Electric field

We start from Coulomb's law and reach Gauss's law, by computing the total flux out of a sphere drawn around a single charge.

Setting up. Put a charge q at the centre of an imaginary sphere of radius r. The sphere is chosen with the charge at the centre because that is the shape the field respects.

Step 1 — the field on the sphere. By Coulomb's law, at every point of the sphere

and this has the same value everywhere on the sphere, because every point is the same distance r away.

Step 2 — the direction of the field. The field of a point charge is radial, and the outward normal d\vec{A} of a sphere is also radial. So the angle between them is zero everywhere:

Step 3 — write the flux integral:

Step 4 — put \cos\theta = 1 and take the constant E outside the integral. We may do this because Step 1 showed E has the same value at every point:

Step 5 — do the remaining integral. \oint dA is just the total surface area of the sphere:

Step 6 — substitute Steps 1 and 5:

Step 7 — cancel 4\pi r^{2}, which appears on top and bottom:

Result.

What it means. r disappeared in Step 7, so the flux is the same through a big sphere as through a small one. Physically, the field weakens as 1/r^{2} but the area grows as r^{2}, and the two effects cancel exactly. Since the same field lines cross every closed surface drawn round the charge, the result holds for a surface of any shape, not just a sphere — which is what makes Gauss's law useful.

Conditions used. The medium is vacuum or air (otherwise \varepsilon_0 becomes \varepsilon); the charge is a point charge at the centre; and the surface is closed.

We start from Gauss's law and reach the field at three places: outside the sphere, at its surface, and inside it.

Setting up. A conducting sphere of radius R carries charge Q. Because it is a conductor, all the charge sits on the outer surface, spread evenly. The surface charge density is \sigma = Q/(4\pi R^{2}).

Case 1 — outside, r > R.

Step 1 — choose the Gaussian surface. Take a sphere of radius r concentric with the charged sphere. The charge distribution is spherically symmetric, so the field must be radial and must have the same magnitude everywhere on this surface.

Step 2 — write the flux. Field parallel to the outward normal, constant magnitude, total area 4\pi r^{2}:

Step 3 — find the enclosed charge. The Gaussian sphere is bigger than the conductor, so it contains all of it:

Step 4 — apply Gauss's law, \Phi = q_{enc}/\varepsilon_0:

Step 5 — divide both sides by 4\pi r^{2}:

So outside, the sphere behaves exactly like a point charge Q at its centre.

Case 2 — at the surface, r = R.

Step 6 — put r = R in the result of Step 5:

Step 7 — rewrite using the surface charge density. From \sigma = Q/(4\pi R^{2}) we get Q = 4\pi R^{2}\sigma. Substituting:

This is the largest value the field reaches anywhere.

Case 3 — inside, r < R.

Step 8 — choose a Gaussian sphere of radius r inside the metal. Its flux expression is the same as before:

Step 9 — find the enclosed charge. All the charge lies on the outer surface, at radius R, which is outside this Gaussian sphere. So it encloses nothing:

Step 10 — apply Gauss's law:

Step 11 — divide by 4\pi r^{2}, which is not zero:

Result.

What it means. The field is zero everywhere inside a charged conductor, jumps suddenly to \sigma/\varepsilon_0 at the surface, then dies away as 1/r^{2}. The zero inside is the principle of electrostatic shielding — why you are safe inside a car in a thunderstorm, and why sensitive instruments are put in metal boxes.

Conditions used. The sphere is a conductor in electrostatic equilibrium (so all charge is on the outer surface and E = 0 within the metal), the charge is spread uniformly, and the surrounding medium is air.

We start from Gauss's law with a cylindrical surface and reach the field at distance r from a long charged wire.

Setting up. Charge is spread along a long straight wire with linear charge density \lambda (coulombs per metre). We want E at perpendicular distance r from the wire.

Step 1 — use the symmetry to guess the field's direction. The wire looks the same from every direction around it and the same from every point along it, so the field can only point straight out from the wire (radially), and its magnitude can depend only on r.

Step 2 — choose the Gaussian surface to match that symmetry. Take a cylinder of radius r and length l, with the wire along its axis.

Step 3 — flux through the two flat ends. There the outward normal points along the wire, while \vec{E} points perpendicular to the wire. Perpendicular vectors give no flux:

Step 4 — flux through the curved surface. Here \vec{E} is parallel to the outward normal everywhere, and has the same magnitude everywhere (Step 1). The curved area is 2\pi r l:

Step 5 — total flux is the sum of Steps 3 and 4:

Step 6 — find the enclosed charge. A length l of wire lies inside the cylinder, and each metre carries \lambda:

Step 7 — apply Gauss's law:

Step 8 — cancel l from both sides. The length of cylinder we chose was arbitrary, so it had better not appear in the answer — and it does not:

Step 9 — divide both sides by 2\pi r:

Step 10 — write it with k = 1/(4\pi\varepsilon_0). Since \dfrac{1}{2\pi\varepsilon_0} = \dfrac{2}{4\pi\varepsilon_0} = 2k:

Result.

What it means. The field falls off as 1/r, not as 1/r^{2}. A line of charge is a "fatter" source than a point charge: as you move away, the wire still stretches out beside you, so the field dies more slowly. It is directed away from the wire if \lambda is positive, towards it if negative.

Conditions used. The wire must be effectively infinite — in practice, r much smaller than the length of the wire, and the point not near either end. Otherwise Step 1's symmetry fails.

We start from Gauss's law with a flat "pill box" and reach the field close to a large charged plate — and then see why a non-conducting sheet gives half as much.

Setting up. A large flat conductor carries surface charge density \sigma (coulombs per square metre). We want E at a point just outside its surface. Because the plate is large and flat, the field there must be perpendicular to the surface and the same at all nearby points.

Step 1 — choose the Gaussian surface. Take a short cylinder (a "pill box") of cross-sectional area A, with one flat face just outside the surface and the other buried inside the metal.

Step 2 — flux through the inner face is zero. Inside a conductor in equilibrium E = 0, so nothing crosses that face:

Step 3 — flux through the curved side is zero. There \vec{E} lies along the surface, parallel to the curved wall, so it crosses nothing:

Step 4 — flux through the outer face. Here \vec{E} is perpendicular to the face, i.e. parallel to its outward normal, and constant across it:

Step 5 — total flux is the sum of Steps 2, 3 and 4:

Step 6 — find the enclosed charge. The pill box cuts out a patch of surface of area A, and each square metre carries \sigma:

Step 7 — apply Gauss's law:

Step 8 — cancel A from both sides:

Now the thin non-conducting sheet.

Step 9 — what changes. A thin insulating sheet has charge on it but no interior in which E must vanish. The field now comes out of both faces of the pill box, so the flux is doubled:

Step 10 — the enclosed charge is the same \sigma A, so Gauss's law gives

Step 11 — cancel A and divide by 2:

Result.

What it means. r never appeared, so the field near a large plate does not weaken as you move away — it is uniform. That is exactly why a parallel-plate capacitor has a uniform field between its plates. The factor of two is not a formula to memorise but a consequence of how many faces the flux comes out of.

Conditions used. The plate is large compared with the distance of the point from it (so edge effects can be ignored), and the charge is spread uniformly.