Home › Sessions › Gauss's law

Gauss's law

Class 12Physics Flux and enclosed chargeSession 7 Free

The total electric flux out of any closed surface equals the charge enclosed divided by ε₀. Charges outside contribute no net flux. With symmetry, the law gives the field of a line, a sheet and a shell in a few steps.

The statement

A closed surface is one with no gaps, enclosing a region of space, such as the surface of a sphere or a box. Flux through a closed surface is counted as positive where the field points outwards and negative where it points inwards.

Gauss’s law states that the total flux out of any closed surface equals the total charge enclosed by it, divided by ε₀:

Φ = q_enclosed / ε₀

Here ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻², the permittivity of free space. The total flux does not depend on the shape or size of the surface, on where the charge is inside it, or on any charges outside it.

Charges outside give no net flux

The field of a charge outside the surface does pass through the surface. But every field line from that charge that enters the closed surface at one place must leave it at another, because there is no charge inside for it to end on.

Where a line enters, the flux is negative. Where it leaves, the flux is positive. These contributions cancel, so the net flux due to an outside charge is zero. A charge inside is different: its field lines start inside and cross the surface outwards, with nothing to cancel them.

⚴ Move a charge in and out › Move a charge inside and outside a closed surface and watch the total flux. It changes only when the charge crosses the surface.

A charge inside a spherical surface gives a certain total flux. The sphere is replaced by an irregular closed surface of a different size, with the same charge inside. How does the total flux change?

It does not change. Gauss’s law depends only on the enclosed charge.

In the field-line picture, every field line from the enclosed charge must cross the surface outwards on its way to infinity. If an irregular surface makes a line cross several times, the crossings go out, in, out: the extra in-and-out pair cancels. The net count is one outward crossing per line, whatever the shape.

For a sphere centred on a point charge Q, this can be checked directly: E = kQ/r² and A = 4πr², so Φ = EA = 4πkQ = Q/ε₀, independent of r.

Using the law: symmetry

Gauss’s law holds for every closed surface. It gives the field easily only when the charge arrangement is symmetric. Then a Gaussian surface can be chosen on which the field has the same size everywhere and is perpendicular to the surface. On such a surface, Φ = EA, and the law gives E = q_enclosed / (ε₀A).

Three results

An infinitely long straight line of charge, with charge λ per unit length. Use a cylinder of radius r around the line. The field points radially outwards: E = λ / (2πε₀r). It falls off as 1/r, more slowly than the 1/r² of a point charge.
An infinite flat sheet of charge, with charge σ per unit area. The field is perpendicular to the sheet: E = σ / (2ε₀). It does not depend on the distance from the sheet. Near a large flat sheet, the field is uniform.
A thin spherical shell of radius R with charge q spread uniformly. Outside (r > R): E = kq/r², the same as a point charge q at the centre. Inside (r < R): a Gaussian sphere encloses no charge, so the flux is zero, and by symmetry E = 0 at every point inside.

“Infinite” is a model. A real line or sheet is finite. The results are good approximations at points much closer to the line or sheet than to its ends or edges.

A hollow metal sphere carries charge on its surface. What electric field is there inside the hollow?

Zero at every point inside.

Take any spherical Gaussian surface inside the hollow, centred on the sphere’s centre. All the charge is on the metal shell, outside this surface, so q_enclosed = 0 and the total flux is zero. By symmetry, the field would have to be the same size at every point of this surface and point radially. A radial field of the same size everywhere with zero total flux must be zero.

Each point inside is surrounded by charge, but the fields from different parts of the shell cancel exactly. A closed metal enclosure shields its interior from external electric fields in a similar way; this is called electrostatic shielding.

The unit in summary

Charge is additive, conserved and quantised. Coulomb’s law gives the force between two point charges, and superposition extends it to many. The electric field is the force per unit charge at each point, drawn with field lines. Flux measures the field through a surface, and Gauss’s law relates the flux through a closed surface to the charge inside. The dipole shows how a field acts on a neutral pair of charges: no net force, but a torque.

Check what you have understood

1Gauss’s law states that the total flux out of a closed surface depends on…

Φ = q_enclosed / ε₀. Only the enclosed charge matters.

2A charge outside a closed surface contributes…

Inward (negative) and outward (positive) contributions cancel.

3A closed surface around a fixed charge is made larger. The total flux…

The enclosed charge is unchanged, so the flux is unchanged. For a sphere: E falls as 1/r² while A grows as r².

4When is Gauss’s law easiest to use for finding a field?

With symmetry, Φ = EA, and E follows in one step.

5The electric field inside a uniformly charged thin spherical shell is…

A Gaussian sphere inside encloses no charge, so by symmetry E = 0.

6The field of an infinite flat sheet of charge, found with Gauss’s law,…

The field of an infinite sheet does not depend on distance.

7An infinite line of charge produces a field that varies with distance r as…

E = λ/(2πε₀r), so E ∝ 1/r.

That completes the unit on electric charges and fields.

‹ Previous: The electric dipole All sessions ›