Fields from Symmetry

Sphere, line and sheet fields in two lines each.

11 minUpdated 22 October 2026Gaussian surfaceline chargesurface charge density

Lesson at a glanceShow the key points

Symmetry fixes the direction of E; choose a surface where E is constant and normal, or parallel.
Uniform sphere: Q/4πε₀r² outside (30.2), Qr/4πε₀R³ inside (30.3).
Long line: E = λ/2πε₀r (30.4).
Large sheet: E = σ/2ε₀, independent of distance (30.5).

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

Use symmetry to find the direction of E⃗. Ask: is there only one direction that isn’t favoured over any other?
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).
Then the flux is just E × (area), set it equal to q_in/ε₀, and solve for E.

Interactive

Choosing a Gaussian surface

Pick a charge shape. Symmetry fixes the direction of E, and we choose a closed surface (gold, dashed) on which E is either constant and perpendicular, or parallel so it adds no flux. Slide the surface outward to trace E against distance on the right.

E\cdot 4\pi r^2 = \frac{q_{in}}{\varepsilon_0}

E = \frac{Qr}{4\pi\varepsilon_0R^3}

0.75(relative)

Charge

Distance of the surface

r

1.50

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.

Outside a uniformly charged sphere(30.2)

E = \frac{Q}{4\pi\varepsilon_0 r^2}

exactly as if all the charge sat at the centre

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

q_{in} = \frac{Q}{\tfrac43\pi R^3}\cdot\tfrac43\pi r^3 = Q\,\frac{r^3}{R^3}

Inside a uniformly charged sphere(30.3)

E\cdot4\pi r^2 = \frac{Qr^3}{\varepsilon_0R^3} ;\Rightarrow; E = \frac{Qr}{4\pi\varepsilon_0R^3}

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.

Worked example

Inside and outside a charged ball

A ball of radius 10 cm carries 2.0 µC spread uniformly through its volume. Find E at 5.0 cm and at 20 cm from the centre.

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:

Field of a long line charge(30.4)

E = \frac{\lambda}{2\pi\varepsilon_0 r}

Worked example

Near a charged wire

A long thin wire carries 3.0 µC per metre. Find the field 9.0 cm from it.

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:

Field of a large charged sheet(30.5)

E = \frac{\sigma}{2\varepsilon_0}

the same at every distance, as long as the sheet is large compared with that distance

Worked example

Flux from a charged sheet

A large sheet in the x–y plane carries 4.0 µC/m². A circle of radius 2.0 cm lies above it (z > 0), with its normal at 60° to the z-axis. Find the flux through the circle.

Charge	Gaussian surface	Field
Sphere, outside	concentric sphere	Q/4πε₀r²
Sphere, inside	concentric sphere	Qr/4πε₀R³
Long line	coaxial cylinder	λ/2πε₀r
Large sheet	pillbox through the sheet	σ/2ε₀

What to remember

1Symmetry gives the direction; a well-chosen surface makes ∮E⃗·dS⃗ = E × area.
2Uniform sphere: Q/4πε₀r² outside (30.2), Qr/4πε₀R³ inside (30.3).
3Line: λ/2πε₀r (30.4). Sheet: σ/2ε₀, independent of distance (30.5).

Check yourself

Moving twice as far from a long line charge, the field becomes…

Where is the field of a uniformly charged ball greatest?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

ReadApplying Gauss's lawOpenStax University Physics Vol. 2 §6.3 ReadField of a charged sphereHyperPhysics ReadField of a line chargeHyperPhysics ReadField of a sheet of chargeHyperPhysics

PreviousGauss's Law Next Charged Shells & Spheres

Fields from Symmetry

Sphere, line and sheet fields in two lines each.

11 minUpdated 22 October 2026Gaussian surfaceline chargesurface charge density

Lesson at a glanceShow the key points

Symmetry fixes the direction of E; choose a surface where E is constant and normal, or parallel.
Uniform sphere: Q/4πε₀r² outside (30.2), Qr/4πε₀R³ inside (30.3).
Long line: E = λ/2πε₀r (30.4).
Large sheet: E = σ/2ε₀, independent of distance (30.5).

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

Use symmetry to find the direction of E⃗. Ask: is there only one direction that isn’t favoured over any other?
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).
Then the flux is just E × (area), set it equal to q_in/ε₀, and solve for E.

Interactive

Choosing a Gaussian surface

E\cdot 4\pi r^2 = \frac{q_{in}}{\varepsilon_0}

E = \frac{Qr}{4\pi\varepsilon_0R^3}

0.75(relative)

Charge

Distance of the surface

r

1.50

A uniformly charged sphere

Outside (r > R)

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

Outside a uniformly charged sphere(30.2)

E = \frac{Q}{4\pi\varepsilon_0 r^2}

exactly as if all the charge sat at the centre

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

q_{in} = \frac{Q}{\tfrac43\pi R^3}\cdot\tfrac43\pi r^3 = Q\,\frac{r^3}{R^3}

Inside a uniformly charged sphere(30.3)

E\cdot4\pi r^2 = \frac{Qr^3}{\varepsilon_0R^3} ;\Rightarrow; E = \frac{Qr}{4\pi\varepsilon_0R^3}

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.

Worked example

Inside and outside a charged ball

A ball of radius 10 cm carries 2.0 µC spread uniformly through its volume. Find E at 5.0 cm and at 20 cm from the centre.

A long line of charge

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:

Field of a long line charge(30.4)

E = \frac{\lambda}{2\pi\varepsilon_0 r}

Worked example

Near a charged wire

A long thin wire carries 3.0 µC per metre. Find the field 9.0 cm from it.

A large plane sheet of charge

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:

Field of a large charged sheet(30.5)

E = \frac{\sigma}{2\varepsilon_0}

the same at every distance, as long as the sheet is large compared with that distance

Worked example

Flux from a charged sheet

A large sheet in the x–y plane carries 4.0 µC/m². A circle of radius 2.0 cm lies above it (z > 0), with its normal at 60° to the z-axis. Find the flux through the circle.

Charge	Gaussian surface	Field
Sphere, outside	concentric sphere	Q/4πε₀r²
Sphere, inside	concentric sphere	Qr/4πε₀R³
Long line	coaxial cylinder	λ/2πε₀r
Large sheet	pillbox through the sheet	σ/2ε₀

What to remember

1Symmetry gives the direction; a well-chosen surface makes ∮E⃗·dS⃗ = E × area.
2Uniform sphere: Q/4πε₀r² outside (30.2), Qr/4πε₀R³ inside (30.3).
3Line: λ/2πε₀r (30.4). Sheet: σ/2ε₀, independent of distance (30.5).

Check yourself

Moving twice as far from a long line charge, the field becomes…

Where is the field of a uniformly charged ball greatest?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

ReadApplying Gauss's lawOpenStax University Physics Vol. 2 §6.3 ReadField of a charged sphereHyperPhysics ReadField of a line chargeHyperPhysics ReadField of a sheet of chargeHyperPhysics

PreviousGauss's Law Next Charged Shells & Spheres