC/ε₀, because of the charge outside
4. Φ = 4 μC/ε₀, adding the sizes of the charges inside

Summary

• Charge densities: λ = Δq/Δl, σ = ΔQ/ΔS, ρ = ΔQ/ΔV
• Gauss’s law (q = charge enclosed): Φ = ∮ E · dS = q / ε₀
• Point charge, any closed surface: Φ = q / ε₀

• Charge on large bodies is described by smooth linear, surface and volume densities λ, σ and ρ.
• Gauss’s law: the net flux through any closed surface is the charge enclosed divided by ε₀, whatever the shape or size of the surface.
• Charges outside the surface give zero net flux, though they do change E on the surface.
• A Gaussian surface must not pass through a point charge.