EG1216 · Physics 2 — Electricity & Magnetism
Theme 1 · Electrostatics — charge & the electric field

Electric flux & Gauss's law

Count how much field "flows through" a surface — the flux — then meet the law that ties that flux straight to the charge trapped inside.

Source: Serway & Jewett, Physics for Scientists and Engineers, 7th ed., §24.1–24.2.

Before you start

What you need first

  • Topic 3 — field & field lines — \(E\) in N/C, and the picture of field lines leaving \(+\) charges.
  • Area and direction — the idea of the normal (the line sticking straight out of a surface) and \(\cos\theta\) from trigonometry.

What you'll be able to do

  • Compute electric flux \(\Phi = EA\cos\theta\).
  • Explain why the net flux out of a closed surface depends only on the charge inside.
  • State and use Gauss's law \(\Phi = Q_{\text{enc}}/\varepsilon_0\).
  • Move between the two constants \(\varepsilon_0\) and \(k\).

The idea

What is electric flux?

Electric flux \(\Phi\) measures how many field lines pass through a surface. More lines through it means more flux.

Everyday picture: hold a net in a river. The flux is how much water flows through the net. Open it wide and face it into the flow → lots gets through. Turn it sideways → almost none.
area A E
Flux = how much of the field \(E\) passes through the area \(A\).

Flux when the field hits straight on

If the field \(E\) passes straight through a flat area \(A\) (perpendicular to it), the flux is simply field times area:

$$ \Phi = E\,A $$
where:
SymbolMeaningSI unit
Φelectric flux through the surfaceN·m²/C
Efield strength at the surfaceN/C
Aarea of the surfacem²
Bigger field or bigger area → more flux. Straight-on is the most flux you can get for a given field and area.

Flux when the field hits at an angle

If the surface is tilted, only the "straight-on" part of the field counts. We measure the angle \(\theta\) from the normal — the line sticking straight out of the surface:

$$ \Phi = E\,A\cos\theta $$
Straight on (\(\theta=0\)): \(\cos0=1\) → most flux. Edge-on (\(\theta=90^\circ\)): \(\cos90^\circ=0\) → no flux at all.
E area A normal θ
\(\theta\) is measured between the field \(E\) and the normal to the area. Only \(E\cos\theta\) passes through.
📐 Worked example 1

Flux straight on, and at an angle

A field \(E = 200\) N/C passes through a flat area \(A = 0.50\) m².

1Straight on (\(\theta=0\)):
$$ \Phi = E\,A = (200)(0.50) = 100\ \text{N·m}^2/\text{C} $$
2Now tilt the surface to \(\theta = 60^\circ\) from the normal (\(\cos60^\circ=0.5\)):
$$ \Phi = E\,A\cos\theta = (200)(0.50)(0.5) = 50\ \text{N·m}^2/\text{C} $$
Tilting the surface to \(60^\circ\) halved the flux, because \(\cos60^\circ=0.5\).

The key step

Flux out of a closed surface

Now wrap a closed surface (imagine a balloon) around some space and add up the flux over the whole thing:

  • Charge inside → field lines poke outward through the surface → net flux out.
  • No charge inside → every line that enters also leaves → net flux \(=0\).
The net flux out of a closed surface tells you the charge trapped inside. That single sentence is Gauss's law.
+ charge inside → net flux out
A \(+\) charge inside the (dashed) closed surface sends flux outward through it.

The law

Gauss's law

In words: the total flux out of any closed surface equals the charge enclosed divided by the constant \(\varepsilon_0\). In symbols (the loop on the integral means "over the whole closed surface"):

$$ \Phi = \oint \vec{E}\cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} $$
where:
SymbolMeaningSI unit
Φtotal flux out of the closed surfaceN·m²/C
Qenctotal charge enclosed (trapped inside the surface)C
ε₀permittivity of free space \(=8.85\times10^{-12}\)C²/(N·m²)
It does not matter what shape the surface is, or how the charge is arranged inside — only the total charge trapped matters. For the symmetric surfaces we choose in Topic 7, the field is constant over the surface and the integral collapses to the simple \(\Phi = EA\).

The two constants \(\varepsilon_0\) and \(k\)

Gauss's law uses \(\varepsilon_0\). It is just a relative of our old friend \(k\) from Coulomb's law:

$$ \varepsilon_0 = 8.85\times10^{-12} \qquad\qquad k = \frac{1}{4\pi\varepsilon_0} = 8.99\times10^{9} $$
Same physics, two costumes. We use \(\varepsilon_0\) inside Gauss's law, and \(k\) in the final field formulas.
📐 Worked example 2

Flux from the enclosed charge

A closed surface traps a charge \(Q_{\text{enc}} = 5.0\ \mu\text{C}\). Find the total flux out of the surface.

1Use Gauss's law, \(\Phi = Q_{\text{enc}}/\varepsilon_0\):
$$ \Phi = \frac{5.0\times10^{-6}}{8.85\times10^{-12}} $$
2Work it out:
$$ \Phi \approx 5.6\times10^{5}\ \text{N·m}^2/\text{C} $$
The answer does not depend on the shape of the surface — only on the charge inside.

✏️ Try it yourself

The total flux out of a closed box is measured to be \(\Phi = 3.0\times10^{5}\) N·m²/C.

(a) Find the charge \(Q_{\text{enc}}\) trapped inside.
(b) If you used a bigger box around the same charge, what would the flux be?

(a) Step 1. Rearrange Gauss's law: \(Q_{\text{enc}} = \Phi\,\varepsilon_0 = (3.0\times10^{5})(8.85\times10^{-12})\). Step 2. \(Q_{\text{enc}} \approx 2.7\times10^{-6}\ \text{C} = 2.7\ \mu\text{C}.\) (b) Exactly the same flux, \(3.0\times10^{5}\) N·m²/C — the flux depends only on the charge enclosed, not on the size or shape of the surface.

Common mistakes

MistakeFix
Measuring \(\theta\) from the surface itself.\(\theta\) is from the normal (the line sticking straight out), not from the surface.
Counting charge that sits outside the surface.Only \(Q_{\text{enc}}\) — charge inside — contributes to the net flux.
Thinking a bigger surface gives more flux.Same enclosed charge → same flux, whatever the surface's size or shape.
Mixing up \(\varepsilon_0\) and \(k\).\(k = 1/(4\pi\varepsilon_0)\). Use \(\varepsilon_0\) in Gauss's law.

Recap — the whole topic on one screen

IdeaWhat you own now
Flux\(\Phi = EA\cos\theta\); \(\theta\) measured from the normal.
Closed surfaceNet flux out depends only on the charge trapped inside.
Gauss's law\(\Phi = \oint \vec{E}\cdot d\vec{A} = \dfrac{Q_{\text{enc}}}{\varepsilon_0}\).
Constants\(\varepsilon_0 = 8.85\times10^{-12}\), \(k = \dfrac{1}{4\pi\varepsilon_0} = 8.99\times10^{9}\).

Next topic

Applying Gauss's law

Gauss's law is most powerful as a shortcut: pick a clever surface that matches the symmetry of the charge, and the field falls out in one line. Next we use it to re-derive the fields of a point charge, a long line, and a flat sheet.

→ Topic 7 · Applying Gauss's law (symmetry)