Gauss's Law

Gauss’s law ties the field on a closed surface to the charge inside it. To state it we first need a way to measure “how much field passes through a surface”. That measure is called flux.

Flux through a small flat surface#

Take a small flat patch of area ΔS in a uniform field E⃗. Pick one side of the patch and draw the perpendicular on that side: this is the positive normal. Let the field make an angle θ with it.

• Field along the normal (θ = 0): ΔΦ = EΔS, the largest possible.
• Field opposite to the normal (θ = π): ΔΦ = −EΔS.
• Field along the surface (θ = π/2): ΔΦ = 0. Nothing passes through.

Think of the field lines: flux is proportional to the number of lines that pierce the surface. Tilt the plate and fewer lines get through, exactly as cos θ says.

Curved surfaces and non-uniform fields#

Flux is a scalar, so the flux through a surface made of pieces is the sum of the fluxes through the pieces. Break any surface into tiny, nearly flat bits, small enough that E barely changes across each, and add:

A closed surface encloses a volume: a sphere, or a cylinder together with its two flat end caps. (A hemisphere, or a cylinder without its caps, is open.) For a closed surface we write a small circle on the integral, $\oint \vec E\cdot d\vec S$, and always take the outward normal as positive. Field leaving the volume counts positive; field entering counts negative.

A plane angle measures how wide a pair of lines opens at a point. A solid angle does the same in three dimensions: how wide a cone opens. It is the tool that turns Coulomb’s law into Gauss’s law.

From plane angles to solid angles#

Join the two ends of a curve to a point O and the curve subtends a plane angle at O. To measure it, draw a circle of any radius r about O and measure the arc length l cut off: θ = l/r.

Now take a surface patch and join every point on its edge to O. The lines form a cone (like a paper cone for roasted peanuts), and we say the patch subtends a solid angle at O. Measure it the same way, with a sphere instead of a circle:

A small tilted patch#

A short segment Δl at distance r, square-on to the line from O, subtends Δθ ≈ Δl/r. If the line from O makes an angle α with the segment’s normal, only the part Δl cos α faces O, so Δθ = Δl cos α / r. The same holds for a small area:

Whole circles and whole spheres#

A full circle subtends θ = 2πr/r = 2π at its centre, and in fact any closed curve subtends 2π at any point inside it. A full sphere subtends Ω = 4πr²/r² = 4π, and any closed surface subtends 4π at any point inside.

What about a point outside? Close a curve around an outside point step by step and the angle it subtends shrinks to zero: the near and far parts of the curve cover the same directions with opposite sense and cancel. So a closed surface subtends zero solid angle at an external point.

Draw any closed surface you like. Add up the flux of the electric field over it. The answer depends on only one thing: how much charge is inside.

The law#

Read the two sides carefully, because they don’t refer to the same charges:

• $\vec E$ on the left is the total field, produced by every charge in the universe, inside the surface or outside.
• q_in on the right is only the net charge inside the surface.

Charges outside do change the field on the surface, but they send as much flux in as out, so they add nothing to the total. The surface you apply the law to is called a Gaussian surface.

Gauss’s law is treated as a fundamental law of nature, checked by experiment. Historically Coulomb’s law came first, and Gauss’s law can be derived from it.

Proof from Coulomb’s law#

A charge inside

Put q at a point O inside a closed surface. At a point P on the surface, at distance r, the field is q/4πε₀r² along OP. If OP makes an angle α with the outward normal of a small patch ΔS there,

The flux through each patch is just q/4πε₀ times the solid angle it subtends at the charge. The 1/r² of Coulomb’s law cancels the r² in the solid angle. Adding all patches, ΣΔΩ = 4π for an inside point, so

A charge outside

The same step gives ΔΦ = (q/4πε₀)ΔΩ, but now the total solid angle of a closed surface seen from outside is 0. So Φ = 0.

Many charges

With charges q₁ … qₙ inside and Q₁ … Q_N outside, the field is the vector sum of their separate fields, so the flux is the sum of their separate fluxes: each inside charge gives qᵢ/ε₀, each outside charge gives 0.

When a charge distribution is very symmetric, Gauss’s law hands you the field in two lines. The trick is to let symmetry fix the field’s direction, then choose a surface where E is constant.

The recipe#

1. Use symmetry to find the direction of E⃗. Ask: is there only one direction that isn’t favoured over any other?
2. Choose a closed surface through the point where, on each part, E⃗ is either perpendicular to the surface with constant size, or parallel to it (no flux).
3. Then the flux is just E × (area), set it equal to q_in/ε₀, and solve for E.

A uniformly charged sphere#

Charge Q is spread evenly through a ball of radius R. Every point at distance r from the centre looks the same as every other, so E is radial and has the same size all over a concentric sphere of radius r. Take that sphere as the Gaussian surface: Φ = E · 4πr².

Outside (r > R)

All of Q is enclosed: E · 4πr² = Q/ε₀.

Inside (r < R)

Only the fraction of the volume inside radius r counts. The charge per unit volume is Q/(4/3 πR³), so the enclosed charge is

The field grows in proportion to r, is zero at the centre (no direction is special there, so it has to be), and the two formulas agree at r = R.

A long line of charge#

A long straight line carries charge λ per unit length. At a point P, the only direction not favoured over any other is perpendicular to the line, so E⃗ is radial (away from the line if λ is positive). Take a coaxial cylinder of radius r and length l, closed by two flat caps.

• Curved side: E⃗ is perpendicular to it and the same size everywhere: flux E · 2πrl.
• Flat caps: E⃗ runs along them: no flux.

The enclosed charge is λl, so E · 2πrl = λl/ε₀:

A large plane sheet of charge#

A large flat sheet carries charge σ per unit area. By symmetry E⃗ is perpendicular to the sheet, away from it on both sides if σ is positive. Take a small cylinder (a “pillbox”) that pierces the sheet, with flat faces of area ΔS at equal distances on each side.

• Each flat face: E⃗ along the outward normal: flux EΔS each, 2EΔS in all.
• Curved side: E⃗ runs along it: no flux.

The enclosed charge is σΔS, so 2EΔS = σΔS/ε₀:

Coulomb’s law and Newton’s law of gravitation have the same inverse-square form, so every result we proved for the gravity of spheres carries straight over to charged spheres.

Shells and solid spheres#

Compare F = q₁q₂/4πε₀r² with F = Gm₁m₂/r²: swap mass for charge and G for 1/4πε₀. For a spherical charge distribution of radius R carrying Q:

• Outside, either one acts exactly like a point charge Q at the centre, for both field and potential.
• Inside a shell there is no field at all, so the potential is the same everywhere inside, equal to its value on the surface.
• Inside a solid sphere the field rises linearly from zero at the centre.

Energy stored in a charged sphere#

The electric potential energy of a charge distribution is the work needed to assemble it, bringing each bit of charge in from infinity.

Uniform solid sphere

Build the sphere layer by layer. When it has reached radius x, with charge density ρ = 3Q/4πR³, it holds q = Qx³/R³ and its surface is at potential

The next layer, from x to x + dx, needs dq = 4πx²dx · ρ = 3Qx²dx/R³. Bringing it in costs dW = V dq, and adding all layers:

Thin shell

With charge q already on a shell, its potential is q/4πε₀R, so adding dq costs q dq/4πε₀R:

Inside a conductor the field is zero. Feed that one fact into Gauss’s law and out come all the rules about where charge sits on a conductor, and why earthing keeps you safe.

Charge lives on the surface#

Once the free electrons in a conductor have settled, E = 0 at every point inside the metal. Draw a closed surface lying entirely inside the metal: the flux through it is zero, so by Gauss’s law it encloses no net charge.

• Every volume inside a conductor is electrically neutral. Any charge you inject moves to the surface.
• If the conductor is hollow with nothing in the cavity, the charge goes to the outer surface, and the field in the cavity is zero (this needs a little more mathematics to prove).

A charge inside a cavity

Now put +q in the cavity. A Gaussian surface through the metal around the cavity still has E = 0 on it, so it must enclose zero charge: −q appears on the inner wall. If the conductor was neutral, the balancing +q appears on the outer surface.

The field just outside a conductor#

A conductor’s surface is an equipotential, so the field just outside is perpendicular to it. Take a pillbox with one face (area ΔS) just outside, the other inside the metal. Only the outside face has flux, EΔS, and the charge enclosed is σΔS:

The field jumps from σ/ε₀ to zero across a skin only 4–5 atomic layers thick; that skin is what we mean by “the surface” of a conductor.

Earthing a conductor#

The earth is a huge conductor. Adding or removing a little charge barely changes its potential, so we take it as zero, the same as a point at infinity (its real surface charge, about −1 nC/m², can be ignored for most purposes). Connect a conductor to the earth and charge flows until the conductor is at zero potential too: it is earthed (grounded).

• A charged metal sphere at Q/4πε₀R, earthed: all its charge flows away.
• A shell with +Q at its centre holds −Q inside and +Q outside, and sits at Q/4πε₀R. Earthed, the outer +Q drains away; the inner −Q stays, held by the central charge.