Applications of Gauss's Law — Spheres and Planes

Walk up to the metal sphere of a Van de Graaff generator at IIT Kanpur after it has been charged up for a minute. At any point outside the sphere, you measure the same field you would get if all the accumulated charge were squeezed down into a point at the sphere's centre. At any point inside the metal shell — impossible to get a probe in there without making a hole, but bear with me — you measure exactly zero field. Outside: a perfect point-charge field. Inside: nothing at all. That is not a coincidence, and it is not an approximation. It is a mathematical consequence of Coulomb's inverse-square law and a single, cleanly derived theorem.

That theorem is Gauss's law, and this article is about what you can do with it.

Gauss's law by itself is not a shortcut. It is always true — for every charge configuration, every surface, every situation — but it is only useful when the geometry is symmetric enough that the flux integral \oint\vec{E}\cdot\mathrm{d}\vec{A} collapses from "a nightmare integral" to "E times area." The good news is that a handful of symmetric geometries — spheres, planes, parallel plates, cylinders, conductors — cover an astonishing fraction of the physics that appears in an electrostatics syllabus, and each one yields its field from Gauss's law in a few lines.

This article works through the five most important symmetric cases that are not cylinders or cavities (those are handled in the next article):

1. Uniformly charged spherical shell — the punchline being that a hollow charged sphere is a point charge on the outside and nothing on the inside.
2. Uniformly charged solid sphere — like the shell on the outside, but with a linearly growing field on the inside.
3. Infinite plane of charge — field uniform and perpendicular, independent of distance.
4. Two parallel oppositely-charged planes — the capacitor geometry; field doubled between the plates, cancelled outside.
5. Conductor in electrostatic equilibrium — why charges go to the surface and why the interior field is always exactly zero.

Each subsection has the same rhythm: identify the symmetry, draw the Gaussian surface, write the flux, apply Gauss's law, read off E. After the third time, the pattern is reflex. The going-deeper section then handles (i) the surprising fact that even a non-uniform spherically symmetric charge gives a point-charge field outside, (ii) the discontinuity in E across a surface charge, and (iii) the electrostatic pressure on a charged conductor.

1. The Gauss's-law recipe, and why symmetry is everything

Gauss's law states that for any closed surface S and any charge configuration,

where Q_{\text{enc}} is the total charge enclosed by S. The left-hand side, the flux, is the quantity you met in the electric flux article. The right-hand side is a number you can usually read off by looking at the charge configuration.

To use Gauss's law to find \vec{E}, you have to be able to do the flux integral. In general, an arbitrary closed surface in an arbitrary field gives a nasty surface integral. The trick is that certain charge distributions force \vec{E} to have a simple form everywhere in space — by symmetry alone. The canonical three symmetries:

• Spherical symmetry. If the charge distribution is invariant under every rotation about a point O, the field at any location must be radial (point away from or toward O) and must have magnitude that depends only on the distance r from O. A spherical Gaussian surface of radius r then has \vec{E}\cdot\hat{n} = E(r) constant over its whole area.
• Planar symmetry. If the charge distribution is invariant under every translation parallel to some plane, the field must be perpendicular to that plane and must depend only on the perpendicular distance from it. A "pillbox" Gaussian surface (a short cylinder with axis perpendicular to the plane) has flux only through its two flat faces.
• Cylindrical symmetry. If the charge distribution is invariant under translation along a line and rotation about that line, the field is radial (perpendicular to the line) and depends only on the perpendicular distance from the line. A cylindrical Gaussian surface co-axial with the line has flux only through its curved side. This case is the subject of the next article.

The upshot: when you see one of these three symmetries, the flux integral on the left of Gauss's law becomes E\times(\text{area}), and the law solves for E immediately.

2. Spherical shell of charge

Setup

A thin spherical shell of radius R carries total charge Q spread uniformly over its surface. By the spherical symmetry of the configuration (rotations about the centre leave it unchanged), the field at any point a distance r from the centre must: (i) point radially — outward if Q > 0, inward if Q < 0; (ii) have magnitude that depends only on r, not on direction.

Write this as \vec{E}(r) = E(r)\,\hat{r}, where E(r) is an unknown scalar function we want to find.

Outside the shell: r > R

Choose a Gaussian surface S to be a sphere of radius r concentric with the shell, with r > R. On this surface, \hat{n} = \hat{r} (outward), and \vec{E}\cdot\hat{n} = E(r) — constant over the whole surface by symmetry.

The flux integral collapses:

Why: because \vec{E}\cdot\hat{n} is constant over S, it pulls out of the integral. The remaining \oint_S\mathrm{d}A is just the area of the Gaussian sphere, 4\pi r^2.

The enclosed charge is the entire shell: Q_{\text{enc}} = Q. Apply Gauss's law:

Solve for E:

Why: this is identical to the field of a point charge Q sitting at the centre. Outside the shell, the field is exactly what you would get from a pinpoint of charge at the shell's centre, regardless of the shell's radius. This is sometimes called the "shell theorem" for electrostatics; it was established — for the gravitational analogue — by Newton himself, who needed to know whether the Earth's gravity behaved like a point mass at its centre. The answer is yes, provided the mass distribution is spherically symmetric.

Inside the shell: r < R

Now choose a Gaussian surface that is a sphere of radius r with r < R — entirely inside the shell, where there is no charge. The same symmetry argument applies: \vec{E}(r) = E(r)\hat{r}, constant on the Gaussian surface.

The flux is again E(r)\cdot 4\pi r^2. But now the enclosed charge is zero (the shell's charge is outside the Gaussian surface). So Gauss's law gives:

Why: this is the remarkable part. Every point inside the shell is being pulled on by every charge element on the shell, but those pulls cancel exactly, at every interior point, leaving zero net field. There is no special "centre" where the field vanishes by symmetry — the cancellation is pointwise, and it holds at every interior location. This is a consequence of the inverse-square form of Coulomb's law; any other power law would not give an identically-zero interior field.

Summary for the shell

The field has a discontinuity at r = R: jumping from zero inside to Q/(4\pi\varepsilon_0 R^2) just outside. The discontinuity is exactly \sigma/\varepsilon_0 with \sigma = Q/(4\pi R^2) the surface charge density — a general feature of any surface charge, as you will see in §6.

3. Solid uniformly charged sphere

Setup

A solid sphere of radius R has total charge Q distributed uniformly throughout its volume. The volume charge density is \rho = Q/(\tfrac{4}{3}\pi R^3) = 3Q/(4\pi R^3).

The spherical symmetry is unchanged from §2, so \vec{E}(r) = E(r)\hat{r} still holds. The new physics comes in when r < R: the Gaussian surface now encloses only part of the charge, and the enclosed charge grows with r.

Outside the sphere: r > R

The calculation is identical to the shell case. Gaussian sphere of radius r encloses the whole charge Q; flux is E(r)\cdot 4\pi r^2; Gauss's law gives

Again, the sphere looks like a point charge at the centre. This is general: any spherically symmetric charge distribution (uniform, nonuniform, shell of shells, it does not matter) looks like a point charge from outside.

Inside the sphere: r < R

Now choose a Gaussian sphere of radius r with r < R. The enclosed charge is the charge in the smaller sphere of radius r:

Why: the density \rho is uniform, so the charge in any sub-volume is just \rho times the sub-volume. Here the sub-volume is \tfrac{4}{3}\pi r^3. The final form Q\,(r/R)^3 shows that the enclosed fraction grows as the cube of the radius ratio — a geometric fact about filling 3D space uniformly.

Gauss's law gives

Solve for E:

Why: the field grows linearly with r inside the sphere, starting from zero at the centre. This is because the enclosed charge grows as r^3 while the Gaussian surface area grows as r^2; their ratio is r, so E \propto r. At the centre, the field vanishes by symmetry (every direction is equivalent, so no preferred direction for the field to point), which matches the formula at r = 0.

Matching at the surface

At r = R from the inside formula: E(R^-) = Q/(4\pi\varepsilon_0 R^2). From the outside formula: E(R^+) = Q/(4\pi\varepsilon_0 R^2). The two match exactly — no jump at r = R.

This is different from the shell case! A solid uniformly charged sphere has a continuous field at its surface, because the charge is spread through the volume rather than concentrated in a thin layer. A shell, by contrast, has a finite surface charge density \sigma, and the field jumps by \sigma/\varepsilon_0. The solid sphere's surface has zero surface density (since all the charge is bulk), so no jump.

Interactive: field profile inside and outside a solid sphere

Drag the radial position r/R and watch where you are on the field curve and what E (in units of kQ/R^2) actually is at that radius.

4. Infinite plane of charge

Setup and symmetry

A flat sheet of uniform surface charge density \sigma (in C/m²) extends to infinity in both directions. Place it in the xy-plane. The configuration is invariant under (i) any translation parallel to the sheet, (ii) reflection through the sheet.

Consequences:

• The field depends only on the perpendicular distance z from the sheet (not on x or y).
• The field must point in the \pm\hat{k} direction (perpendicular to the sheet) — any transverse component would pick a special direction, violating translational symmetry.
• By reflection symmetry, \vec{E} on the +z side is the mirror image of \vec{E} on the -z side: \vec{E}(+z) = -\vec{E}(-z) where the minus sign is the reflection.

So \vec{E} is +E(z)\hat{k} above the sheet and -E(z)\hat{k} below (for \sigma > 0; signs flip for \sigma < 0).

Gaussian surface: the pillbox

Choose a Gaussian surface shaped like a short right cylinder — a "pillbox" — with axis along \hat{k} and flat end-caps parallel to the sheet. Let each end-cap have area A; let the pillbox straddle the sheet, so half of it sits above the sheet and half below.

Flux through the two end-caps. Each end-cap has outward normal \hat{n} = \pm\hat{k} (top cap outward is +\hat{k}; bottom cap outward is -\hat{k}). On the top cap, \vec{E} = +E\hat{k}, so \vec{E}\cdot\hat{n} = +E. On the bottom cap, \vec{E} = -E\hat{k}, so \vec{E}\cdot\hat{n} = (-E)(- 1) = +E. Both caps contribute positively to the outward flux. Total:

Why: the field points out of the pillbox on both sides, so each cap contributes +EA to the outward flux. If the pillbox were entirely above the plane, both caps would be on the same side and their contributions would be opposite in sign.

Flux through the side wall. On the side wall, the outward normal points horizontally — perpendicular to \hat{k}. Since \vec{E} is purely vertical, \vec{E}\cdot\hat{n}_{\text{wall}} = 0 everywhere on the wall. The wall contributes zero flux.

Enclosed charge. The pillbox cuts a piece of the infinite plane of area A, so the charge enclosed is Q_{\text{enc}} = \sigma A.

Apply Gauss's law

Solve for E:

Why: the factor of 2 in the denominator comes from the fact that the field emerges on both sides of the plane; the flux \sigma A/\varepsilon_0 is split equally between the two faces of the pillbox. This is one of the most useful formulas in electrostatics — the field of an infinite plane does not depend on distance at all. It is uniform everywhere on each side.

The field points away from the plane on both sides if \sigma > 0, and toward the plane on both sides if \sigma < 0. The magnitude |\sigma|/(2\varepsilon_0) is the same at one millimetre from the plane as it is at one kilometre. Of course, a real charged sheet is finite and eventually the infinite-plane approximation breaks down at distances comparable to the sheet's size.

Sanity-check numbers

For a Diwali polythene streamer rubbed vigorously against a silk saree on a dry December evening, typical surface charge densities are around \sigma \sim 10^{-6}\,\text{C/m}^2. The resulting field is

About 56 kV/m — substantial, enough to pull visible hairs and dust particles from a centimetre away. Below air's breakdown strength (about 3\times 10^{6} V/m in dry air), so no spark — but enough to attract lint and fibres with enthusiasm.

5. Two parallel planes — the capacitor field

Place two infinite parallel planes, separated by a distance d, with the left plate carrying +\sigma and the right plate carrying -\sigma. This is the idealised model of a parallel-plate capacitor.

By superposition — and because Gauss's law is linear — the total field is the sum of the fields from each plane separately. Each plane, on its own, produces E = \sigma/(2\varepsilon_0) directed away from itself (for +\sigma) or toward itself (for -\sigma).

Between the plates (say, a point in the gap): the +\sigma plate produces a field pointing away from it, i.e., to the right. The -\sigma plate (on the right) produces a field pointing toward it, also to the right. Both fields point the same way; their magnitudes add:

Why: the two plates reinforce each other's field in the gap. This doubling is why a parallel-plate capacitor is such a good concentrator of electric field — the total is exactly twice the "single plane" field.

Outside the plates (say, to the left of the +\sigma plate): the +\sigma plate produces a field to the left (away from itself). The -\sigma plate produces a field to the right (toward itself). The two fields point in opposite directions and have equal magnitudes; they cancel:

Why: superposition cancels them exactly. This is why a charged capacitor does not leak field to the outside world — in the ideal infinite-plate limit, there is no field outside. Real capacitors with finite plates have small "fringing" fields at the edges, but in the interior of the plates this is a good approximation.

Two parallel planes with the same sign

For completeness: if both plates have +\sigma (no minus), then outside the plates the fields reinforce (E = \sigma/\varepsilon_0), and between the plates they cancel (E = 0). The cancellation and reinforcement are swapped. You can work out all four plate-sign combinations by the same superposition argument; the key is to remember that a single plate's field points away (for +\sigma) or toward (for -\sigma) itself on both sides.

6. Conductors in electrostatic equilibrium

A conductor is a material (like a metal) in which some electrons are free to move. In equilibrium — after all the transient motion has settled — these free electrons have rearranged themselves until there is no net electric force on them from other electrons or external sources. This single statement has three immediate consequences, which Gauss's law unpacks in turn.

Consequence 1: the field inside a conductor is zero

If the field inside the conductor were nonzero at some point, the free electrons there would feel a force and accelerate. They would redistribute until the field vanished. In equilibrium, no such redistribution is happening, so the field must be zero:

This is an equilibrium statement, not a property of the material. Under a changing external field, charges do accelerate briefly; but after a tiny time (microseconds, for most metals), equilibrium is restored and the interior field is zero again.

Consequence 2: all excess charge sits on the outer surface

Apply Gauss's law to a small Gaussian surface entirely inside the bulk of the conductor. Inside the conductor, \vec{E} = \vec{0}, so the flux is zero. Gauss's law gives Q_{\text{enc}} = 0. But this must hold for every Gaussian surface that lies entirely in the bulk — so the charge density must be zero everywhere in the bulk.

Any excess charge on the conductor must therefore reside on the surface. On the outer surface, it can freely produce a field in the space outside the conductor, without violating the "zero field inside" condition.

(If the conductor has a cavity inside it, things get richer — that is the subject of the cylinders-and-cavities article.)

Consequence 3: the field just outside the surface is E = \sigma/\varepsilon_0, perpendicular to the surface

Let the local surface charge density at some point on the conductor be \sigma. Draw a Gaussian pillbox straddling the surface, with one end-cap inside the conductor and the other just outside. Let the cap area be A, small enough that the surface is essentially flat over that area.

• Inside cap. Field there is zero (inside the conductor), so flux contribution is zero.
• Side wall. The field just outside the conductor must be perpendicular to the surface (any tangential component would push charges along the surface; in equilibrium, there is no such motion, so the tangential component is zero). Therefore \vec{E}\cdot\hat{n}_{\text{wall}} = 0 on the side wall.
• Outside cap. The field is perpendicular to the surface, so \vec{E}\cdot\hat{n}_{\text{outside}} = E. Flux contribution is EA.
• Enclosed charge. Q_{\text{enc}} = \sigma A.

Apply Gauss's law:

Why: the factor of 2 from the infinite-plane case has disappeared — because on one side of the surface (the inside), the field is zero by the "field inside conductor" rule, so only one cap contributes to the flux, not two. All the flux squeezes out the single outer cap, doubling the number you would see for an isolated plane of the same density.

The field's magnitude just outside the conductor is \sigma/\varepsilon_0 (a factor of 2 larger than an infinite plane with the same \sigma), and its direction is normal to the surface.

Example: charged conducting sphere

A solid conducting sphere of radius R carries total charge Q. In equilibrium, all the charge sits on the outer surface with uniform density \sigma = Q/(4\pi R^2). So:

• Inside the conductor (r < R): E = 0.
• On the surface (r = R): E = \sigma/\varepsilon_0 = [Q/(4\pi R^2)]/\varepsilon_0 = Q/(4\pi\varepsilon_0 R^2).
• Outside (r > R): by the shell theorem (spherical symmetry), E = Q/(4\pi\varepsilon_0 r^2).

Why: the conducting sphere behaves exactly like the charged spherical shell of §2, because in equilibrium the charge distributes uniformly on the outer surface (by spherical symmetry) and the interior (conductor) must have zero field. The single extra physics beyond §2 is the reason the charge ends up uniformly on the surface — it is forced there by the requirement that the field inside the conductor vanishes.

This is the Van de Graaff generator from the opening: a charged metal sphere, whose exterior field is indistinguishable from a point charge at its centre, and whose interior field is zero. Put a probe inside a charged metal ball, and no matter how much charge is on the outside, the probe reads zero field.

7. Common confusions

• "Gauss's law gives the field only outside the Gaussian surface." No — the field E that appears in Gauss's law is the field on the Gaussian surface itself. Once you have solved for E at radius r, you know the field at every point at that radius (by symmetry).

• "The field inside a charged conductor is zero only if the conductor is isolated." No — it is zero whenever the conductor is in electrostatic equilibrium, regardless of what other charges or external fields exist outside. If there are external charges, the conductor's surface charges redistribute until the interior field (from external plus surface charges) is zero. The interior of a metal box is shielded from any exterior electrostatic configuration — a phenomenon called electrostatic shielding or the Faraday cage effect (treated in detail in the cylinders-and-cavities article).

• "The shell theorem (no field inside the shell) fails if the shell is not closed." Yes, and significantly. A charged ring or disc is not a closed shell, so the theorem does not apply — the field inside a ring, for instance, does depend on position. Only a completely closed spherical surface of charge produces zero interior field. The theorem is a spherical-symmetry consequence; break the symmetry and you break the theorem.

• "The formula E = \sigma/(2\varepsilon_0) applies to a conductor." No — the factor-of-2 formula is for an isolated plane of charge (charge floating in space, with field emerging on both sides). For a conductor, the formula is E = \sigma/\varepsilon_0, because one side of the surface is inside the conductor (field zero there). Half the flux simply does not exist; it all comes out the outside.

• "The capacitor formula E = \sigma/\varepsilon_0 also applies to an isolated plane." No — \sigma/\varepsilon_0 is for the field between two oppositely-charged infinite plates (where the two single-plate fields reinforce). For a lone infinite plane, the field is \sigma/(2\varepsilon_0). Which formula to use depends on whether you are dealing with a single plane or a pair.

• "Inside a solid charged sphere, the field is zero." No — the field is zero only inside a charged shell (hollow). Inside a solid uniformly charged sphere, the field grows linearly with r, from zero at the centre to the surface value at r = R. The distinction between "solid" and "shell" is essential.

Where this leads next

• Applications of Gauss's Law — Cylinders and Cavities — Extends the same "pick a Gaussian surface that respects the symmetry" machinery to infinite lines and cylinders, then handles conductors with hollow cavities (including the Faraday cage).
• Gauss's Law — The law behind everything in this article. Read there for the general statement and its derivation from Coulomb + superposition.
• Electric Flux — The object \oint\vec{E}\cdot\mathrm{d}\vec{A} whose value Gauss's law pins down.
• Electric Field — The field \vec{E} itself — the protagonist of this whole subject.
• Superposition of Electrostatic Forces — The linearity that lets the parallel-plate field in §5 be computed as a sum of two single-plane fields.