Flux and Gauss's law

The integrals at the end of the previous lesson were hard for a reason that has nothing to do with the physics, which is that they tracked the direction of a field whose direction symmetry had already settled.

This lesson builds the tool that exploits that. It needs the field of a point charge, $E = k Q / r^{2}$ , and nothing else from what has gone before. The result is the first of the four equations that will eventually be called Maxwell's, and it is worth saying now that it is not a new law: it is Coulomb's law rewritten, and every step of the rewriting is reversible.

Flux

Imagine the field as something streaming through space, and ask how much of it crosses a given surface. For a uniform field $E$ crossing a flat area $A$ face on, the natural measure is the product $E A$ . Tilt the surface by an angle $θ$ away from face on and the field crosses a smaller effective area, $A \cos θ$ , so the measure becomes $E A \cos θ$ . The quantity

Φ_{E} = E A \cos θ

is the electric flux, in units of newton metres squared per coulomb. Nothing is actually flowing; the name is inherited from fluid mechanics, where the same integral counts real litres per second, and the analogy is close enough to be useful and loose enough to be worth distrusting.

For a curved surface in a varying field, chop it into patches small enough to be flat and uniform, and add:

Φ_{E} = \int E \cos θ d A

The angle $θ$ is measured from the normal to the patch, which for a closed surface is taken to point outwards by convention. With that convention flux is signed: field leaving the surface counts positive, field entering counts negative, and a field that goes in one side and out the other contributes zero net.

From Coulomb to Gauss

Put a point charge $q$ at the centre of a sphere of radius $r$ . Two features make this the easiest surface in physics: the field has the same magnitude $k q / r^{2}$ everywhere on it, and it is everywhere perpendicular to it, so $\cos θ = 1$ . The flux is then the field times the area:

Φ_{E} = \frac{k q}{r^{2}} \times 4 π r^{2} = 4 π k q = \frac{q}{ε_{0}}

using $k = 1 / (4 π ε_{0})$ . This is the payoff promised in the first lesson for carrying $4 π$ around in Coulomb's law: the $r^{2}$ of the inverse square cancels against the $r^{2}$ of the sphere's area, and the $4 π$ cancels too, leaving a result with no geometry in it whatsoever. The flux does not depend on the radius. Doubling the sphere quarters the field and quadruples the area.

That cancellation is a fact about three dimensions and the exponent 2, and it is the whole content of what follows. Now generalise in two steps.

First, the surface need not be a sphere. Take any closed surface around the charge and imagine a narrow cone of field lines spreading out from the charge. It crosses the sphere in a patch and the odd surface in some other patch, further away by a factor $s$ and tilted by some angle. Being further reduces the field by $s^{2}$ ; the larger patch that the cone cuts, and the tilt, together increase the effective area by exactly $s^{2}$ . The flux through the two patches is identical. Since every part of the odd surface is covered by some such cone, the total flux is the same as for the sphere.

Second, a charge outside the surface contributes nothing. A cone of lines from an external charge pierces the closed surface an even number of times, entering and leaving alternately, and the entering and leaving contributions have opposite signs and equal magnitudes by the same argument as before. They cancel in pairs.

Put the two together, add up over however many charges there are (superposition again), and the result is Gauss's law:

Φ_{E} = \int E \cos θ d A = \frac{q_{enc}}{ε_{0}}

The flux through any closed surface equals the charge enclosed divided by $ε_{0}$ . Charges outside the surface still contribute to $E$ at every point of it; what they do not contribute to is the total.

Example. A charge of $3.0$ μC sits at the centre of a cube. What is the flux through one face?

Gauss's law gives the total flux out of the whole cube as $q / ε_{0} = 3.0 \times 10^{- 6} / 8.854 \times 10^{- 12} = 3.39 \times 10^{5}$ N m² C⁻¹. The cube has six faces and the charge sits symmetrically, so each face takes an equal share: $Φ = 5.65 \times 10^{4}$ N m² C⁻¹. Doing this by integrating the field over a square face is a genuinely unpleasant calculation, and symmetry has just replaced it with a division by six.

Now you. A charge of $8.0$ μC sits at the centre of a cube. What is the flux through one face?

Answer

Total flux is $q / ε_{0} = 8.0 \times 10^{- 6} / 8.854 \times 10^{- 12} = 9.04 \times 10^{5}$ N m² C⁻¹, and one sixth of that is $1.51 \times 10^{5}$ N m² C⁻¹.

Using it: three classic distributions

Gauss's law is always true and only sometimes useful. It becomes a method when the symmetry is good enough that $E$ can be pulled out of the integral, which requires a surface on which the field magnitude is constant and the angle is either 0 or 90 degrees. Three symmetries qualify: spherical, cylindrical and planar. Learning to see which one applies is most of the skill.

A uniformly charged sphere. Take total charge $Q$ spread evenly through a ball of radius $R$ . Outside it, choose a spherical surface of radius $r > R$ . By symmetry the field is radial and constant in magnitude on that surface, so $E \times 4 π r^{2} = Q / ε_{0}$ and

E = \frac{Q}{4 π ε_{0} r^{2}} = \frac{k Q}{r^{2}}

Outside, the ball is indistinguishable from a point charge at its centre. That is the electrical version of the shell theorem from mechanics, which Newton needed a geometrical tour de force to prove; here it takes two lines, and it is the theorem that licensed treating charged spheres as points in the first lesson.

Inside, at radius $r < R$ , the enclosed charge is only the fraction $r^{3} / R^{3}$ of the total, so $E \times 4 π r^{2} = Q r^{3} / (ε_{0} R^{3})$ and $E = k Q r / R^{3}$ . The field rises linearly from zero at the centre to $k Q / R^{2}$ at the surface, then falls as the inverse square outside. Both expressions agree at $r = R$ , which is the check worth doing every time.

A long line. For a line with $λ$ coulombs per metre, take a cylinder of radius $r$ and length $L$ coaxial with it. No flux passes through the flat ends, since the field is radial and runs parallel to them; the curved side has area $2 π r L$ with the field perpendicular to it everywhere. So $E (2 π r L) = λ L / ε_{0}$ , and the $L$ cancels:

E = \frac{λ}{2 π ε_{0} r}

which is the result the previous lesson obtained by integrating over the rod and taking a limit. Compare the labour.

A large flat sheet. For a sheet carrying $σ$ coulombs per square metre, take a small cylinder poking through it, with flat ends of area $A$ parallel to the sheet on either side. By symmetry the field points straight out of the sheet on both sides, so nothing crosses the curved wall and each end contributes $E A$ . The enclosed charge is $σ A$ , so $2 E A = σ A / ε_{0}$ and

E = \frac{σ}{2 ε_{0}}

with no $r$ in it at all. The field of an infinite sheet does not fall off with distance. This is the limiting case of the pattern noticed for the line: the more of the source stays in view as you retreat, the slower the decline, and a plane keeps all of it in view. It is also the reason a parallel plate capacitor, two such sheets with opposite charge, has a uniform field between its plates, which is the subject of the fifth lesson.

Example. A ball of radius 10.0 cm carries 50 nC spread uniformly through its volume. Find the field at 5.0 cm and at 20.0 cm from the centre.

At 5.0 cm we are inside, so $E = k Q r / R^{3} = (8.9876 \times 10^{9}) (50 \times 10^{- 9}) (0.050) / (0.100)^{3} = 2.25 \times 10^{4}$ N/C. At 20.0 cm we are outside, so the ball counts as a point: $E = k Q / r^{2} = (8.9876 \times 10^{9}) (50 \times 10^{- 9}) / (0.200)^{2} = 1.12 \times 10^{4}$ N/C. The field at the surface, from either formula, is $4.49 \times 10^{4}$ N/C, and it is the maximum.

Now you. A ball of radius 6.0 cm carries 24 nC spread uniformly. Find the field at 3.0 cm and at 12.0 cm from the centre.

Answer

Inside: $E = k Q r / R^{3} = (8.9876 \times 10^{9}) (24 \times 10^{- 9}) (0.030) / (0.060)^{3} = 3.00 \times 10^{4}$ N/C. Outside: $E = k Q / r^{2} = (8.9876 \times 10^{9}) (24 \times 10^{- 9}) / (0.120)^{2} = 1.50 \times 10^{4}$ N/C.

What a conductor does

A conductor contains charges free to move through it. Wait long enough after any disturbance and the charges stop moving, which is electrostatic equilibrium, and that single condition forces several consequences that Gauss's law makes quick.

If charges have stopped moving, the field inside the conducting material must be zero. Any field there would push the free charges, and they would still be moving. This is not a property of the metal but a consequence of equilibrium: the charges rearrange themselves precisely until their own field cancels whatever was applied, and the arrangement that achieves it is what equilibrium means.

Now take any closed surface drawn entirely within the metal. The field is zero everywhere on it, so the flux is zero, so the enclosed charge is zero. Since the surface can be drawn anywhere, all excess charge on a conductor sits on its surface. Shrink the surface to hug the outer boundary and the same argument says any net charge lives in a layer at most a few atoms deep.

Just outside the surface, take a small flat pillbox with one end in the metal and one end just outside. The inner end contributes nothing since the field there is zero, the walls contribute nothing since the field just outside a conductor is perpendicular to it (any parallel component would drive a surface current), and the outer end contributes $E A$ . So

E = \frac{σ}{ε_{0}}

just outside a conductor, twice the field of an isolated sheet with the same charge density, because a conductor's charge produces field on one side only.

Hollow out the conductor and put nothing inside, and Gauss's law applied to a surface in the metal surrounding the cavity gives zero enclosed charge, and a further argument rules out equal and opposite charges on the cavity wall: the field in the cavity is exactly zero regardless of what happens outside. That is the Faraday cage, and it is why a car is a reasonable place to be in a thunderstorm and why sensitive instruments live inside metal boxes. Faraday made the point in 1843 with an ice pail: a charged sphere lowered inside a metal container induced exactly its own charge on the outer surface, and touching it to the inside transferred the charge completely, leaving the sphere neutral.

Put a charge $q$ inside the cavity, however, and the enclosed-charge argument runs the other way: a surface in the metal encloses $q$ plus whatever is on the cavity wall, and the flux is zero, so the cavity wall must carry exactly $- q$ . Charge conservation then puts $+ q$ on the outer surface, and it distributes itself according to the outer shape alone, knowing nothing about where the charge sits inside.

Example. A charge of $+ 4.0$ nC sits at the centre of a hollow, uncharged conducting shell with inner radius 5.0 cm and outer radius 8.0 cm. What charge sits on each surface, and what is the field at 10.0 cm from the centre?

A Gaussian sphere drawn inside the metal, say at 6.5 cm, encloses the central charge plus the inner surface and must enclose zero net charge, so the inner surface carries $- 4.0$ nC. The shell as a whole is uncharged, so the outer surface carries $+ 4.0$ nC. At 10.0 cm the enclosed charge is $+ 4.0 - 4.0 + 4.0 = + 4.0$ nC, giving $E = k q / r^{2} = (8.9876 \times 10^{9}) (4.0 \times 10^{- 9}) / (0.100)^{2} = 3.60 \times 10^{3}$ N/C, pointing outwards. The shell is invisible from outside.

Now you. A charge of $- 6.0$ nC sits at the centre of a hollow conducting shell that itself carries a net charge of $+ 2.0$ nC. What charge sits on each surface, and what is the field at 15.0 cm from the centre?

Answer

The inner surface must cancel the central charge, so it carries $+ 6.0$ nC. The shell's total is $+ 2.0$ nC, so the outer surface carries $2.0 - 6.0 = - 4.0$ nC. At 15.0 cm the enclosed charge is $- 4.0$ nC, so $E = (8.9876 \times 10^{9}) (4.0 \times 10^{- 9}) / (0.150)^{2} = 1.60 \times 10^{3}$ N/C, pointing inwards.

The null experiment

The shielding result is the sharpest test of Coulomb's law there is, and the reason was flagged at the end of the first lesson. Every step from the inverse square to "the field inside a closed conductor is zero" used the exact cancellation of $r^{2}$ against $r^{2}$ . If the exponent were $2 + q$ for any $q$ other than zero, the cancellation would be imperfect, flux would depend on the shape and size of the surface, and a charge would appear inside a hollow conductor.

That converts a hard measurement into an easy one. Instead of measuring a force to one part in $10^{16}$ , which is impossible, charge a hollow conductor heavily and look inside for any charge at all, which needs only a sensitive electrometer and a good null. Cavendish did it in 1773 with concentric spheres and got $| q | < 0.02$ . Maxwell, who edited and published Cavendish's unpublished papers, repeated it in 1873 and reached about $5 \times 10^{- 5}$ . Williams, Faller and Hill in 1971 used five nested shells, an alternating potential of 10 kV and a lock-in amplifier, and reached $3 \times 10^{- 16}$ .

Null experiments are worth noticing as a technique. They convert a question about the value of a quantity into a question about whether something is exactly zero, and zero is the one number that can be measured to arbitrary precision, because the answer does not depend on knowing any calibration.

What Gauss's law does not do

The law is completely general and it is not a general method. Given an arbitrary blob of charge, Gauss's law is still true, and it gives one equation for a field that varies over the surface in an unknown way, which is one equation and infinitely many unknowns. Symmetry is what collapses the unknowns to one, and only three symmetries do it.

It is also worth being clear about what enclosed means. The flux depends only on the charge inside, and the field at each point of the surface depends on every charge in the universe. Bring an external charge near a Gaussian sphere and the field on the near side rises while the far side falls, in such a way that the total is untouched. Students often read Gauss's law as saying the outside charge has no effect, and it says something much narrower and much stranger than that.

The deeper limitation is that this lesson has produced a vector field from a vector integral, and the second lesson's complaint stands: adding vectors is laborious, and for a distribution with no symmetry neither method helps. There is a third route, which uses the fact that the electric force, being an inverse square central force, is conservative in exactly the sense mechanics gave that word. That makes the whole field derivable from a single scalar function, and scalars add without components. It is the potential, and it is next.